Math 651 Extra Problems Autumn, 2005 · 2017-12-22 · Math 651 Extra Problems Autumn, 2005 X7. Let...

Math 651 Extra Problems Autumn, 2005

X1. Let X be any set. Define d : X ×X → [0,∞) by

d(x, x′) =

1 if x 6= x′,

0 if x = x′.

Verify that d is a metric on X and determine which subsets of X are open with respect to D .

X2. Let (X, ρ) be a metric space. Define σ : X ×X → [0, 1) by due 1Th

σ(x, x′) =ρ(x, x′)

1 + ρ(x, x′).

(a) Let a, b ∈ (−1,∞). Prove that

a < b if and only ifa

1 + a<

b

1 + b.

Hint: Do not use calculus. Do not cross multiply. Just notice what

1− 1

1 + c

is equal to when c ∈ R \ −1 .(b) Prove that σ is a metric on X .(c) Prove that each open ball for ρ is also an open ball for σ .(d) Prove that each open ball for σ is either an open ball for ρ or is all of X .(e) Prove that for each G ⊆ X , G is open with respect to ρ iff G is open with respect to σ .

Remark. The significance of problem X2 is that it shows that any metric has the same open sets as somebounded metric.

X3. Let X be a non-empty set, let (Y, d) be a metric space, and let Z be the set of bounded functions due 1Th

from X into Y . Define a function D on Z × Z by

D(f, g) = sup

d(

f(x), g(x))

: x ∈ X

.

(a) Prove that D is a metric on Z . (Warning: Do not “calculate” with sup. Any inequalities youassert involving sup must be justified based on the definition of sup. Also, please do not use proofby contradiction when it is not needed and does not shorten the argument.)

(b) If f ∈ Z and if (fn) is a sequence in Z which converges to f with respect to the metric D , thenclearly (fn) converges pointwise to f because for each x ∈ X , we have d

(

f(x), fn(x))

≤ D(f, fn).However, the converse is false in general. Pointwise convergence does not imply convergencewith respect to the metric D . To see this, consider the special case where X = Y = R andd(y, y′) = |y − y′| for all y, y′ ∈ R . Let (fn) be the sequence in Z defined by

fn(x) =

1 if x > n,

0 if x ≤ n.

Prove that (fn) converges pointwise to the zero function but does not converge to any element ofZ with respect to the metric D .

(c) Now consider the general case again. Suppose that Y has at least two points. Prove that eachpointwise convergent sequence in Z is convergent with respect to the metric D if and only if theset X is finite.

Remark. In Chapter 7 of the text, the notion of uniform convergence of sequences of functions is introducedand studied. Let f be a bounded function from X to Y and let (fn) be a sequence of bounded functionsfrom X to Y . In the notation of problem X3, fn → f uniformly on X iff D(f, fn) → 0. In other words,fn → f uniformly on X iff (fn) converges to f in the metric space (Z,D). This provides an example toillustrate why we introduced the notion of a general metric space. There are many important metric spacesbesides R , C , and Rk . Some of the most important ones are metric spaces of functions. The space (Z,D)is one of the simplest examples of a metric space of functions.


Let (X, d) be a metric space. If (xn) is a sequence in X , then to say that (xn) is Cauchy means thatfor each ε > 0, there exists N ∈ N such that for all m,n ∈ N with m,n > N , we have d(xm, xn) < ε . It iseasy to see that if (xn) is a convergent sequence in X , then (xn) is Cauchy. To say that (X, d) is complete(as a metric space) means that each Cauchy sequence in X is convergent in X . Thus in a complete metricspace, a sequence is convergent if and only if it is Cauchy.

As we know, each non-empty subset of R which is bounded above has a least upper bound in R . (Thisproperty of R is sometimes called Dedekind completeness.) Using this fact, we shall soon show that R isalso complete as a metric space, when we give it its usual metric, namely d(x, y) = |x − y| . For now, youmay take it for granted that R is complete in this sense.

X4. As in problem X3, let X be a non-empty set, let (Y, d) be a metric space, let Z be the set of bounded due 2Th

functions from X into Y , and let D be the metric on Z defined by

D(f, g) = sup

d(

f(x), g(x))

: x ∈ X

.

Prove that if the metric space (Y, d) is complete, then so is the metric space (Z,D).

Let (X, ρ) and (Y, σ) be metric spaces. An isometry from X into Y is a map f : X → Y such thatσ(

f(x), f(x′))

= ρ(x, x′) for all x, x′ ∈ X . Informally, an isometry is a map that preserves distances betweenpoints.

X5. Let (X, d) be a metric space. Give R its usual metric. Let Z be the set of bounded functions from due 2Th

X into R . Give Z the metric defined by

D(f, g) = sup |f(ξ)− g(ξ)| : ξ ∈ X .

By problem X4, the metric space (Z,D) is complete, because R with its usual metric is complete as a metricspace. Fix x0 ∈ X . For each x ∈ X , define fx : X → R by

fx(ξ) = d(ξ, x)− d(ξ, x0).

(a) Prove that for each x ∈ X , we have fx ∈ Z .(b) Define Φ: X → Z by Φ(x) = fx . Prove that Φ is an isometry from (X, d) into (Z,D).

Let (X, d) be a metric space and let X1 ⊆ X . Let d1 be the restriction of d to X1 ×X1 . Then clearlyd1 is a metric on X1 . We call d1 the subspace metric that X1 inherits from (X, d) and we say that themetric space (X1, d1) is a subspace of the metric space (X, d).

Remark. The result of problem X5 shows that any metric space is isometric to a subspace of a completemetric space.

Let A and B be sets. To say that A is equinumerous to B means that there is a bijection1 from Ato B . Recall that ω denotes the set 0, 1, 2, . . . . If n ∈ ω , then to say that A has n elements meansthat either n = 0 and A is empty or n ∈ N and A is equinumerous to 1, . . . , n . To say that A is finitemeans that there exists n ∈ ω such that A has n elements. To say that A is infinite means that A is notfinite. To say that A is countably infinite means that A is equinumerous to N . To say that A is countablemeans that A is finite or countably infinite.2 It is easy to show that if B is an infinite subset of N , thenB is equinumerous to N . It follows that A is countable if and only if A is equinumerous to a subset ofN . To say that A is uncountable means that A is not countable. Cantor (1873) pointed out that the set ofrational numbers is countable and proved that the set of real numbers is uncountable.

X6. Let X be a set. Prove that X contains a countably infinite subset if and only if X is equinumerousto a proper subset of itself. (Do not use the axiom of choice.)

1 A bijection from A to B is a one-to-one map from A onto B . Another name for a bijection from A to B is a one-to-onecorrespondence between A and B .

2 Warning: The definition of countable that I have given is the one accepted by most mathematicians, but you should watchout for the fact that Rudin uses the term countable to mean what I have chosen to call countably infinite. So for Rudin, a finiteset is not countable. To me, that is just ridiculous, so I will not follow Rudin’s use of the term countable.

2


X7. Let X be an infinite set. Prove that X contains a countably infinite subset.3

X8. Let B be a set, let C be a countable subset of B , and let A = B \C . Suppose that A has a countablyinfinite subset. Prove that A is equinumerous to B . (Do not use the axiom of choice.)

Example. Let Σ = 0, 1 N be the set of infinite binary sequences. Let C be the subset of Σ consistingof binary sequences that end with an infinite string of ones. Then C is countable. Let A = Σ \ C . ThenA has a countably infinite subset. For instance, if an is the element of Σ that has a one in the n-th placeand zeroes in all the other places, then a1, a2, a3, . . . is a countably infinite subset of A . Therefore, byproblem X8, A is equinumerous to Σ.

X9. Let (X, d) be a complete metric space. Suppose (An) is a sequence of non-empty subsets of X suchthat An ⊇ An+1 for each n and diam(An) → 0 as n→ ∞ . Show that

⋂

nAn contains exactly one point.

Notation. The following notation is not standard but will be convenient to use in the next few exercises.For each n ∈ N , let Sn = 0, 1 n , the set of finite binary sequences of length n . Let S =

⋃

n∈N Sn , theset of non-empty finite binary sequences. Finally, let Σ = 0, 1 N , the set of infinite binary sequences.

Reminder. To say that x is a dyadic rational number means that there exist an integer k and a naturalnumber ℓ such that x = k/2ℓ . For instance 0, 1, 1/2, 1/4, and 3/4 are dyadic rationals but 2/3 is not.The number 15/24 is a dyadic rational because it is equal to 5/8.

X10. Let Sn , S , and Σ be as above. For each n ∈ N and each s = (s1, . . . , sn) ∈ Sn , let x(s) =due 3Threvised26 Oct 05

∑nk=1 sk2

−k , let y(s) = x(s) + 2−n , and let I(s) be the closed interval [x(s), y(s)] . Thus I(0) = [0, 1/2],I(1) = [1/2, 1], I(0, 0) = [0, 1/4], I(0, 1) = [1/4, 1/2], I(1, 0) = [1/2, 3/4], I(1, 1) = [3/4, 1], I(0, 0, 0) =[0, 1/8], and so on. For each σ = (s1, s2, s3, . . . ) ∈ Σ, the sequence of sets

I(s1, . . . , sn), n = 1, 2, 3, . . .

is a decreasing sequence of non-empty closed subsets of R with diameters tending to zero, so by problemX9, the intersection of this sequence of sets contains exactly one point which we shall denote by f(σ) andwhich by definition is the point represented in binary by

0.s1s2s3 · · · .

The parts below will lead you through a proof that each number in [0, 1] has at least one such binaryrepresentation, that each dyadic rational in [0, 1) has exactly two such representations, one ending withrepeating zeroes and one ending with repeating ones, and that each other number in [0, 1) has exactly onesuch representation and that this unique representation does not end with repeating zeroes nor does it endwith repeating ones. Of course 1 has the two binary representations 0.111 · · · and 1.000 · · · but we are notconsidering the latter representation here.

(a) Prove that f maps Σ onto [0, 1].

(b) Let σ = (s1, s2, s3, . . . ) ∈ Σ. Prove that f(σ) = 1 if and only if sn = 1 for all n .

(c) Let σ = (s1, s2, s3, . . . ) ∈ Σ and let τ = (t1, t2, t3, . . . ) ∈ Σ. Suppose that σ 6= τ . Then sn 6= tnfor some n . Consider the least such n and suppose (without loss of generality) that sn = 0 andtn = 1. Prove that f(σ) = f(τ) if and only if for each k > n , we have sk = 1 and tk = 0.

Now let C be the subset of Σ consisting of binary sequences that end with an infinite string of ones. LetA = Σ \ C and let g be the restriction of f to A .

(d) Prove that g is a one-to-one map from A onto the interval [0, 1).

(e) Deduce that Σ is equinumerous to [0, 1].

3 You will have to use the axiom of choice to prove this, though not the full strength of the axiom of choice. The simplestproof uses the principle of dependent choice. The principle of dependent choice is the principle that justifies constructing asequence by induction, where at each stage one must select the next term of the sequence from a non-empty set of possiblechoices which depends on the terms that have already been chosen. There is a slightly more complicated proof that uses onlythe countable axiom of choice. The countable axiom of choice states that each countable family of non-empty sets has a choicefunction. The countable axiom of choice is strictly weaker than the principle of dependent choice.

3


Let A be a subset of a topological space X . To say that A is dense in X means that the closure of Ain X is all of X . For instance, the set of rational numbers is dense in R . To say that A is nowhere dense inX means that the closure of A in X has empty interior in X . For instance, the set of integers is nowheredense in R . To say that p is an isolated point of A means that p ∈ A and there exists a nhd U of p in Xsuch that U ∩A = p . For instance, if X = R and A = 0 ∪

n−1 : n ∈ N

, then for each n ∈ N , n−1

is an isolated point of A but 0 is not an isolated point of A . If p ∈ X , then to say that p is a limit pointof A means that for each nhd U of p , U ∩ (A \ p ) is non-empty. It is not hard to see that the closureof A is the disjoint union of the set of isolated points of A and the set of limit points of A . If p ∈ X , thento say that p is a condensation point of A means that for each nhd U of p , U ∩A is uncountable. Clearlyeach condensation point of A is a limit point of A . But if X = R and A = 0 ∪

n−1 : n ∈ N

, then 0is a limit point of A but not a condensation point of A .

X11. Let Sn , S , and Σ be as above. For each n ∈ N and for each s = (s1, . . . , sn) ∈ Sn , let a(s) =due 3Threvised13 Oct 05

∑nk=1 2sk3

−k , let b(s) = a(s) + 3−n , and let J(s) be the interval [a(s), b(s)] . Notice that a(s, 0) = a(s),b(s, 0) = a(s) + 3−(n+1) , a(s, 1) = a(s) + 2 · 3−(n+1) , and b(s, 1) = b(s). Thus when we remove the openmiddle third from J(s), we get J(s, 0) ∪ J(s, 1). Let C0 = [0, 1]. For each n ∈ N ,

Cn =⋃

s∈Sn

J(s).

Thus C1 = [0, 1/3]∪ [2/3, 1], C2 = [0, 1/9]∪ [2/9, 1/3]∪ [2/3, 7/9]∪ [8/9, 1], and in general, for each n ∈ N ,Cn consists of the union of the 2n closed intervals, each of length 3−n , that are obtained by removing theopen middle third from each of the 2n−1 disjoint closed intervals that make up Cn−1 . Let

C =⋂

n∈N

Cn.

The set C is called the Cantor set.(a) Prove that C is closed and has empty interior in R . Thus C is nowhere dense in R .

Now for each σ = (s1, s2, s3, . . . ) in Σ, let f(σ) be the unique point belonging to⋂

n∈N J(s1, . . . , sn). (Wecan do this, by problem X9.)

(b) Prove that f is a one-to-one map from Σ onto C .(c) Deduce that C is equinumerous to [0, 1]. This is surprising, since by part (a), C is a topologically

small subset of [0, 1]. (Benoit Mandelbrot calls C the “Cantor dust” because its points are sosparsely scattered in the interval [0, 1].)

(d) Prove that each point of C is a condensation point of C . In particular, C has no isolated points.(Hint: Prove that for each x ∈ C and each nhd U of x , U ∩ C contains a “scale model” of C .)

Remark. Let Cn and C be as in problem X11. Since Cn is the union of 2n disjoint intervals each ofwhich has length 3−n , the total length of Cn is (2/3)n . Since C ⊆ Cn for each n , and since (2/3)n → 0as n → ∞ , it seems reasonable to say that the total length of C is 0. (In spring quarter, we’ll develop atheory of length that makes this rigorous.) This is another sense in which C is a very small set and anotherreason why it is surprising that C is equinumerous to [0, 1].

Remark. By modifying the construction of the Cantor set, it is possible to obtain other interesting Cantor-like sets, including ones which, while they still are nowhere dense in R , have strictly positive total length.With this as motivation, in the next exercise we shall consider these Cantor-like sets in a very general setting.First we need a definition.

Definition. Let Sn , S , and Σ be as above. If (X, d) is a metric space, then by a Cantor scheme4 in (X, d),we shall mean a family (A(s))s∈S of non-empty subsets of X , indexed by S , such that:

(a) A(0) and A(1) are disjoint;(b) For each n ∈ N and each s ∈ Sn , A(s1, . . . , sn, 0) and A(s1, . . . , sn, 1) are disjoint and for i = 0, 1,

A(s1, . . . , sn) ⊇ A(s1, . . . , sn, i);

(c) For each σ = (s1, s2, s3, . . . ) ∈ Σ, diam(A(s1, . . . , sn)) → 0 as n→ ∞ .

4 This is not standard terminology.

4


X12. Let Sn , S , and Σ be as above. Let (X, d) be a complete metric space and let (A(s))s∈S be aCantor scheme in (X, d). For each n ∈ N , let Bn =

⋃

s∈SnA(s). Let K =

⋂

n∈NBn . For each σ =(s1, s2, s3, . . . ) ∈ Σ, let f(σ) be the unique point in

⋂

nA(s1, . . . , sn). (We can do this by problem X9.)(a) Prove that h is a one-to-one map from Σ onto K .(b) Deduce that K is equinumerous to the [0, 1].

Notation. Let (X, d) be a metric space. For each non-empty set A ⊆ X and each x ∈ X , the distancefrom x to A is

d(x,A) = inf d(x, a) : a ∈ A ,

by definition. (Warning: There need not exist a ∈ A with d(x,A) = d(x, a).)

X13. Let (X, d) be a metric space, let A be a non-empty subset of X , and let E be the closure of A .Prove that E = x ∈ X : d(x,A) = 0 . In particular, A is closed if and only if A = x ∈ X : d(x,A) = 0 .X14. Let (X, d) be a metric space. Let A be a non-empty subset of X and define f : X → [0,∞) byf(x) = d(x,A).

(a) Prove that for all x, y ∈ X , we have

f(x) ≤ d(x, y) + f(y).

(Warning: Do not “calculate” with inf . Any inequalities you assert involving inf must be justifiedbased on the definition of inf . Also, please do not use proof by contradiction when it is not neededand does not shorten the argument.)

(b) Deduce from part (a) that for all x, y ∈ X , we have |f(x)− f(y)| ≤ d(x, y).

Definitions. Let (X, d) and (M,ρ) be metric spaces and let g : X → M . To say that C is a Lipschitzconstant for g means that C ∈ [0,∞) and that for all x, y ∈ X , we have ρ(g(x), g(y)) ≤ C d(x, y). To saythat g is a Lipschitz function means that there exists a Lipschitz constant for g .

Remark. Part (b) of problem X14 says that the function x 7→ d(x,A) is a Lipschitz function, with Lipschitzconstant 1.

X15. Let (X, d) be a metric space, let A be a non-empty subset of X , let r > 0, and let

G = x ∈ X : d(x,A) < r and H = x ∈ X : d(x,A) ≤ r .

Prove that G is open and H is closed.

Definitions. Let X be a topological space. To say that A is a Gδ set in X means that there exists asequence (Gn) of open subsets of X such that A =

⋂∞n=1Gn . To say that A is an Fσ set in X means

that there exists a sequence (Fn) of closed subsets of X such that A =⋃∞

n=1 Fn . Less formally, a Gδ set isa countable intersection of open sets and an Fσ set is a countable union of closed sets.

Example. The set of irrational numbers is a Gδ set in R because it is equal to⋂

q∈Q(R\ q ), a countableintersection of open sets. It can be shown that the set of irrational numbers is not an Fσ set. (This followseasily from the Baire category theorem, for those of you who know that theorem. We will discuss it later.)

X16. Let (X, d) be a metric space and let E be a closed subset of X . Prove that E is a Gδ set in X . (ByDe Morgan’s laws, it follows that if U is an open subset of X , then U is an Fσ set in X .)

X17. Let a and b be non-negative real numbers. The geometric mean of a and b is√ab . The arithmetic

mean of a and b is (a+ b)/2. Prove that these two means satisfy the inequality

√ab ≤ a+ b

2,

with equality if and only if a = b . (Do not use calculus. Just use simple algebra. If your proof is longerthan one line, you are making it too complicated.)

5


X18. Let α ∈ (0,∞). Define f : (0,∞) → (0,∞) by due 4Th

f(x) =1

2

(

x+α

x

)

.

(a) Prove that for each x ∈ (0,∞), we have f(x) ≥ √α , with equality if and only if x =

√α .

Now let x1 ∈ (√α,∞) and define x2, x3, x4, . . . by the recursion formula xn+1 = f(xn).

(b) Prove that (xn) is a strictly decreasing sequence in (√α,∞) and that xn → √

α .(c) Let εn = xn −√

α , let β = 2√α , and show that for each n ,

εn+1 =ε2n2xn

<ε2nβ.

By induction, deduce that for each n ,

εn+1 < β

(

ε1β

)2n

.

(d) This is a good algorithm for computing square roots, because the recursion formula is simple andthe convergence is swift. For example, suppose α = 3 and x1 = 2. Show that ε1/β < 10−1 andthat therefore

ε5 < 4 · 10−16 and ε6 < 4 · 10−32.

Definition. Let X be a topological space and let (xn) be a sequence in X . To say that p is a clusterpoint of (xn) in X means that p ∈ X and that for each N ∈ N and for each nhd U of p in X , there existsn ≥ N such that xn ∈ U .

Remark. Informally, p is a cluster point of (xn) if and only if each tail of (xn) meets each nhd of p .

Reminder. Each metric space is first countable.

X19. Let X be a topological space. Let (xn) be a sequence in X and let p ∈ X .(a) Suppose (xn) has a subsequence (yk) =

(

xnk

)

which converges to p . Prove that p is a clusterpoint of (xn).

(b) Conversely, suppose p is a cluster point of (xn). Prove that if, in addition, X is first countable,then (xn) has a subsequence which converges to p .

Definition. Let X be a set and let B and C be filter bases on X . To say that C is finer than B meansthat for each B ∈ B , there exists C ∈ C such that C ⊆ B .

Example. Let X be a set, let (xn) be a sequence in X , and let (yk) =(

xnk

)

be a subsequence of (xn).Then the filter base of tails of (yk) is finer than the filter base of tails of (xn). In this sense, the notion offiner filter base generalizes the notion of subsequence.

Example. Let X be a topological space, let p ∈ X , and let B be a nhd base at p for X . Then B is afilter base on X . Let C be another filter base on X . Then C converges to p if and only if C is finer thanB .

Definition. Let X be a topological space and let B be a filter base on X . To say that p is a clusterpoint for B in X means that p ∈ X and that for each B ∈ B and for each nhd U of p in X , B ∩ U isnon-empty.

Example. Let X be a topological space and let (xn) be a sequence in X . Then clearly p is a cluster pointfor the filter base of tails of (xn) if and only if p is a cluster point for (xn).

Remark. Let X be a topological space, let B be a filter base on X , and let C be the set of cluster pointsfor B in X . A moment’s thought reveals that

C =⋂

B : B ∈ B

,

so C is closed because the intersection of any non-empty collection of closed sets is closed. In particular,the set of cluster points of any sequence is a closed set, because it is the same as the set of cluster points ofthe filter base of tails of the sequence.

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The next exercise generalizes problem X19.

X20. Let X be a topological space, let B be a filter base on X , and let p ∈ X . Prove that p is a clusterpoint for B in X if and only if there exists a filter base C on X such that C is finer than B and Cconverges to p .

Reminder. Let (xn) be a sequence in [−∞,∞] . For each n , let

an = infm≥n

xm and bn = supm≥n

xm.

Then an ≤ an+1 and bn ≥ bn+1 for all n . Hence each of (an) and (bn) has a limit in [−∞,∞] . Bydefinition,

lim infn→∞

xn = limn→∞

an and lim supn→∞

xn = limn→∞

bn.

Note that this is not the definition given in Rudin, Principles of Mathemtical Analysis, but I recommendthat you use it because it is a more convenient working definition than the one in Rudin.

X21. Let (xn) and (yn) be bounded sequences of real numbers.(a) Prove that for each n , we have

supm≥n

(xm + ym) ≤ supm≥n

xm + supm≥n

ym

and

infm≥n

(xm + ym) ≥ infm≥n

xm + infm≥n

ym.

(Warning: Do not “calculate” with sup or inf . Any inequalities you assert involving sup mustbe justified based on the definition of sup. Similarly for inf . Also, please do not use proof bycontradiction when it is not needed and does not shorten the argument.)

(b) Prove that

lim sup(xn + yn) ≤ lim supxn + lim sup yn.

and

lim inf(xn + yn) ≥ lim inf xn + lim inf yn.

(c) Prove that

lim inf(xn + yn) ≤ lim supxn + lim inf yn ≤ lim sup(xn + yn)

and

lim inf(xn + yn) ≤ lim inf xn + lim sup yn ≤ lim sup(xn + yn).

(Hint: These inequalities follow from the ones in part (b) by routine calculations. For instance,try applying the second inequality in part (b) to the sequences (−xn) and (xn + yn) instead of(xn) and (yn) and see what you get.)

(d) Use parts (b) and (c) to prove that if (xn) is convergent, then

lim sup(xn + yn) = limxn + lim sup yn

and

lim inf(xn + yn) = limxn + lim inf yn.

X22. Give an example of two bounded sequences in R for which all of the inequalities in parts (a), (b),and (c) of problem X21 are strict. (Hint: There is an example in which each sequence just repeats its firstfour terms over and over.)

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X23. Let (xn) be a sequence in R . For each n , let sn = (x1 + · · ·+ xn)/n be the average of x1, . . . , xn . due 4Th

Let a = lim inf xn , A = lim inf sn , B = lim sup sn , and b = lim supxn .(a) Prove that B ≤ b . (Hint: To show that B ≤ b , it suffices to show that for each c ∈ (b,∞), we

have B ≤ c . This is even valid when b = ∞ , since in that case there is nothing to show anyway!)Similarly, a ≤ A . (Or this can be deduced by applying part (a) to the sequence (−xn) instead of (xn).)

(b) Let L ∈ [−∞,∞] . Deduce from part (a) that if xn → L , then sn → L too.

X24. Give an example of a sequence (xn) in R such that lim supxn = ∞ and lim inf xn = −∞ but(x1 + · · ·+ xn)/n→ 0.

Definition. Let (xn) be a sequence in R and let L ∈ R . To say that (xn) is Cesaro convergent to Lmeans that (x1 + · · ·+ xn)/n→ L .

Remark. By problem X23(b), ordinary convergence implies Cesaro convergence. By problem X24, theconverse does not hold in general.

I hope the last few exercises have convinced you that the definitions for lim inf and lim sup givenabove are good working definitions. The next exercise, in combination with problem X19, shows that thesedefinitions are equivalent to the corresponding ones given in Rudin, Principles of Mathemtical Analysis.

X25. Let (xn) be a sequence in [−∞,∞] .(a) Let a = lim inf xn and let b = lim supxn . Prove that a and b are cluster points for (xn) in

[−∞,∞] .(b) Let c be a cluster point for (xn) in [−∞,∞] . Prove that a ≤ c ≤ b .

Thus a is the smallest cluster point for (xn) and b is the largest cluster point for (xn).

Definition. Let B be a filter base on [−∞,∞] . Then

lim inf B = sup inf B : B ∈ B and lim supB = inf supB : B ∈ B ,

by definition.

Example. Let (xn) be a sequence in [−∞,∞] and let B be the filter base of tails of (xn). Then clearly

lim inf B = lim inf xn and lim supB = lim supxn.

The next exercise is a generalization of problem X25.

X26. Let B be a filter base on [−∞,∞] .(a) Let a = lim inf B and let b = lim supB . Prove that a and b are cluster points for B in [−∞,∞] .(b) Let c be a cluster point for B in [−∞,∞] . Prove that a ≤ c ≤ b .

Thus a is the smallest cluster point for B and b is the largest cluster point for B .

Definition. Let (X, d) be a metric space and let B be a filter base on X . To say that B is Cauchy meansthat for each ε > 0, there exists B ∈ B such that diam(B) < ε .

Remark. Let (X, d) be a metric space, let (xn) be a sequence in X , and let B be the filter base of tailsof (xn). Then clearly B is Cauchy if and only if (xn) is Cauchy.

The result in the next exercise may be viewed as a refinement of problem X9 and may be proved by asimilar method.

X27. Let (X, d) be a complete metric space, so that each Cauchy sequence in X converges. Prove thateach Cauchy filter base on X converges.

X28. Let (an) be a sequence of strictly positive real numbers. According to Theorem 3.37 in Rudin, do before MT

Principles of Mathematical Analysis, Third Edition,

lim infan+1

an≤ lim inf a1/nn and lim sup a1/nn ≤ lim sup

an+1

an.

(a) Rudin gives a self-contained proof of these inequalities. Give a short alternative proof, based onproblem X23.

(b) Compare the proof of Theorem 3.37 in Rudin with the solution of problem X23(a).

8


The next exercise really belongs right after the definition of lim inf and lim sup for a filter base on[−∞,∞] .

X29. Let B be a filter base on [−∞,∞] and let L ∈ [−∞,∞] . Prove that B → L if and only iflim inf B = L = lim supB .

Reminder. Let X be a topological space and let A ⊆ X . To say that p is a limit point of A in X meansthat p ∈ X and for each nhd U of p in X , U ∩ (A \ p ) is non-empty. It is clear that p is a limit pointof A in X if and only if p belongs to the closure of A \ p in X . Consequently, p is a limit point of Ain X if and only if there exists a filter base B on A \ p such that B → p .

X30. Let X be a first countable topological space.5 Let A ⊆ X and let p ∈ X . Prove that p is a limitpoint of A in X if and only if there is a sequence (an) in A such that an 6= p for each n and an → p asn→ ∞ .

Definition. Let X be a topological space. To say that X is T1 means that for each x ∈ X and eachy ∈ X , if x 6= y , then there exists an open set G in X such that x ∈ G and y /∈ G .

Obviously each Hausdorff space is T1 . In particular, each metric space is T1 .

X31. Let X be a topological space. Prove that the following are equivalent:(a) X is T1 .(b) Each singleton in X is closed.(c) Each finite subset of X is closed.

X32. Let X be a first countable T1 space.6 Let A ⊆ X and let p be a limit point of A in X . Prove thatthere is a sequence (an) of distinct points in A \ p such that an → p as n → ∞ . Deduce that for eachnhd U of p in X , U ∩A is infinite.

The next three exercises may be solved with the help of problem X10.

X33. Prove that [0, 1]2 is equinumerous to [0, 1].

X34. Let n ∈ N . Prove that [0, 1]n is equinumerous to [0, 1].

X35. Prove that [0, 1]N is equinumerous to [0, 1].

Remark. In late 1873, Georg Cantor asked whether it was possible to define a one-to-one correspondencebetween [0, 1] and N . He believed that this would not be possible and within a few days, he succeededin proving this. In doing so, he became the first person in history to realize that just because two setsare both infinite, it does not follow that they have the same number of elements. About a month later,he asked whether it was possible to define a one-to-one correspondence between [0, 1]2 and [0, 1]. Again,he believed that this would not be possible. This time it took him about three and a half years to answerthe question and the answer that he found astonished him: As you were asked to show in problem X33, hefound that such a one-to-one correspondence can be defined! This initially led him to believe that he hadshown that the concept of dimension was meaningless. However, Cantor’s friend Richard Dedekind pointedout that the one-to-one correspondence that Cantor had found was not continuous and conjectured thata one-to-one correspondence would preserve dimension if both it and its inverse were continuous. In theyears that followed, many mathematicians, including Cantor himself, tried to prove this. The first correctproof was published by the Dutch mathematician L. E. J. Brouwer in 1911, some 34 years after Dedekindformulated the conjecture.

X36. Let z ∈ C with z 6= 1. For each n ∈ N , let Sn =∑n−1

k=0 zk , let Tn =

∑nk=1 kz

k , and let do before MT

Un =∑n

k=1 k2zk . As we know,

Sn =1− zn

1− z.

This is a simpler expression for Sn than the expression that we used to define Sn , in the sense that itinvolves far fewer arithmetic operations when n is large.

5 Remember that each metric space is first countable.6 Remember that each metric space is first countable and T1 .

9


(a) Find a simpler expression7 for Tn . Then show that if |z| < 1, then the infinite series∑∞

k=1 kzk

converges absolutely and find its sum.(b) Find a simpler expression8 for Un . Then show that if |z| < 1, then the infinite series

∑∞k=1 k

2zk

converges absolutely and find its sum.

Example. Here is an application of the results of problem X36. Suppose you toss a coin repeatedly. Letp be the probability that the coin comes up heads on any given toss. Assume that p ∈ (0, 1), but don’tassume that p = 1/2. (For instance, perhaps there is a lump of clay stuck to one side of the coin.) Letq = 1− p . Thus q is the probability that the coin comes up tails on any given toss. Let N be the numberof tosses until the coin first comes up heads. For each k ∈ N , P (N = k) denotes the probability that Nhas the value k . We have P (N = 1) = p , P (N = 2) = qp , P (N = 3) = q2p , and so on. In general,P (N = k) = qk−1p , for each k ∈ N . Using the formula for the sum of a geometric series, you can easilycheck that

∑

k∈N P (N = k) = 1. So with probability 1, the coin comes up heads after only a finite numberof tosses. Now E(N) denotes the expected value of N , or in other words, the average value of N . We haveE(N) =

∑

k∈N kP (N = k). Using the result from problem X36(a), you can easily check that E(N) = 1/p .(For instance, if p = 1/2, the average number of tosses until the coin first comes up heads is 2.) Similarly,E(N2) =

∑

k∈N k2P (N = k). Using the result from problem X36(b), you can easily find the value of E(N2).Let ν = E(N). Then E[(N−ν)2] is called the variance of N . The square root of the variance of N is calledthe standard deviation of N . It is a measure of how much N deviates from its average value, on the average.We have E[(N − ν)2] = E(N2 − 2νN + ν2) = E(N2) − 2νE(N) + ν2 = E(N2) − 2ν2 + ν2 = E(N2) − ν2 .You should find that the variance of N is q/p2 , so the standard deviation of N is

√q/p . (For instance, if

p = 1/2, then q = 1/2 too and the variance of N is 2, so the standard deviation of N is√2.)

X37. Let (ck) be a sequence in (0,∞). Suppose∑∞

k=1 ck converges. do before MT

(a) (The limit comparison test for convergence.) Suppose (ak) is a sequence in C and

lim supk→∞

|ak|ck

<∞.

Prove that∑∞

k=1 ak converges absolutely.9

(b) Let bk = ck/√rk , where rk =

∑∞ℓ=k cℓ . Prove that

limk→∞

bkck

= ∞

but that10∑∞

k=1 bk converges.11

X38. Let (dk) be a sequence in (0,∞). Suppose∑∞

k=1 dk diverges. do before MT

(a) (The limit comparison test for divergence.) Suppose (ak) is a sequence in [0,∞) and

lim infk→∞

akdk

> 0.

Prove that∑∞

k=1 ak diverges.12

7 Hint: Consider Tn − zTn and use it to express Tn in terms of Sn . Another way involves the derivative dTn/dz , butplease don’t use that method. It is not shorter and it is less elementary. Remember, we are pretending for the time being thatwe don’t know calculus!

8 Hint: Consider Un − zUn and use it to express Un in terms of Tn and Sn .9 Colloquially, this says that if the terms of a given series tend to zero as fast as the terms of some absolutely convergent

series, then the given series is itself absolutely convergent.10 Hint: Show that bk < 2(

√rk −√

rk+1) .11 Colloquially, this implies that given any convergent series with positive terms, there is another convergent series whose

terms are positive and tend to zero more slowly than the terms of the given series. In this sense, there is no most slowlyconvergent series with positive terms.

12 Colloquially, this implies that if the terms of a given series are positive and tend to zero more slowly than the terms ofsome divergent series with positive terms, then the given series is itself divergent.

10


(b) Let bk = dk/sk , where sk =∑k

ℓ=1 dℓ . Prove that

limk→∞

bkdk

= 0

but that13∑∞

k=1 bk diverges.14

13 Hint: Show that if n > m , then

bm+1 + · · ·+ bn ≥ 1− sm

sn.

14 Colloquially, this implies that given any divergent series with positive terms tending to zero, there is another divergentseries whose terms are positive and tend to zero faster than the terms of the given series. In this sense, there is no most slowlydivergent series with positive terms.

11


X39. Imagine a pile of n identical books stacked on a table in such a way that the top book overhangs due 6Th

the one below by as much as possible (clearly the overhang would be half a booklength), the next book fromthe top overhangs the one below it by as much as possible, and so on, and the bottom book overhangs theedge of the table by as much as possible. Let Sn be the amount (in booklengths) by which the top bookoverhangs the edge of the table.

(a) Find a general formula for Sn .(b) Determine how large n must be so that no part of the top book is over the table.(c) Show that in principle, if we can use as many books as we like, then we can arrange things so that

the top book overhangs the edge of the table by as many booklengths as we like.

X40.(a) Prove that the series

∞∑

k=3

1

(log k)log k

converges.(b) Prove that the series

∞∑

k=3

1

(log k)log log k

diverges. (Hint: The inequality log x ≤ x/e may help.15 )

X41. (The monotone convergence theorem for series.) Suppose that for each k ∈ N , we have due 6Th

0 ≤ a(1, k) ≤ a(2, k) ≤ a(3, k) ≤ · · ·

and a(j, k) → b(k) as j → ∞ . Prove that

∞∑

k=1

a(j, k) →∞∑

k=1

b(k) as j → ∞.

(If∑∞

k=1 b(k) < ∞ , then this could be deduced from Tannery’s theorem. But please give a self-containedproof that works whether or not

∑∞k=1 b(k) is finite. Warning: Do not take it for granted that if n ∈ N ,

then∑n

k=1 a(j, k) →∑n

k=1 b(k) as j → ∞ . This is not covered by what we did in class, because b(k) maybe ∞ for some k , and even a(j, k) may be ∞ for some j and k . So if you want to use this fact, you shouldprove it in detail. Note also that if ε ∈ (0,∞), you cannot conclude that b(k)− ε < b(k) unless b(k) <∞ .)

Remark. Suppose f(j, k) ≥ 0 for all j, k ∈ N . Then

∞∑

j=1

∞∑

k=1

f(j, k) =

∞∑

k=1

∞∑

j=1

f(j, k) (1)

whether or not either side is finite. We have seen one proof of this in class, based on Tannery’s theorem. Foranother proof, we can apply problem X41. Let a(m, k) =

∑mj=1 f(j, k) and let b(k) =

∑∞j=1 f(j, k). Then

0 ≤ a(1, k) ≤ a(2, k) ≤ a(3, k) ≤ · · · and a(m, k) → b(k) as m→ ∞ . Hence

∞∑

k=1

a(m, k) →∞∑

k=1

b(k)

as m → ∞ , by problem X41. But∑∞

k=1 b(k) =∑∞

k=1 f(j, k) and for each m ∈ N ,∑∞

k=1 a(m, k) =∑∞

k=1

∑mj=1 f(j, k) =

∑mj=1

∑∞k=1 f(j, k). Letting m→ ∞ , we get (1).

15 The proof of this inequality relies on calculus, so will justify it fully later. For now let’s give a preview of the way to proveit. The function x 7→ log x is concave down. Hence its graph lies below each of its tangent lines. But the line with equationy = x/e is tangent to the graph of y = logx at the point (e, 1).

12


X42.(a) (Fatou’s lemma for series.) Suppose c(j, k) ∈ [0,∞) for all j, k ∈ N . Let b(k) = lim infj→∞ c(j, k).

Prove that∞∑

k=1

b(k) ≤ lim infj→∞

∞∑

k=1

c(j, k).

(Hint: Apply problem X41 with a(j, k) = infj′≥j c(j′, k).)

(b) (A generalization of Tannery’s theorem.) Suppose that

a(j, k) → b(k) in C as j → ∞ for each k ∈ N,

|a(j, k)| ≤ L(j, k) for all j, k ∈ N,

∞∑

k=1

L(j, k) <∞ for each j ∈ N,

L(j, k) →M(k) as j → ∞ for each k ∈ N,

∞∑

k=1

M(k) <∞,

and∞∑

k=1

L(j, k) →∞∑

k=1

M(k) as j → ∞.

Prove that as j → ∞ ,∞∑

k=1

|a(j, k)− b(k)| → 0

and∞∑

k=1

a(j, k) →∞∑

k=1

b(k).

(Hint: Note that L(j, k) +M(k)− |a(j, k)− b(k)| ≥ 0 and as j → ∞ ,

L(j, k) +M(k)− |a(j, k)− b(k)| → 2M(k).

Apply part (a). You may find that problem X21(b) helps.)

Example. Let us mention an application of problem X42(b) to probability theory. For each j ∈ N , let(p(j, k))k∈N be a probability distribution on N . This means that p(j, k) ≥ 0 and

∑∞k=1 p(j, k) = 1.

Suppose that p(j, k) → π(k) as j → ∞ . Then∑∞

k=1 π(k) ≤ 1. (This obviously follows from problemX42(a).) Suppose that

∑∞k=1 π(k) = 1, so that (π(k))k∈N is also a probability distribution on N . Then as

j → ∞ ,∞∑

k=1

|p(j, k)− π(k)| → 0

and

supA⊆N

∣

∣

∣

∣

∣

∑

k∈A

p(j, k)−∑

k∈A

π(k)

∣

∣

∣

∣

∣

→ 0.

It is clear that the second conclusion follows from the first. As for the first, it follows from problem X42(b)if we take a(j, k) = L(j, k) = p(j, k) and b(k) =M(k) = π(k).

Remark. Suppose 0 ≤ a(j, k) ≤ b(k) and as j → ∞ , a(j, k) → b(k). Then∑∞

k=1 a(j, k) →∑∞

k=1 b(k) asj → ∞ . This may be viewed as a generalization of problem X41. It obviously follows from problem X42(a).

Remark. Suppose 0 ≤ a(j, k) ≤ b(k) and as j → ∞ , a(j, k) → b(k). Suppose also that∑∞

k=1 a(j, k) hasa limit in [0,∞) as j → ∞ . Then

∑∞k=1 b(k) <∞ . This obviously follows from problem X42(a).

13


X43. Suppose that∑∞

k=0 ak and∑∞

ℓ=0 bℓ both converge absolutely. For m = 0, 1, 2, . . . , let

cm =

m∑

k=0

akbm−k.

Prove that∑∞

m=0 cm converges absolutely.

X44. Suppose that∑∞

k=1 ak converges absolutely and that (bk) is bounded. Prove that∑∞

k=1 akbk con-verges.

X45. (Toeplitz’s Lemma.) Let a(j, k) ∈ C for all j, k ∈ N . Suppose that for each k ∈ N , due 7Th

a(j, k) → 0 as j → ∞. (2)

Suppose also that

M = supj∈N

∞∑

k=1

|a(j, k)| <∞. (3)

(a) Let (wk) be a sequence in C . Suppose that wk → 0 as k → ∞ . Prove that

∞∑

k=1

|a(j, k)wk| → 0 as j → ∞

and therefore∞∑

k=1

a(j, k)wk → 0 as j → ∞.

(Hint: Don’t try to use Tannery’s theorem. It does not apply here. Instead, write a proof fromscratch.)

(b) Now suppose in addition that α ∈ C and that

∞∑

k=1

a(j, k) → α as j → ∞. (4)

Let z ∈ C and let (zk) be a sequence in C such that zk → z as k → ∞ . Prove that

∞∑

k=1

a(j, k)zk → αz as j → ∞.

(Hint: This is a simple corollary of part (a).)

Remark. Suppose that for each j ∈ N , (a(j, k))k∈N is a probability distribution on N . Then (3) holdswith M = 1. Hence if (2) also holds, then whenever zk → z in C , we have

∞∑

j=1

a(j, k)zk → z as j → ∞.

Colloquially, we may say that weighted averages of the zk’s tend to z as long as the weights for each fixedk tend to 0. Obviously, this generalizes part of our earlier homework problem on Cesaro averages (problemX23).

Remark. Not only does Tannery’s theorem not apply in problem X45, but in fact, part (b) this exerciseprovides an example where the limit of the sum is not the sum of the limit (unless α = 0 or z = 0) andindeed we would not want it to be.

14


X46. (Kronecker’s Lemma.) Let (bk) be an increasing sequence in (0,∞) such that bk → ∞ as k → ∞ .Let (yk) be a sequence in C and suppose that the series

∞∑

k=1

ykbk

converges. Prove that

1

bn

n∑

k=1

yk → 0 as n→ ∞.

(Hint: Let z =∑∞

k=1 yk/bk , let z0 = 0, and for n ≥ 1, let zn =∑n

k=1 yk/bk . Then yk = bk(zk − zk−1).Use this to rewrite (1/bn)

∑nk=1 yk and thereby show, using Toeplitz’s lemma, that it tends to z − z .)

Remark. A particular case of Kronecker’s lemma is that if

∞∑

k=1

ykk

converges, then

1

n

n∑

k=1

yk → 0 as n→ ∞.

Kronecker’s lemma, and especially this particular case of it, is used in probability theory, in the proof of thestrong law of large numbers. Other particular cases of Kronecker’s lemma are used in the proofs of resultsabout the rate of convergence in the strong law of large numbers.

X47. (A variation on Toeplitz’s Lemma.) Let a(j, k) ∈ [0,∞) for all j, k ∈ N . Suppose that for eachk ∈ N ,

a(j, k) → 0 as j → ∞.

Suppose also that for each j , we have∑∞

k=1 a(j, k) <∞ . Let

α = lim infj→∞

∞∑

k=1

a(j, k) and β = lim supj→∞

∞∑

k=1

a(j, k)

and suppose that α, β ∈ [0,∞). Let (xk) be a bounded sequence in R and let

a = lim infk→∞

xk and b = lim supk→∞

xk.

For each j , let

sj =

∞∑

k=1

a(j, k)xk.

(Be sure to explain why this sum is defined for each j .) Let

A = lim infj→∞

sj and B = lim supj→∞

sj .

Prove that B ≤ bβ . Similarly aα ≤ A .

Remark. Evidently, problem X47 is another generalization of part of our earlier homework problem onCesaro averages (problem X23).

15


X48. Let (am) be a sequence in C and let A ∈ C . Let (Mn) be a sequence of integers satisfying

0 =M1 < M2 < M3 < · · · .

For each n , letcn = max

∣

∣aMn+1 + · · ·+ am∣

∣ :Mn + 1 ≤ m ≤Mn+1

.

Prove that the seriesa1 + a2 + a3 + . . .

converges and its sum is A if and only if cn → 0 and the series(

a1 + · · ·+ aM2

)

+(

aM2+1 + · · ·+ aM3

)

+(

aM3+1 + · · ·+ aM4

)

+ · · ·

converges and its sum is A . In other words, letting

bn =

Mn+1∑

m=Mn+1

am,

prove that the series∑∞

m=1 am converges and its sum is A if and only cn → 0 and the series∑∞

n=1 bnconverges and its sum is A .

X49. Consider the series

S = 1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8+− · · ·

and its rearrangement

T = 1 +1

3− 1

2+

1

5+

1

7− 1

4+ +− · · · . (5)

In class, we showed informally that T = 3S/2. Prove this rigorously. (Hint: The result of problem X48should help. By definition,

S =

∞∑

m=0

(−1)m

m+ 1.

Show that we also have

S =

∞∑

n=0

(

1

2n+ 1− 1

2n+ 2

)

(6)

Use this to get an expression for S/2. Add this expression for S/2 to the expression for S in(6). You shouldget

3S

2=

∞∑

n=0

(

1

4n+ 1+

1

4n+ 3− 1

2n+ 2

)

.

Explain how it follows from this that the series in (5) converges and its sum is 3S/2.)

X50. Let E be a set, let f, g : E → C , let B be a filter base on E , and let a, b, c ∈ C . Prove the following: due 7Th

(a) If f → a along B , then cf → ca along B .(b) If f → a along B and g → b along B , then f + g → a+ b along B .(c) If g → b along B , then there exists B ∈ B such that g is bounded on B .(d) If f → 0 along B and if there exists B ∈ B such that g is bounded on B , then fg → 0 along

B .(e) If f → a along B and g → b along B , then fg → ab along B .(f) Suppose g → b along B and b 6= 0. Let E1 = x ∈ E : g(x) 6= 0 and let

B1 = B ∩ E1 : B ∈ B .

Then B1 is a filter base on E1 and

1

g→ 1

balong B1.

16


X51. Let X and Y be topological spaces, let S be a dense subset of X , and let f : S → Y be continuous. due 8Th

Let g be the set of all points (p, q) belonging to X × Y such that for each nhd V of q in Y , there existsa nhd U of p in X such that f [U ∩ S] ⊆ V . Let D be the set of all p ∈ X such that there exists q ∈ Ysuch that (p, q) ∈ g .

(a) For each p ∈ X , let Up be the collection of nhds of p in X , let Bp = U ∩ S : U ∈ Up , andlet Cp = f [B] : B ∈ Bp . Since S is dense in X , each p ∈ X belongs to the closure of S , soBp is a filter base on S , and hence Cp is a filter base on Y . Verify that g is the set of all points(p, q) ∈ X × Y such that Cp → q , or in other words, such that f(x) → q as x runs along Bp .

(b) Prove that S ⊆ D and that if Y is Hausdorff, then g is a function16 from D to Y and g is anextension of f .

(c) Suppose Y is Hausdorff and regular.17 Prove that g is continuous.(d) Suppose Y is a metric space, with metric d say. Then of course Y is Hausdorff and regular, so

the conclusions of parts (b) and (c) hold. Let

E = p ∈ X : the filter base Cp is Cauchy .For each n ∈ N , let

Gn = p ∈ X : p has a nhd U in X such that diam(f [U ∩ S]) ≤ 1/n .Prove that D ⊆ E , that each Gn is open, and that E =

⋂∞n=1Gn . In particular, E is a Gδ set.

(e) Suppose Y is a complete metric space, with metric d . Prove that D = E . In particular, D is aGδ set.

(f) Continue to suppose that Y is a complete metric space, with metric d . Now suppose in additionthat X is also a metric space, with metric ρ say, and that f is uniformly continuous.18 Provethat D = X and that g is uniformly continuous.

Remark. It is worth noticing that in problem X51(c), g is the maximal continuous extension of f , in asense that we shall now explain. Suppose that S ⊆ T ⊆ X and h : T → Y is a continuous extension of f .Then T ⊆ D and g is an extension of h . In particular, for each T satisfying S ⊆ T ⊆ D , the restriction ofg to T is the unique continuous extension of f to T .

Proof. Let p ∈ T . Then h is continuous at p . We wish to show that p ∈ D and g(p) = h(p). Let q = h(p)and let V be a nhd of q in Y . Since h is continuous at p , there exists a nhd N of p in T such thath[N ] ⊆ V . But N = U ∩ T for some nhd U of p in X . Let x ∈ U ∩ S . Then x ∈ U ∩ T = N , soh(x) ∈ V . But h(x) = f(x), because x ∈ S and h is an extension of f . Thus f(x) ∈ V . This holds foreach x ∈ U ∩ S . Therefore f [U ∩ S] ⊆ V . We have shown that for each nhd V of q in Y , there exists anhd U of p in X such that f [U ∩ S] ⊆ V . Therefore (p, q) ∈ g . In other words, p ∈ D and g(p) = q . Butq = h(p). Therefore g(p) = h(p).

X52. Let E ⊆ R and let f : E → R be increasing.19 Let I = f [E] . Suppose that I is an interval. Prove due 8Th

that f is continuous. Base your proof directly on the definition of a continous function. (Warning: E neednot be an interval. Hint: Let x0 ∈ E . We wish to show that f is continuous at x0 . Let y0 = f(x0). Noweither y0 is in the interior of I , or y0 is the left endpoint of I , or y0 is the right endpoint of I . You mayfind that it clarifies matters to consider these three cases separately. Of course, if I does not contain its leftendpoint, then the second case cannot occur, but you need not make special allowances for this possibility.You just need to make sure you cover all cases that might occur. Whether or not they actually do occur isnot important. So likewise, you need not make special allowances for the possibility that I does not containits right endpoint.)

16 As is commonly done, for the purposes of this problem, we consider a function to be the same thing as the set of orderedpairs that is its graph.

17 To say that a topological space Z is regular means that each point in Z has a neighbourhood base consisting of closedsets. For instance, if Z is a metric space, then Z is regular, because for each z ∈ Z , the closed balls B[z, r] , 0 < r < ∞ , forma neighbourhood base at z consisting of closed sets.

18 To say that f is uniformly continuous means that for each ε > 0, there exists δ > 0 such that for all x, x′ ∈ S , ifρ(x, x′) < δ , then d

(

f(x), f(x′))

< ε .19 To say that f is increasing means that for all x, x′ ∈ E , if x ≤ x′ , then f(x) ≤ f(x′) . And by the way, to say that f

is strictly increasing means that for all x, x′ ∈ E , if x < x′ , then f(x) < f(x′) . So for instance, even a constant function isincreasing, though not strictly increasing of course.

17


Remark. In general, a function whose range is an interval need not be continuous, as simple examples show.The assumption in problem X52 that f is increasing matters. Of course it would also work to assume thatf is decreasing.

Remark. Let a ∈ (1,∞). From Holder’s theorem on ordered groups, we know that there is a uniqueincreasing function f : (0,∞) → R such that f(a) = 1 and for all x, x′ ∈ (0,∞), f(xx′) = f(x) + f(x′).Furthermore, the range of f is all of R and f is strictly increasing. By definition, for each x ∈ (0,∞),loga x is f(x). In view of problem X52, the function x 7→ loga x is continuous on (0,∞). Since f is strictlyincreasing, f is one-to-one, so f−1 is a function. Since the domain of f is (0,∞) and the range of f isR , the domain of f−1 is R and the range of f−1 is (0,∞). Since f is strictly increasing, so is f−1 . Itis not hard to check that f−1 is the unique increasing function g : R → R such that g(1) = a and for ally, y′ ∈ R , g(y + y′) = g(y)g(y′). By definition, for each y ∈ R , ay is f−1(y). In view of problem X52, thefunction y 7→ ay is continuous on R . Of course if a ∈ (0, 1) instead, then by definition, for each x ∈ (0,∞),loga x is − log1/a x , and for each y ∈ R , ay is (1/a)−y . Trivially, if a = 1, then by definition, for eachy ∈ R , ay = 1. (And of course, if a = 1, then loga is undefined.)

X53. Let I be an interval in R and let f : I → R . Suppose that f is midpoint-convex.20

(a) Prove for each dyadic rational number21 t between 0 and 1, and for all x, x′ ∈ I , we have

f((1− t)x+ tx′) ≤ (1− t)f(x) + tf(x′). (7)

(Hint: For n = 0, 1, 2, . . . , let Dn be the set of numbers of the form k/2n where k ∈ 0, 1, . . . , 2n .Prove by induction on n that for each integer n ≥ 0, for each t ∈ Dn , for all x, x

′ ∈ I , (7) holds.)

(b) Now suppose in addition that f is continuous. Prove that f is convex.22

Remark. Let a ∈ (0,∞). Recall that for all u, v ∈ [0,∞),√uv ≤ (u+v)/2. Hence a(x+x′)/2 = (axax

′

)1/2 ≤(ax+ ax

′

)/2. Thus the function x 7→ ax is midpoint convex on R . But as we have seen above, this functionis continuous on R . Therefore, by problem X53, it is convex on R .

Definition. Let X be a topological space, let p ∈ X , and let E ⊆ X . To say that p is near E means thatp belongs to the closure of E .

Example. Let X be a metric space, with metric d . Let p ∈ X and let E ⊆ X . Then p is near E iffd(p,E) = 0.

X54. Let X and Y be topological spaces, let f : X → Y , and let p ∈ X . Prove that f is continuous at pif and only if for each subset E ⊆ X , if p is near E , then f(p) is near f [E] .

Let X be a topological space and let f : X → [−∞,∞] . To say that f is lower semicontinuous meansthat for each y ∈ R , the set f > y = x ∈ X : f(x) > y is open in X . To say that f is uppersemicontinuous means that for each y ∈ R , the set f < y = x ∈ X : f(x) < y is open in X . It is easyto see that f is continuous if and only if f is both lower and upper semicontinous.

Let X be a set and let A ⊆ X . The indicator function for A is the function 1A on X defined by

1A(x) =

1 if x ∈ A,

0 if x ∈ X \A.

X55. Let X be a topological space and let A ⊆ X .

(a) Prove that 1A is lower semicontinuous iff A is open.

(b) Prove that 1A is upper semicontinuous iff A is closed.

(c) Let D = p ∈ X : 1A is not continuous at p . Prove that D is the frontier of A .

20 To say that f is midpoint-convex means that for all x, x′ ∈ I , f((x + x′)/2) ≤ (f(x) + f(x′))/2.21 To say that t is a dyadic rational number means that t = k/2n for some integer k and some integer n ≥ 0.22 To say that f is convex means that for each t ∈ [0, 1] , for all x, x′ ∈ I , f((1 − t)x+ tx′) ≤ (1− t)f(x) + tf(x′) .

18


Let X be a set, let f : X → [−∞,∞] , and let B be a filter base on X . By definition, the limit inferiorof f along B is

lim infB

f = supB∈B

infx∈B

f(x)

and the limit superior of f along B is

lim supB

f = infB∈B

supx∈B

f(x).

Sometimes we may write lim infx→B f(x) for lim infB f and lim supx→B f(x) for lim supB f . The notationlim infx→B f(x) may be read “the limit inferior of f(x) as x runs along B .” The notation lim supx→B f(x)may be read “the limit superior of f(x) as x runs along B .”

X56. Let X be a topological space, let S be a dense subset of X , and let f : S → [−∞,∞] . For eachp ∈ X , let Up be the collection of nhds of p in X and let Bp = U ∩ S : U ∈ Up . Define [−∞,∞]-valuedfunctions f∗ and f∗ on X by

f∗(p) = lim infBp

f and f∗(p) = lim supBp

f.

(a) Verify that f∗ ≤ f∗ on X and that f∗ ≤ f ≤ f∗ on S .(b) Prove that f∗ is lower semicontinuous and that f∗ is upper semicontinuous.(c) Let S ⊆ T ⊆ X , let ϕ : T → [−∞,∞] be lower semicontinuous, and suppose that ϕ ≤ f on S .

Prove that ϕ ≤ f∗ on T . Thus the restriction of f∗ to T is the largest lower semicontinuousfunction on T which is less than or equal to f on S .

(d) Let S ⊆ T ⊆ X , let ψ : T → [−∞,∞] be upper semicontinuous, and suppose that f ≤ ψ on S .Prove that f∗ ≤ ψ on T . Thus the restriction of f∗ to T is the smallest upper semicontinuousfunction on T which is greater than or equal to f on S .

(e) Verify that the set of all p ∈ S such that f is continuous at p is equal to

p ∈ S : f∗(p) = f∗(p)

.

(f) Suppose that f is continuous on S . Let D =

p ∈ X : f∗(p) = f∗(p)

and let g be the restrictionof f∗ (or equivalently, f∗ ) to D . Prove that D and g are the same as in problem X51 in thespecial case where the topological space Y there is [−∞,∞] .

Terminology. In the notation of problem X56, in the special case where S = X , f∗ is called the lowerregularization of f and f∗ is called the upper regularization of f .

X57. Let X be a topological space, let A ⊆ X , and let f = 1A . Prove that

f∗ = 1A and f∗ = 1A,

where f∗ and f∗ are the lower and upper regularizations of f and where A is the interior of A and A isthe closure of A .

Reminder. Let B and C be filter bases. To say that B is finer than C means that for each C ∈ C , thereexists B ∈ B such that B ⊆ C .

X58. Let A , B , and C be filter bases. Suppose that A is finer than B and B is finer than C . Provethat A is finer than C .

Definition. Let B and C be filter bases. To say that B and C are equally fine means that each is finerthan the other.

X59. Let X be a set and let Φ be the set of all filter bases on X . Define a binary relation ∼ on X byA ∼ B iff A and B are equally fine. Prove that ∼ is indeed an equivalence relation on Φ.

X60. Let X be a topological space, let B and C be equally fine filter bases on X , and let p ∈ X .(a) Prove that B → p iff C → p .(b) Prove that p is a cluster point for B iff p is a cluster point for C .

19


X61. Let X and Y be sets, let f : X → Y , and let B and C be equally fine filter bases on X . Prove thatfB and fC are equally fine filter bases on Y .

X62. Let B be a filter base. Let C ⊆ B such that for each B ∈ B , there exists C ∈ C such that C ⊆ B .(a) Prove that C is a filter base.(b) Prove that the filter bases B and C are equally fine.

Example. Let X be a topological space, let p ∈ X , let U be the filter base of all nhds of p in X , and letB be a nhd base at p for X . Then B is a filter base too and the filter bases B and U are equally fine.

X63. Let B be a filter base. Suppose there is a countable filter base D such that B and D are equallyfine. Prove that there is a countable filter base C such that B and C are equally fine and C ⊆ B .

Definition. Let B be a filter base and let (xn) be a sequence. To say that (xn) is cofinal in B meansthat for each B ∈ B , there exists N such that for each n ≥ N , we have xn ∈ B .

Example. Let X be a topological space, let B be a nhd base at p for X , and let (xn) be a sequence inX . Then (xn) is cofinal in B iff xn → p .

X64. Let (xn) be a sequence and let B be a filter base. Let T be the filter base of tails of (xn). Provethat (xn) is cofinal in B iff T is finer than B .

X65. Let B and C be equally fine filter bases. Let (xn) be a sequence. Prove that (xn) is cofinal in Biff (xn) is cofinal in C .

X66. Let E be a set, let Y be a topological space, let f : E → Y , let B be a filter base on E , and letq ∈ Y .

(a) Prove that if f → q along B , then for each sequence (xn) which is cofinal in B , we havef(xn) → q .

(b) Suppose that B and some countable filter base are equally fine and that for each sequence (xn)which is cofinal in B , we have f(xn) → q . Prove that f → q along B .

X67. Let X and Y be topological spaces. Let f : X → Y and let p ∈ X .(a) Prove that if f is continuous at p , then for each sequence (xn) in X such that xn → p , we have

f(xn) → f(p).(b) Suppose that there exists a countable nhd base at p for X and that for each sequence (xn) in X

such that xn → p , we have f(xn) → f(p). Prove that f is continuous at p .

20


X68. Let X be a connected metric space with at least two points. Prove that the cardinality of X isgreater than or equal to the cardinality of R .

Remark. Let X be a topological space. It is obvious that if A and B are separated subsets of X , then Aand B are disjoint. It is also obvious that if E and F are disjoint closed subsets of X , then E and F areseparated. But in general, disjoint sets need not be separated. For instance, in R , the sets Q and R \Qare disjoint but not separated.

X69. Let X be a topological space and let U and V disjoint open subsets of X . Prove that U and V areseparated.

X70. Let X be a metric space, with metric d . Let A and B be separated subsets of X . Prove that thereexist disjoint open subsets U and V in X such that A ⊆ U and B ⊆ V . (Hint: To avoid trivialities,suppose that A and B are non-empty. Consider the functions f and g on X defined by f(x) = d(x,A)and g(x) = d(x,B). Use f and g to define U and V somehow.)

X71. Let X be an infinite set. Let C be the collection of all subsets C ⊆ X such that X \C is finite. LetG = C ∪ Ø .

(a) Prove that G is a topology on X . (G is called the cofinite topology on X .)

For the remainder of this exercise, let X be endowed with the topology G .

(b) Prove that X is T1 but not Hausdorff. (To review the basic facts about T1 spaces, see problemX32.)

(c) Give an example of two disjoint closed sets A and B in X for which there do not exist disjointopen sets U and V with A ⊆ U and B ⊆ V . Contrast this with what we saw in problem X70.

X72. Let X be a metric space and let A be a non-empty bounded subset of X . Define a function f on due 9Tu

X by f(x) = sup d(x, a) : a ∈ A .(a) Verify that f(x) ≥ 0 for each x ∈ X .

(b) Verify that f(x) <∞ for each x ∈ X .

(c) Prove that for all x, x′ ∈ X , |f(x) − f(x′)| ≤ d(x, x′). In particular, f is a Lipschitz function.Still more particularly, f is continuous. (Warning: Do not “calculate” with sup. Any inequalitiesyou assert involving sup must be justified based on the definition of sup. Also, please do not useproof by contradiction when it is not needed and does not shorten the argument.)

(d) Now suppose that each closed, bounded subset of X is compact. Let R be the set of all r ∈ [0,∞)such that A ⊆ B[x, r] for some x ∈ X , where B[x, r] is the closed ball with center x and radius r .Let r0 = inf R . Prove that r0 ∈ R . Thus there exists a closed ball in X that is “circumscribed”about A . (Hint: Apply part (c).)

X73. Let A be a non-empty bounded subset of Rd . By problem X72, there exists a closed ball in Rd thatis “circumscribed” about A . Prove that this ball is unique. (Hint: The proof depends on special propertiesof the metric on Rd .)

Let X be a topological space and let A ⊆ X . To say that A is nowhere dense means that the closureof A has empty interior. To say that A is meager means that A is a countable union of nowhere densesets. For instance, any countable subset of R is meager in R . So is any closed subset of R which hasempty interior. For instance, the Cantor set is meager in R . You should think of meager sets as sets whichare small in the sense of topology. Meager sets are also known as sets of the first category. Sets which arenot meager are also known as sets of the second category. To say that X is a Baire space means that nonon-empty open subset of X is meager.

X74. Let X be a topological space. Prove that the following are equivalent:

(a) X is a Baire space.

(b) For each meager set A ⊆ X , the set X \A is dense in X .

(c) For each sequence (Gn) of dense open subsets of X , we have⋂∞

n=1Gn is dense in X .

21


X75. Let X be a complete metric space. Prove that X is a Baire space. (Terminology: This result isknown as the Baire category theorem. Hint: Let (Gn) be a sequence of dense open subsets of X and letH =

⋂∞n=1Gn . We wish to show that H is dense in X . Let x ∈ X and let r > 0. Since G1 is dense,

B(x, r) ∩ G1 6= Ø. Let x1 ∈ B(x, r) ∩ G1 . Since B(x, r) ∩ G1 is open, there exists r1 ∈ (0, 1) suchthat B[x1, r1] ⊆ B(x, r) ∩ G1 . Since G2 is dense, B(x1, r1) ∩ G2 6= Ø. Let x2 ∈ B(x1, r1) ∩ G2 . SinceB(x1, r1) ∩ G2 is open, there exists r2 ∈ (0, 1/2) such that B[x2, r2] ⊆ B(x1, r1) ∩ G2 . Continuing in thisway, construct a Cauchy sequence (xn) which converges to a point in B(x, r) ∩H .)

X76. Let A be a dense Gδ set in R . Prove that A is uncountable. (Hint: Suppose A is countable. Usethe Baire category theorem to get a contradiction.)

Remark. In problem X76, R could be replaced by any non-empty complete metric space without isolatedpoints.

X77. Prove that the set of rational numbers is not a Gδ set in R .

Let X be a topological space. To say that B is a base for the topology of X means that B is acollection of open subsets of X and each open subset of X is the union of some subcollection of B . Sincethe union of any collection of open sets is open, if B is a base for the topology of X , then the collection ofopen subsets of X is precisely the collection of all possible sets that can be expressed as the union of somesubcollection of B . To say that X is second countable means that there exists a countable base for thetopology of X . To say that A is dense in X means that A is a subset of X whose closure is all of X .For instance, Q is dense in R . More generally, Qd is dense in Rd . To say that X is separable means thatsome countable subset of X is dense in X . For instance, Rd is separable, because Qd is countable anddense in Rd .

X78. Let X be a separable metric space. Prove that X is second countable.

Remark. It follows from problem X78 that Rd is second countable. Another way to see this is to noticethat the collection of sets of the form

∏dk=1(ak, bk), where ak, bk ∈ Q and ak < bk , is a countable base for

the usual topology on Rd .

X79. Let X be a second countable topological space and let X1 be a subspace of X . Prove that X1 issecond countable.

X80. Let X be a second countable topological space. Prove that X is separable and so is each subspaceof X .

X81. For each point z = (x, y) ∈ R2 and each r > 0, let U(z, r) be the rectangle [x, x+ r)× [y, y+ r). LetG be the collection of all subsets G ⊆ R2 such that for z ∈ G , there exists r > 0 such that U(z, r) ⊆ G .

(a) Prove that G is a topology on R2 .Let Z denote R2 endowed with the topology G , rather than the usual topology. Let

Z1 = (x,−x) : x ∈ R .

Endow Z1 with the subspace topology it inherits from Z .(b) Prove that Z is first countable.(c) Prove that Z is separable.(d) Prove that Z1 is discrete. In other words, prove that every subset of Z1 is open in Z1 .(e) Prove that Z1 is not separable. Thus a subspace of a separable topological space need not be

separable.(f) Prove that Z is not second countable. Thus a separable first countable space need not be second

countable.

22


X82. Let Y = R . Give an example of a subset X ⊆ R2 and a one-to-one Lipschitz function f from Xonto Y such that f−1 : Y → X is discontinuous at each point in Y .

X83. Give an example of subsets X,Y ⊆ R and a one-to-one continuous map f from X onto Y such thatf−1 : Y → X is discontinuous at each point in Y . (Hint: Consider an enumeration of the rationals.)

X84. Let X be a totally bounded metric space.(a) Prove that X is separable.(b) Prove that X is second countable.

In particular, a compact metric space is separable and second countable.

There are a couple of concepts in topology that are related to compactness, namely countable compact-ness and sequential compactness. Countable compactness and sequential compactness are less importantconcepts than compactness, but by exploring these three notions and the connections between them, as youare asked to do in some of the following exercises, you may improve your understanding of compactness.

Reminder. To say that a topological space is countably compact means that each countable open cover ofthe space has a finite subcover. Obviously each compact space is countably compact. It is not hard to giveexamples of countably compact spaces which are not compact, though these are not particularly importantin analysis.

X85. Let X be a topological space. In class, we have seen that if X is countably compact, then eachsequence in X has a cluster point. Prove the converse of this.

X86. Let X be a non-empty topological space. In class, we have seen that if X is countably compact, due 10Th

then each upper semicontinuous function on X achieves a maximum. Prove the converse of this.

X87. Let X be a topological space. Prove that X is compact iff each filter base on X has a cluster pointin X .

Definition. Let X be a topological space. To say that X is sequentially compact means that each sequencein X has a convergent subsequence.

X88. Let X be a topological space. Suppose X is sequentially compact. Prove that X is countablycompact.

It is not hard to give examples of countably compact spaces which are not sequentially compact, but assuch spaces are not particularly important in analysis, we shall not bother to do so.

X89. Let X be a topological space. Suppose X is countably compact and first countable. Prove that Xis sequentially compact.

Remark. Since each compact space is countably compact, it follows from problem X89 that each compactfirst countable space is sequentially compact.

X90. Let X be a topological space. Prove that X is compact iff for each filter base B on X , there existsa finer filter base C on X which converges to some point in X .

Reminder. Let X be a metric space. In class, we saw that X is totally bounded iff each sequence in Xhas a Cauchy subsequence.

X91. Let X be a metric space. Prove that the following are equivalent:(a) X is compact.(b) X is countably compact.(c) X is sequentially compact.

Remark. Let X be a topological space. By problem X87 and problem X90, the following are equivalent:(d) X is compact(e) Each filter base on X has a cluster point.(f) For each filter base B on X , there exists a finer filter base C on X which converges to some

point in X .

X92. Let Γ be a set of filter bases. Suppose Γ is upwards directed by fineness. In other words, supposethat for each A ∈ Γ and each B ∈ Γ, there exists C ∈ Γ such that C is finer than each of A and B . LetF =

⋃

Γ. Prove that F is a filter base.

23


Definition. Let X be a set. To say that M is a maximal filter base on X means that M is a filter baseon X and for each filter base F on X , if F is finer than M , then M and F are equally fine.

Remark. Some of you may know about filters and ultrafilters. If so, you may be wondering about therelationship between maximal filter bases and ultrafilters. If you don’t care about this relationship, or if youdon’t know about these topics, then you may disregard the rest of this remark, because it is enough just toknow about filter bases and maximal filter bases.

Let X be a set. To say that F is a filter on X means that F is a non-empty collection of non-emptysubsets of X such that F is closed under finite intersections and for each F ∈ F and each A ⊆ X , ifA ⊇ F , then A ∈ F . Evidently a filter on X is also a filter base on X , but not conversely. If B is acollection of subsets of X and

F = F ⊆ X : F ⊇ B for some B ∈ B

then F is a filter on X iff B is a filter base. In this case, B and F are equally fine filter bases and we saythat B is a base for the filter F and that F is the filter on X generated by B . If A and B are filterbases on X and if E and F are the filters on X generated by A and B respectively, then B is finer thanA iff F ⊇ E , and A and B are equally fine iff E = F . To say that U is an ultrafilter on X means thatU is a filter on X which is not properly contained in any other filter on X . If M is a filter base on X andU is the filter on X generated by M , then M is a maximal filter base on X iff U is an ultrafilter on X .

One reason why I have chosen to emphasize filter bases rather than filters is that filter bases are moreconcrete objects than filters. For instance, if (xn) is a sequence in R , then the filter base of tails of (xn) is

a countable set of countable subsets of R , whereas the filter that this generates on R has cardinality 22ℵ0

and contains many uncountable sets. Another reason is that the notion of the image of a filter base under amap is simpler than the notion of the image of a filter under a map.

X93. Let X be a set and let E be a filter base on X . Prove that there exists a maximal filter base M onX such that M is finer than E . (Hint: Let Φ be the set of all filter bases A on X such that E ⊆ A .Partially order Φ by inclusion. Use Zorn’s lemma to show that Φ has an element M which is maximal withrespect to inclusion. Since E ⊆ M , it is obvious that M is finer than E . Prove that M is maximal in thesense of filter bases.)

X94. Let X be a topological space, let M be a maximal filter base on X , and let p ∈ X . Prove thatM → p iff M clusters at p .

X95. Let X be a topological space. Prove that X is compact iff each maximal filter base on X converges.

X96. Let X be a set, let A be a filter base on X , let E ⊆ X , let B = E ∩ A : A ∈ A , let Ec = X \E ,and let C = Ec ∩ A : A ∈ A . Prove that at least one of B and C is a filter base.

X97. Let X be a set and let M be a filter base on X . For each E ⊆ X , let us say that E is big iff Ehas a subset belonging to M .

(a) Prove that for each E ⊆ X , at most one of E and X \ E is big.(b) Prove that M is a maximal filter base on X iff for each E ⊆ X , either E is big or X \E is big.

X98. Let X and Y be sets, let f : X → Y , and let M be a maximal filter base on X . Prove that fM is a maximal filter base on Y . (Hint: Use problem X97.)

Example. If U and V are open intervals in R , then the rectangle U × V is open in the Cartesian planeR2 = R×R . Most open subsets of R2 are not rectangles, of course. For instance, the unit disk

D =

(x, y) ∈ R×R : x2 + y2 < 1

is open but is not a rectangle. But a subset of R2 is open iff it is a union of some collection (usually aninfinite collection) of open rectangles. This motivates the general definition of the product topology on aCartesian product of topological spaces.

In the next exercise, the product topology is treated for the Cartesian product of any family (even aninfinite family) of topological spaces. But before that, let us review the notion of the Cartesian product ofa family of sets.

24


Notation and Terminology. An indexed family (xα)α∈A is just a function whose domain is the set A andwhose value at α is xα . The term family is just a shorter name for an indexed family. A family x = (xα)α∈A

is not the same thing as the set xα : α ∈ A . Rather, this set is the range of the function x . (If we thinkof a function as being the same thing as its graph, then a family x = (xα)α∈A is the same thing as the set (α, xα) : α ∈ A .) A family of sets (Xα)α∈A is just a family with the property that for each α ∈ A , Xα

is a set. A choice function for a family of sets (Xα)α∈A is a function x whose domain is A and which hasthe property that for each α ∈ A , we have x(α) ∈ Xα . In other words, a choice function for a family of sets(Xα)α∈A is an family (xα)α∈A such that for each α ∈ A , we have xα ∈ Xα . Informally, a choice functionfor a family of sets chooses one element from each of the sets in the family. One form of the axiom of choicestates that each family of non-empty sets has at least one choice function. If (Xα)α∈A is a family of sets,then the Cartesian product

∏

α∈AXα of the family is the set of all choice functions for the family. In otherwords,

∏

α∈A

Xα = (xα)α∈A : xα ∈ Xα for each α ∈ A .

For each α ∈ A , the projection map πα : X → Xα is defined as follows: if x = (xα)α∈A ∈ X , thenπα(x) = xα . In the special case where Xα = B for each α ∈ A , the Cartesian product

∏

α∈AXα is usuallydenoted by BA . Thus BA is the set of all functions from A into B .

X99. Let (Xα)α∈A be a family of topological spaces and let X be the Cartesian product∏

α∈AXα of thisfamily. Let U be the collection of all sets of the form

∏

α∈A Uα , where Uα is open in Xα for each α ∈ Aand Uα = Xα for all but finitely many α ∈ A . Let G be the set of all G ⊆ X such that G is the union ofsome subcollection of U .

(a) Prove that U is closed under finite intersections.

(b) Prove that G is a topology on X .

We call G the product topology. Endow X with the topology G . For each α ∈ A , let πα denote theprojection map from X to Xα .

(c) Prove that for each α ∈ A , πα : X → Xα is continuous.

(d) Let E be a topological space and let f : E → X . For each α ∈ A , let fα = πα f . Evidently, wehave f(u) = (fα(u))α∈A for each u ∈ E . Let p ∈ E . Prove that f is continuous at p iff for eachα ∈ A , fα is continous at p .

Remark. Unless otherwise stated, the Cartesian product of a family of topological spaces is understood tobe endowed with the product topology.

X100. Let (Xα)α∈A be a family of topological spaces and let X =∏

α∈AXα . For each α ∈ A , let παdenote the projection map from X to Xα . Let B be a filter base on X and let x = (xα)α∈A be an elementof X . Prove that B → x iff for each α ∈ A , παB → xα .

Remark. Since convergence of filter bases characterizes a topology, the property of the product topologydescribed in problem X100 characterizes the product topology.

Remark. Let (Xα)α∈A be a countable family of sequentially compact topological spaces and let X =∏

α∈AXα . Then X is sequentially compact. This is almost trivial when A is finite. When A is countablyinfinite, it is proved in Theorem 7.23 in Rudin, Principles of Mathematical Analysis, Third Edition, in thespecial case where each Xα = C , and the proof for the general case is essentially the same. It is worthmentioning that the proof is by a “diagonal subsequence argument” similar to the argument that we usedin class to show that in a totally bounded metric space, each sequence has a Cauchy subsequence.

X101. (Tychonoff’s theorem, 1930, 1935.) Let (Xα)α∈A be a family of compact topological spaces and letX =

∏

α∈AXα . Prove that X is compact. (Hint: Prove that each maximal filter base on X converges.)

Remark on the Proof of Tychonoff’s Theorem. The proof of Tychonoff’s theorem via maximal filterbases is transparently clear and is a natural generalization of the proof that the product of countably manysequentially compact spaces is sequentially compact. There is another common proof of Tychonoff’s theoremwhich is based a result called the Alexandroff subbase theorem. This approach does not require the theoryof filter bases, so it has the advantage of being short and self-contained. Its disadvantage is that it is lessmotivated than the approach via the theory of filter bases.

25


Historical Remarks. It is fair to say that it was Tychonoff’s theorem that clinched the modern definitionof compactness. Prior to Tychonoff’s theorem, countable compactness and sequential compactness weretaken seriously as competitors for the distinction of being the “right” definition of compactness. But neithercountable compactness nor sequential compactness behaves as well with respect to formation of productspaces as compactness does. By the remark immediately preceding problem X101, the product of countablymany sequentially compact spaces is sequentially compact. The product of continuum many copies of [0, 1] iscompact, by Tychonoff’s theorem, but it is not sequentially compact. The behaviour of countable compact-ness with respect to formation of product spaces is even worse: a product of two countably compact spacesneed not be countably compact. For a more detailed discussion of the history of notions of compactness, seethe historical notes in Willard, General Topology, Addison-Wesley, 1970, pages 303 to 305.

Remark. Tychonoff’s theorem depends on the axiom of choice through Zorn’s lemma, which we used inproblem X93. But more is true: It is not hard to show that Tychonoff’s theorem is actually equivalent tothe axiom of choice.

Let X and Y be topological spaces. To say that h is a homeomorphism from X to Y means that his a one-to-one continuous map from X onto Y and h−1 : Y → X is also continuous. To say that X ishomeomorphic to Y means that there exists a homeomorphism from X to Y . Informally, X and Y arehomeomorphic when they look the same as topological spaces.

X102. Let Σ = 0, 1 N . Note that Σ is the product of countably many copies of the two-point discretespace 0, 1 . Give Σ the corresponding product topology.

(a) Explain why Σ is compact.Now let C be the Cantor set, and let f be the one-to-one map from Σ onto C defined in problem X11.

(b) Prove that f : Σ → C is continuous.(c) Prove that f−1 : C → Σ is continuous. (Hint: Σ is compact and C is Hausdorff.)

Thus Σ is homeomorphic to C . (In particular, Σ is not discrete, even though it is a product of discretespaces.)

To say that a topological space is metrizable means that its topology arises from some metric. A metricspace is more special than a metrizable space because a metric space has a metric already given on it. Thetopology on a metrizable space can always be induced by any one of an infinite number of metrics, providedthat the space has at least two points.

X103. Prove that the product of countably many metrizable spaces is metrizable. (Hint: For a product due 10Th

of finitely many spaces, this is easy, so let’s consider the countably infinite case. Suppose (Xn)n∈N is asequence of metrizable spaces. Let X =

∏

n∈NXn . For each n , let dn be a metric on Xn which inducesthe given topology on Xn . By problem X2, may suppose that each dn ≤ 1. Define d : X ×X → [0, 1] by

d(x, y) =∑

n∈N

2−ndn(xn, yn).

Prove that d is a metric on X and that d induces the product topology on X .)

X104. Let (X, ρ) be a metric space. Let σ = ρ/(1+ ρ). Then by problem X2, σ is a metric on X and thetwo metrics ρ and σ induce the same topology on X .

(a) Prove that the identity map on X is uniformly continuous from (X, ρ) to (X, σ).(b) Prove that the identity map on X is uniformly continuous from (X, σ) to (X, ρ).(c) Let (xn) be a sequence in X . Prove that (xn) is ρ-Cauchy iff (xn) is σ-Cauchy.(d) Prove that X is ρ-complete iff X is σ-complete.

To say that a topological space is completely metrizable means that its topology arises from somecomplete metric. For instance, R is completely metrizable, because the usual metric on R is complete. Tomention another example, (0, 1) is completely metrizable, even though the usual metric on it is not complete,because (0, 1) is homeomorphic to R .

X105. Prove that the product of countably many completely metrizable spaces is completely metrizable.

X106. Prove that the product of countably many second countable spaces is second countable.

X107. Prove that the product of countably many separable spaces is separable. (Warning: If A1, A2, A3, . . .are countable, it does not follow that

∏

n∈NAn is countable. For instance, 0, 1 N has the cardinality ofthe continuum.)

26

Math 652 Extra Problems Winter, 2006

These problems are numbered consecutively with the ones from last quarter. (That is why the first of themis not numbered X1.)

X108. (A discrete analog of part of l’Hospital’s rule.) Let (uk) be a sequence in R and let (vk) be a strictlyincreasing sequence in (0,∞) such that vk → ∞ as k → ∞ . Then

lim supj→∞

ujvj

≤ lim supk→∞

uk − uk−1

vk − vk−1

and

lim infj→∞

ujvj

≥ lim infk→∞

uk − uk−1

vk − vk−1.

(Hint: Let u0 = v0 = 0. For all j, k ∈ N , let

a(j, k) =

(vk − vk−1)/vj if k ≤ j,

0 if k > j,

and

xk =uk − uk−1

vk − vk−1.

Apply problem X47.)

X109. Let f : I → R be twice-differentiable, where I = (a,∞) and a ∈ [−∞,∞).(a) Suppose A,C ∈ [0,∞) such that for each x ∈ I , |f(x)| ≤ A and |f ′′(x)| ≤ C . Prove that for each

x ∈ I and each h > 0,

|f ′(x)| ≤ A

h+ Ch.

(Hint: Use Taylor’s formula to express f(x+2h) in terms of f(x), f ′(x), and f ′′(ξ) for a suitableξ between x and x+ 2h .)

(b) Suppose A and C are as in part (a). Deduce from part (a) that for each x ∈ I ,

|f ′(x)|2 ≤ 4AC.

(c) Suppose f ′′ is bounded and f(x) → 0 as x → ∞ . Prove that f ′(x) → 0 as x → ∞ . (Hint: Letα ∈ I and apply part (b) with I replaced by (α,∞). Let α→ ∞ .)

X110. Let f : I → R , where I is an interval in R . Suppose f is convex on I . Let I be the interior due 2Th

of I . Then as we have seen in class, for each x ∈ I , the left derivative DLf(x) and the right derivativeDRf(x) exist and satisfy −∞ < DLf(x) ≤ DRf(x) <∞ . As we have also seen in class, for all x1, x2 ∈ I

with x1 < x2 , we have DRf(x1) ≤ DLf(x2).(a) Prove that DRf is right-continuous on I and DLf is left-continuous on I .(b) Prove that for all but countably many x ∈ I , f is differentiable at x .

By the way, you should not need (a) to prove (b) and you should not need (b) to prove (a).

Although we have not yet begun to discuss integration in this course, I think that should not hinderyou in solving the next two problems.

X111. Let f : [a, b] → R , where a, b ∈ R with a < b . Suppose f is continuous on [a, b] and f ′ and f ′′

exist on (a, b). Let m = (a+ b)/2, the midpoint of the interval [a, b] . Prove that there exists c ∈ (a, b) suchthat

∫ b

a

f(x)dx = (b − a)f(m) +1

24f ′′(c)(b− a)3

(Hint: Let F be an antiderivative for f . Expand F about m using Taylor’s formula to get suitableexpressions for F (a) and F (b). The fact that f ′′ has the intermediate value property, by Darboux’stheorem, is relevant too.)

27


Remark. The result of problem X111 is the basis for the error estimate in the midpoint rule for numericalintegration.

Remark. Let T ∈ (0,∞), let L ∈ R , let g, h : [0, T ] → R be continuous, and suppose h is differentiableon (0, T ) and h′(t) = g(t) + Lh(t) for all t ∈ (0, T ). In any introductory course on differential equations,students are taught how to solve for h in terms of g and h(0) by the method of integrating factors, as

follows: h′(s) − Lh(s) = g(s), so e−Lsh′(s) − Le−Lsh(s) = e−Lsg(s), so(

e−Lsh(s))′

= e−Lsg(s), so

e−Lth(t)− e−L0h(0) =∫ t

0e−Lsg(s) ds , so h(t) = eLt

(

h(0) +∫ t

0e−Lsg(s) ds

)

.

X112. Let T ∈ (0,∞) and let L ∈ [0,∞). due 2Th

(a) (Gronwall’s Inequality). Let f, g : [0, T ] → R be continuous, and suppose that for each t ∈ [0, T ] ,

f(t) ≤ g(t) + L

∫ t

0

f(s) ds. (8)

Prove that for each t ∈ [0, T ] ,

f(t) ≤ g(t) + L

∫ t

0

eL(t−s)g(s) ds. (9)

(Hint: Let h(t) =∫ t

0 f(s) ds . Then h′(t) = f(t), so by (8), h′(t) ≤ g(t) + Lh(t). Adapt themethod illustrated in the remark that precedes this exercise to derive an inequality for h(t) fromwhich (9) follows.)

(b) Let F : [0, T ]×Rd → Rd be continuous and satisfy the following Lipschitz condition:

|F (t, x) − F (t, y)| ≤ L|x− y| for all x, y ∈ Rd and all t ∈ [0, T ].

Suppose u, v : [0, T ] → Rd are continous and are differentiable on (0, T ) and satisfy u′(t) =F (t, u(t)) and v′(t) = F (t, v(t)) for all t ∈ (0, T ). Prove that |u(t) − v(t)| ≤ |u(0) − v(0)|eLt

for all t ∈ [0, T ] . In particular, if u(0) = v(0), then u(t) = v(t) for all t ∈ [0, T ] . (Hint:

Notice that u(t) = u(0) +∫ t

0F (s, u(s)) ds and v(t) = v(0) +

∫ t

0F (s, v(s)) ds for all t ∈ [0, T ] , by

the fundamental theorem of calculus (remember to verify its hypotheses!). Apply part (a) withf(t) = |u(t)− v(t)| and g(t) ≡ γ where γ = |u(0)− v(0)| . You will need the triangle inequality for

integrals of vector-valued functions:∣

∣

∫ t

0 ϕ(s) ds∣

∣ ≤∫ t

0 |ϕ(s)| ds if ϕ : [0, t] → Rd is continuous. Wewill prove this triangle inequality in class in due time. For now, feel free to use it without provingit.)

Remark. In problem X112(b), Gronwall’s inequality is used to derive an estimate on how much the solutionof an initial value problem

y′ = F (t, y),

y(0) = y0,

changes as the initial value y0 changes. Gronwall’s inequality has other important applications in the theoryof differential equations. For instance, if F is a function not just of t and y but also of a parameter w , thenunder suitable conditions, Gronwall’s inequality can be used to derive estimates on how much the solutionof the initial value problem

y′ = F (t, y;w),

y(0) = y0,

changes as the parameter w changes.

28


X113. Let f : [a, b] → R , where a, b ∈ R with a < b . Suppose that f ′ is continuous on [a, b] and that f ′′

exists on (a, b). Suppose in addition that f ′(a) = 0 = f ′(b). Prove that there exists a number c in (a, b)such that

|f ′′(c)| ≥ 4

(b− a)2|f(b)− f(a)|.

X114. Let Φ be the set of all twice continuously differentiable functions f : [−1, 1] → R such that |f ′′(x)| ≤1 for all x ∈ [−1, 1] and f(−1) = f ′(−1) = f(1) = f ′(1) = 0. Let E = f(0) : f ∈ Φ . Find supE .

X115. Let f : [a, b] → R , where a, b ∈ R with a < b . Let c ∈ (a, b).(a) Prove that if f is Riemann-integrable over [a, b] , then f is Riemann-integrable over each of [a, c]

and [c, b] and∫ b

a

f(x) dx =

∫ c

a

f(x) dx +

∫ b

c

f(x) dx.

(Hint: If P is a subdivision of [a, b] , then either c is one of the points of P , or it is not. In thelatter case, if we add c to P , we get a finer subdivision of [a, b] .)

(b) Prove that if f is Riemann-integrable over each of [a, c] and [c, b] , then f is Riemann-integrableover [a, b] .

Review of the Integral Test. Let h : (0,∞) → [0,∞) be decreasing. According to the integral test, whichyou will have learned in calculus, since h is decreasing and non-negative, the series

∑∞k=1 h(k) converges

iff the improper integral∫∞

1 h(x) dx converges. Let us review why this is true and discuss some relatedmatters.

Since h is monotone, h is Riemann-integrable over [a, b] for all a, b ∈ (0,∞) with a < b . As a decreases,

or as b increases,∫ b

a h(x) dx increases, because h ≥ 0. Hence if α, β ∈ [0,∞] with α < β , and if α = 0 orβ = ∞ (or both), then the improper Riemann integral

∫ β

α

h(x) dx = lima↓α, b↑β

∫ b

a

h(x) dx

exists in [0,∞] . Since h is decreasing, if k ∈ N , then whenever 0 < x1 ≤ k ≤ x2 < ∞ , we haveh(x1) ≥ h(k) ≥ h(x2), so

∫ k

k−1

h(x) dx ≥∫ k

k−1

h(k) dx = h(k) =

∫ k+1

k

h(k) dx ≥∫ k+1

k

h(x) dx.

Hence for all m,n ∈ N with m ≤ n ,

∫ n

m−1

h(x) dx =

n∑

k=m

∫ k

k−1

h(x) dx ≥n∑

k=m

h(k) ≥n∑

k=m

∫ k+1

k

h(x) dx =

∫ n+1

m

h(x) dx.

It follows that for each m ∈ N ,

∫ ∞

m−1

h(x) dx ≥∞∑

k=m

h(k) ≥∫ ∞

m

h(x) dx. (10)

In particular, if∑∞

k=1 h(k) < ∞ , then∫∞

1h(x) dx < ∞ and conversely, if

∫∞

1h(x) dx < ∞ , then

∑∞k=2 h(k) < ∞ , so

∑∞k=1 h(k) < ∞ . Furthermore, if we wish to estimate the value of

∑∞k=1 h(k) to

a specified degree of precision, then (10) gives us a way to determine how large m should be so that thedifference

∑∞k=m+1 h(k) between

∑∞k=1 h(k) and the partial sum

∑mk=1 h(k) is as small as we like. Fur-

thermore, it often happens that the difference between the upper and lower bounds on this difference thatare furnished by (10) differ from each other by substantially less than the remainder

∑∞k=m+1 h(k) itself, in

which case the summ∑

k=1

h(k) +

∫ ∞

m+1

h(x) dx

gives a much better approximation to∑∞

k=1 h(k) than the partial sum∑m

k=1 h(k) alone. For example, thisis the case if h(x) = 1/x2 .

29


Definition. Let X be a set, let f, g : X → C \ 0 , and let B be a filter base on X . To say that f isasymptotic to g along B (denoted f(x) ∼ g(x) as x→ B ) means that

f(x)

g(x)→ 1 as x→ B.

For instance, if X ⊆ R and X is not bounded above, then f(x) ∼ g(x) as x → ∞ iff f(x)/g(x) → 1 asx → ∞ . (This should tell you what you need to know about this definition to solve the next exercise, evenif you do not know about filter bases.)

X116. Let a ∈ (1,∞). Define f : (0,∞) → (0,∞) by due 3Th

f(x) =

∞∑

k=1

1

x+ ka.

(a) Let b = 1/a . Prove that there exists C(a) ∈ (0,∞) such that

f(x) ∼ C(a)xb−1 as x→ ∞

and find an expression for C(a) in the form of a certain improper integral.23 (Hint: Compare fwith the function g defined by

g(x) =

∫ ∞

0

1

x+ tadt.

The review above of the integral test should help you understand how to compare f and g .)

(b) Find limx→∞

log f(x)

log x. (Hint: Write f(x) as g(x)

f(x)

g(x), where g(x) = C(a)xb−1 .)

Example. Let a = 7/5 and let f be as in problem X116. Then f(x) ∼ C(a)x−2/7 as x → ∞ . You mayfind it enlightening to compare this with problem 6 on the analysis qualifying from Autumn, 2004.

X117. Let ϕ be a continuous function from R to C such that ϕ(0) = 1 and ϕ(t1 + t2) = ϕ(t1)ϕ(t2) forall t1, t2 ∈ R . Since ϕ is continuous, we may fix δ > 0 such that for each t ∈ [0, δ] , |ϕ(t) − ϕ(0)| ≤ 1/2.

Let c =∫ δ

0 ϕ(t) dt .(a) Prove that c 6= 0. (Make sure your proof does not assume that ϕ is real-valued! This goes for all

parts of this problem.)(b) Define F : R → C by F (x) =

∫ x

0 ϕ(t) dt . Prove that for each x ∈ R , F (x + δ) = F (x) + cϕ(x).(c) Prove that ϕ is differentiable on R .(d) Let k = ϕ′(0). Prove that ϕ′ = kϕ .(e) Prove that ϕ(t) = ekt for all t ∈ R . (Please give a detailed proof that ϕ(t) = ekt . Please do not

just quote a theorem on uniqueness of solutions of differential equations. And please do not give anargument that involves logϕ(t). Since ϕ is complex-valued, the meaning of logϕ(t) would haveto be clarified to make such an argument work and this would be the hard way to do the proof.)

Remark. It is not difficult to extend the result of problem X117 to the case where ϕ takes values in Cd×d ,the set of d× d matrices with complex entries. Then k ∈ Cd×d and ekt is defined as the sum of the series∑∞

n=0(kt)n/n! .

Reminder. If f : [a, b] → R , where a, b ∈ R with a < b , then the upper Riemann integral of f over [a, b]is the infimum of all U(f, P ), where P varies over all subdivisions of [a, b] .

Definition. Let f : [a, b] → Rd , where a, b ∈ R with a < b . To say that f is Riemann null means thatthe upper Riemann integral of |f | over [a, b] is 0.

23 You need not evaluate the integral, but be sure to prove that it is finite and strictly positive, so that C(a) ∈ (0,∞) asstated.

30


X118. Let C be the Cantor set and define f : [0, 1] → R by

f(x) =

1 if x ∈ C,

0 if x ∈ [0, 1] \ C.

Prove that f is Riemann null.

X119. Let f : [a, b] → R , where a, b ∈ R with a < b . Prove that f is Riemann null iff f is Riemann-

integrable and∫ b

a |f(x)| dx = 0.

X120. Let f : [a, b] → Rd , where a, b ∈ R with a < b . Suppose that f is Riemann null and that f iscontinuous. Prove that f = 0. In other words, prove that for each x ∈ [a, b] , f(x) = 0.

X121. Let f1, . . . , fd : [a, b] → R , where a, b ∈ R with a < b . Define f : [a, b] → Rd by f(x) =(f1(x), . . . , fd(x)) for all x ∈ [a, b] . Prove that f is Riemann null iff f1, . . . , fd are all Riemann null.

X122. Let f : [a, b] → Rd , where a, b ∈ R with a < b . Suppose f is Riemann null. Let E be the range due 3Th

of f , let ϕ : E → R be bounded, and suppose that for each ε > 0, there exists δ > 0 such that for eachy ∈ E , if |y| < δ , then |ϕ(y)| < ε . Let h = ϕ f . Prove that h is Riemann null. (Hint: There is a proof ofthis which is similar to but simpler than the proof of a result we discussed in class, namely a variation onTheorem 6.11 in Rudin, Principles of Mathematical Analysis, Third Edition.)

Notation. Let R[a, b] be the vector space of Riemann integrable complex-valued functions on [a, b] , wherea, b ∈ R with a < b . (This is not standard notation.) Let p ∈ [1,∞). Then for each f ∈ R[a, b] , we definethe p-norm of f to be

‖f‖p =

(

∫ b

a

|f(x)|p dx)1/p

.

This is a slight abuse of terminology, since if f ∈ R[a, b] and ‖f‖p = 0, it does not follow that f = 0. It isobvious that if f ∈ R[a, b] and C ∈ C , then ‖Cf‖p = |C|‖f‖p .X123. Let f ∈ R[a, b] , where a, b ∈ R and a < b . Let p ∈ (1,∞). Prove that ‖f‖p = 0 iff f is Riemannnull.

X124. Let f ∈ C[a, b] , where a, b ∈ R and a < b . Let p ∈ (1,∞). Prove that ‖f‖p = 0 iff f = 0.

X125.(a) Let B = h ∈ R[a, b] : ‖h‖p ≤ 1 . Show that B is a convex subset of the vector space R[a, b] .

In other words, for all h1, h2 ∈ B , for each t ∈ [0, 1], we have (1− t)h1 + th2 ∈ B . (Hint: Use thefact that since p ∈ [1,∞), the function y 7→ yp is convex on [0,∞).)

(b) (Minkowski’s Inequality.) Let p ∈ [1,∞) and let f, g ∈ R[a, b] . Prove that ‖f+g‖p ≤ ‖f‖p+‖g‖p .(Hint: This can be deduced from (a). Let a = ‖f‖p and b = ‖g‖p . Notice that if a 6= 0, thenf/a ∈ B . Similarly, if b 6= 0, then g/b ∈ B . Note: Soon we shall consider Holder’s inequality.Many books give a proof of Minkowski’s inequality that is based on Holder’s inequality. That is amuch less enlightening proof. Please do not do it that way.)

Notice that Minkowski’s inequality is the triangle inequality for p-norms. The next result develops somemore theory related to convex functions and applies it to prove another famous inequality involving p-norms,namely Holder’s inequality.

X126.(a) Let I be an interval in R and let f : I → R be convex. Let x1, . . . , xn ∈ I and let t1, . . . , tn ∈ [0, 1]

with∑n

k=1 tk = 1. Prove in two ways that

ϕ(∑n

k=1 tkxk) ≤∑n

k=1 tkϕ(xk).

(Hint: For your first proof, use induction on n , starting from the definition of what it means for ϕto be convex. For your second proof, proceed as follows. Let X = x1, . . . , xn , let a = minX ,let b = maxX , and let x =

∑nk=1 tkxk . To avoid trivialities, assume that a < b and that each

31


tk > 0. Check that a < x < b , so x is an interior point of I . Then apply the fact that the graphof a convex function has a support line at each interior point of the interval in which the functionis defined.)

(b) Let y1, . . . , yn ∈ [0,∞) and let t1, . . . , tn ∈ [0, 1] with∑n

k=1 tk = 1. Let G =∏n

k=1 ytkk and let

A =∑n

k=1 tkyk . (Note that G is a weighted geometric mean of y1, . . . , yn and A is a weightedarithmetic mean of y1, . . . , yn .) Prove that G ≤ A . (Hint: If some yk = 0, then G = 0, so thereis nothing to prove. In the case where each yk > 0, apply the fact that the function x 7→ ex isconvex on R .)

(c) Let u, v ∈ [0,∞) and let p, q ∈ (1,∞) with

1

p+

1

q= 1.

Prove that

uv ≤ up

p+vq

q.

(This follows from (b) with n = 2. The analogous statement for general n is true, but for the sakeof focus, we refrain from considering it.)

(d) (Holder’s Inequality.) Again, let p, q ∈ (1,∞) with

1

p+

1

q= 1.

Let f, g ∈ R[a, b] . Prove that‖fg‖1 ≤ ‖f‖p‖g‖q.

Remark. We can generalize problem X126(d) as follows: Let p1, . . . , pn ∈ (1,∞) with

1

p1+ · · ·+ 1

pn= 1.

Let f1, . . . , fn ∈ R[a, b] . Then‖f1 · · · fn‖1 ≤ ‖f1‖p1

· · · ‖fn‖pn.

The proof is just like that of problem X126(d), but it uses the generalization of problem X126(c) to generaln that we mentioned but did not pursue.

Remark. In problem X126(d), if p = 2, then q = 2. In this case, Holder’s inequality, in conjunction withthe triangle inequality for integrals, implies that if f ∈ R[a, b] , then

∣

∣

∣

∣

∣

∫ b

a

f(x)g(x) dx

∣

∣

∣

∣

∣

≤ ‖f‖2‖g‖2.

This is a particular case of Schwartz’s inequality from the theory of inner product spaces.

32


X127. Let f : (1,∞) → R be differentiable and define g, h : (1,∞) → R by g(x) = f ′(x)/x and h(x) =f(x)/x . Suppose g is bounded. Prove that h is uniformly continuous.

X128. Let f be an n times differentiable real-valued function on [a, b] , where a, b ∈ R with a < b . Supposethat the n-th derivative of f satisfies f (n)(x) > 0 for each x ∈ [a, b] . Prove that f has at most n zeroes in[a, b] .

X129. Let f : [0,∞) → R be Riemann integrable over [0, T ] for each T ∈ (0,∞). For each T ∈ (0,∞), let

I(T ) =1

T

∫ T

0

f(t) dt.

Let a = lim inft→∞ f(t), A = lim infT→∞ I(T ), B = lim supT→∞ I(T ), and b = lim supt→∞ f(t).(a) Prove that B ≤ b .

Similarly, a ≤ A . (Or this can be deduced by applying part (a) to the function −f instead of f .)(b) Let L ∈ [−∞,∞] . Deduce from part (a) that if f(t) → L as t→ ∞ , then I(T ) → L as T → ∞ .

X130. Give an example of a continuous function f : [0,∞) → R such that lim supt→∞ f(t) = ∞ andlim inft→∞ f(t) = −∞ but I(T ) → 0 as T → ∞ , where I(T ) is as in problem X129.

X131. Let f ∈ R[−1, 1] and suppose f is continuous at 0. Prove thatdue 4Th; puta simpler onefirst next time

∫ 1

−1

uf(x)

u2 + x2dx→ πf(0) as u→ 0+.

X132. Let ϕ : R → C be 1-periodic24 and N times continuously differentiable, where N ∈ N . Prove that due 4Th

there is a constant C ∈ [0,∞) such that for each k ∈ Z \ 0 ,∣

∣

∣

∣

∫ 1

0

e2πikxϕ(x) dx

∣

∣

∣

∣

≤ C

|k|N .

X133. Let f : [a, b] → Rd , where a, b ∈ R with a < b . Suppose f is Riemann integrable over [a, b] . DefineF : [a, b] → Rd by F (x) =

∫ x

af(t) dt . Prove that F is continuous.

X134. Let f : [a, b] → R be bounded, where a, b ∈ R with a < b , and suppose that f is Riemann integrableover (α, β) for all α, β ∈ (a, b) with α < β . Prove that f is Riemann integrable over [a, b] . In particular, iff is bounded on [a, b] and continuous on (a, b), then f is Riemann integrable over [a, b] .

X135. Let I = [0, 1], let f : I → R , let C be the Cantor set, and let D = I \ C . Suppose f is boundedon I and continuous at each point in D . Prove that f is Riemann-integrable over I . (Hint: From thedefinition of C , you should be able to see that for each γ > 0, C can be covered by a finite number ofintervals whose total length is less than γ .)

Notation. Let X be a set and let f, g : X → R . Then the join, or maximum, of f and g is the functionf ∨ g on X defined by (f ∨ g)(x) = max f(x), g(x) . The meet, or minimum, of f and g is the functionf ∧ g on X defined by (f ∧ g)(x) = min f(x), g(x) . The positive part of f is f ∨ 0. The negative part off is (−f) ∨ 0. (Yes, according to this definition, the negative part of f is positive.) Notice that f+ ≥ 0,f− ≥ 0, f = f+ − f− , and |f | = f+ + f− . Also, if f = g − h , where g, h : X → [0,∞] , then f+ ≤ g andf− ≤ h .

X136. Let X be a set and let f, g : X → R . Prove that f +g = (f ∨g)+(f ∧g), |f −g| = (f ∨g)− (f ∧g),

f ∨ g =f + g + |f − g|

2and f ∧ g =

f + g − |f − g|2

.

X137. Let f, g : [a, b] → R be Riemann integrable, where a, b ∈ R . Prove that f∨g and f ∧g are Riemannintegrable.

24 To say that ϕ is 1-periodic means that for each x ∈ R , ϕ(x+ 1) = ϕ(x) .

33


X138. Let f, g : [a, b] → Rd be Riemann null and let c ∈ R . Prove that f + g and cf are both Riemannnull.

X139. Let f : [a, b] → R , where a, b ∈ R with a < b . Prove that f is Riemann null iff f+ and f− areboth Riemann null.

Definition. Let f, g : [a, b] → Rd , where a, b ∈ R with a < b . To say that f = g essentially means thatf − g is Riemann null.

Exercise. Let f, g, h : [a, b] → Rd , where a, b ∈ R with a < b . Suppose that f = g essentially and g = hessentially. Prove that f = h essentially.

Notation. Let f : [a, b] → R , where a, b ∈ R with a < b . Let U(f, [a, b]) denote the upper Riemann integralof f over [a, b] and let L(f, [a, b]) denote the lower Riemann integral of f over [a, b] . If no confusion islikely to result, we may write U(f) for U(f, [a, b]) and L(f) for L(f, [a, b]) .

X140. Let f, g : [a, b] → R , where a, b ∈ R with a < b .(a) Prove that U(f + g) ≤ U(f) + U(g).

Similarly L(f) + L(g) ≤ L(f + g).(b) Prove that for each c ∈ (0,∞), we have U(cf) = cU(f) and L(cf) = cL(f).

Similarly, for each c ∈ (−∞, 0), we have U(cf) = cL(f) and L(cf) = cU(f). Obviously U(0 ·f) = 0 ·U(f) =L(0 · f) = 0 · L(f) = 0.

Notation and Terminology. Let a, b ∈ R with a < b and E ⊆ [a, b] . Let E ⊆ [a, b] . Then the outercontent of E is γ(E) = U(1E , [a, b]) and the inner content of E is γ•(E) = L(1E , [a, b]) (It is not hard tocheck that these quantities do not depend on which interval [a, b] containing E we consider, but we shallnot need this fact.)

X141. Let X be a set and let f, g : X → R . Prove that f = (f ∧ g)+ (f − g)+ and g = (f ∨ g)− (f − g)+ .

X142. Let f : [a, b] → Rd . Prove that f is Riemann null iff f is bounded and for each ε > 0, we haveγ(|f | > ε) = 0. (Hints: Since only |f | is involved, we may as well assume that f : [a, b] → [0,∞). This willlighten the notation because then we can just write f instead of |f | . For the forward implication, noticethat if ε > 0 then ε1f>ε ≤ f . For the reverse implication, notice that M = sup f <∞ and that if ε > 0,then f = (f ∧ ε) + (f − ε)+ , 0 ≤ f ∧ ε ≤ ε , and (f − ε)+ ≤M1f>ε .)

The next exercise furnishes an example of a Riemann null function on [0, 1] which is strictly positive ona dense subset of [0, 1].

X143. Let D = Q ∩ [0, 1]. As we know, D is dense in [0, 1]. For each n ∈ N , let

En =

0

n,1

n, . . . ,

n

n

.

Notice that⋃

n∈NEn = D . For each x ∈ D , let N(x) = min n ∈ N : x ∈ En . Define f : [0, 1] → R by

f(x) =

N(x)−1 if x ∈ D,

0 if x ∈ [0, 1] \D.

Prove that f is Riemann null. (Hint: This is very easy to do if you use problem X142. Or with a only littlebit more work, it can be done directly from the definition of Riemann null.)

The next result is a variation on problem X122 and for most practical purposes is a satisfactory substitutefor the result that exercise. The proof we suggest for it is different from the proof we suggested for problemX122 and you may feel that it is simpler than that proof.

X144. Let f, g : [a, b] → Rd be bounded, where a, b ∈ R with a < b . Let h : Rd → R be continuous.Suppose that f = g essentially. Prove that h f = h g essentially. (Hint: Let K be the closure of theunion of the ranges of f and g . Since f and g are bounded, K is a compact subset of Rd , so h is uniformlycontinuous on K . Use problem X142.)

34


Example. Let f1, f2, g1, g2 : [a, b] → C be bounded and suppose that f1 = g1 essentially and f2 = g2essentially. Define f, g : [a, b] → C2 by f(x) = (f1(x), f2(x)) and g(x) = (g1(x), g2(x)). By problem X121,f = g essentially. Then applying problem X144 with h(z, w) = zw , we find that f1f2 = g1g2 essentially.

Definition. Let f, g : [a, b] → R , where a, b ∈ R with a < b . To say that f ≤ g essentially means that(f − g)+ is Riemann null.

X145. Let f, g : [a, b] → R , where a, b ∈ R with a < b . Prove that the following are equivalent:(a) f ≤ g essentially.(b) f = f ∧ g essentially.(c) g = f ∨ g essentially.

X146. Let f, g : [a, b] → R , where a, b ∈ R with a < b . Prove that f = g essentially iff f ≤ g essentiallyand g ≤ f essentially.

X147. Let f, g, h : [a, b] → R , where a, b ∈ R with a < b . Suppose that f ≤ g essentially and g ≤ hessentially. Prove that f ≤ h essentially.

X148. Let f, g : [a, b] → R , where a, b ∈ R with a < b . Suppose that for each ε > 0, f ≤ g+ ε essentially.Prove that f ≤ g essentially.

X149. Let f1, g1, f2, g2 : [a, b] → R , where a, b ∈ R with a < b . Suppose that f1 ≤ g1 essentially andf2 ≤ g2 essentially. Prove that f1 + f2 ≤ g1 + g2 essentially.

Notation. Let a, b ∈ R with a < b . For each f ∈ R[a, b] , we define the ∞-norm of f to be

‖f‖∞ = inf M ∈ [0,∞) : |f | ≤M essentially .Evidently ‖f‖∞ ≤ sup |f | . It is not hard to see that if f ∈ R[a, b] and c ∈ C , then ‖cf‖∞ = |c|‖f‖∞ .

X150. Let f : [a, b] → C , where a, b ∈ R with a < b . Suppose f is continuous. Prove that ‖f‖∞ = sup |f | .X151. Let f ∈ R[a, b] , where a, b ∈ R with a < b . Let L = ‖f‖∞ . Prove that |f | ≤ L essentially.

X152. Let f, g ∈ R[a, b] , where a, b ∈ R with a < b . Prove the following “limiting case” of Minkowski’sinequality:

‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.(Note: Even though we used the expression “limiting case,” the most natural solution of this exercise doesnot involve taking a limit of p-norms as p→ ∞ . But read on to learn what can be said about such a limit.)

Remark. In problem X126(d), on Holder’s inequality, one can easily check that q = p/(p − 1). Thus asp ↓ 1, q ↑ ∞ . The next exercise justifies the notation ‖g‖∞ .

X153. Let g ∈ R[a, b] , where a, b ∈ R with a < b . Prove that ‖g‖q → ‖g‖∞ as q → ∞ . (Hints: Since only|g| is involved, we may as well assume that g : [a, b] → [0,∞). This will lighten the notation because thenwe can just write g instead of |g| . Suppose ∞ > t > ‖g‖∞ . Then g ≤ t essentially, so by problem X144 andproblem X145, for each q ∈ [1,∞), we have gq ≤ tq essentially. Use this to prove that lim supq→∞ ‖g‖q ≤ t .Next, suppose 0 ≤ s < ‖g‖∞ . Then (g − s)+ is not Riemann null. In other words, U((g − s)+) > 0. LetE = g > s and let M = sup g . Then (g − s)+ ≤ M1E , so U(M1E) > 0, so U(1E) > 0. Now g ≥ s1E .Use this to prove that lim infq→∞ ‖g‖q ≥ s .)

X154. Let f, g ∈ R[a, b] , where a, b ∈ R with a < b . Prove the following “limiting case” of Holder’sinequality:

‖fg‖1 ≤ ‖f‖1‖g‖∞.(Note: Even though we used the expression “limiting case,” the most natural solution of this exercise doesnot involve taking a limit as p ↓ 1 and q = p/(p− 1) → ∞ .)

X155. Let g : [0, 1] → R be continuous. Suppose that∫ 1

0

xg(x) dx = 1 and

∫ 1

0

g(x) dx = 0.

Prove that |g(x)| ≥ 4 for some x ∈ [0, 1].

35


Reminder. If E is a bounded subset of R , then γ(E) denotes the outer content of E and γ•(E) denotesthe inner content of E .

X156. Let f : [a, b] → R , where a, b ∈ R with a < b .(a) Let y, z ∈ R with y < z . Obviously f ≥ z ⊆ f > y , so γ(f ≥ z) ≤ γ(f > y). But if f

is Riemann integrable, then we can do better than this. Suppose that f is Riemann integrable.Prove that γ(f ≥ z) ≤ γ•(f > y). (Hint: Let h : R → R be a continuous function such thath = 0 on (−∞, y] , 0 ≤ h ≤ 1 on [y, z] , and h = 1 on [z,∞). For instance, on [y, z] , we couldtake h(t) = (t− y)/(z − y). Then 1f≥z ≤ h f ≤ 1f>y .)

(b) Continue to suppose that f is Riemann integrable. Prove that γ•(f > y) = γ(f ≥ y) for all butcountably many y ∈ R . (Hint: This follows from part (a).) In particular, the set of all y ∈ R forwhich this equality holds is dense in R .

X157. Let f : [0, 1] → R be continuously differentiable. Suppose that f(0) = 0. Show that

∫ 1

0

|f(x)|2 dx ≤ 1

2

∫ 1

0

|f ′(x)|2 dx.

Remark. We know that∫ n

11/t dt = logn . Hence we expect that

∑nk=1 1/k should be approximately logn .

The next exercise makes this precise.

X158. Prove that there is a unique real number γ such that for each n ∈ N ,

γ + logn <

n∑

k=1

1

k<

1

2n+ γ + logn.

Furthermore, 1/2 < γ < 1. (Hint: Let

ak =1

k−∫ k+1

k

1

tdt.

Thenn−1∑

k=1

1

k=

n−1∑

k=1

ak + logn.

Use the fact that the function t 7→ 1/t is strictly decreasing and strictly convex on (0,∞) to show that

1

2

(

1

k− 1

k + 1

)

< ak <1

k− 1

k + 1.

You should find that γ =∑∞

k=1 ak works. Be sure to explain why this series converges. By the way, there is apicture that should help you understand how to do the proof. Notice that ak is the area of an approximatelytriangular region of width 1 and height 1/k− 1/(k+1). If you slide each of these regions leftward, you canmake them all fit, without overlapping, into the unit square [0, 1]2 . It is visually clear that they then fill upa little more than half of this square.)

Remark. The number γ in problem X158 is called Euler’s constant. To ten decimal places, its value isγ = 0.5772156649 · · · . It is not known whether γ is rational or irrational.

Remark. The function x 7→ ex is convex on R , so its graph lies above each of its tangent lines. Now theline with equation y = 1 + x is tangent to the graph of x 7→ ex at the point (x, y) = (0, 1). Therefore

ex ≥ 1 + x for each x ∈ R. (11)

You should already be familiar with the inequality (11). This is just a reminder. Now let us observe somerelated inequalities that you may not have encountered before. The function x 7→ 2x is convex on R , so its

36


graph on an interval [a, b] lies below the chord joining the points (a, 2a) and (b, 2b). But when a = 0 andb = 1, this chord is a portion of the line with equation y = 1 + x . Thus

1 + x ≥ 2x for each x ∈ [0, 1]. (12)

Similarly, the function x 7→ 4−x is convex on R , so its graph on an interval [a, b] lies below the chord joiningthe points (a, 4−a) and (b, 4−b). When a = 0 and b = 1/2, this chord is a portion of the line with equationy = 1− x . Therefore

1− x ≥ 4−x for each x ∈ [0, 1/2]. (13)

X159. Suppose that ak ∈ [0,∞) for each k ∈ N . Let pn =∏n

k=1(1 + an) for each n ∈ N . Notice that1 ≤ p1 ≤ p2 ≤ p3 ≤ · · · < ∞ . Let p = limn→∞ pn and let s =

∑∞k=1 ak . Obviously p ∈ [1,∞] and

s ∈ [0,∞] . Prove that p <∞ iff s <∞ . (Hint: Use (11) and (12).)

X160. Suppose that bk ∈ [0, 1) for each k ∈ N . Let qn =∏n

k=1(1 − bn) for each n ∈ N . Notice that1 ≥ q1 ≥ q2 ≥ q3 ≥ · · · > 0. Let q = limn→∞ qn and t =

∑∞k=1 bk . Obviously q ∈ [0, 1] and t ∈ [0,∞] .

Prove that q > 0 iff t <∞ . (Hint: Use (11) and (13).)

X161. Let P be the set of prime numbers. Prove that due 5Th

∑

p∈P

1

p= ∞.

Hint: Let n ∈ 2, 3, 4, . . . . Let Pn be the set of all p ∈ P such that there exists k ∈ 2, . . . , n such thatp divides k . Now 2n > n . Hence for each k ∈ 1, . . . , n , there exist numbers mp ∈ 0, 1, 2, . . . , n suchthat k =

∏

p∈Pnpmp . Therefore

n∑

k=1

1

k≤ ∏

p∈Pn

(1 + p−1 + p−2 + · · ·+ p−n) ≤ ∏

p∈Pn

1

1− p−1=

(

∏

p∈Pn

(

1− p−1)

)−1

.

Now use this and (13) to show that∑

p∈Pn

1p is greater than or equal to a certain quantity which tends to

infinity as n tends to infinity.

Remark. For each n ∈ N , let pn be the n-th prime number, so that p1 = 2, p2 = 3, p3 = 5, and so on.In problem X161, you were asked to show that

∑∞n=1 1/pn = ∞ . This shows that pn tends to infinity fairly

slowly. Now there is a famous, difficult result, called the prime number theorem, which gives much moreexact information about pn . The usual formulation of the prime number theorem is that

π(x) ∼ x

log xas x→ ∞, (14)

where π(x) denotes the number of primes which are less than or equal to x . In the next exercise, you areasked to show that an equivalent statement is that

pn ∼ n logn as n→ ∞. (15)

It is interesting to remark that (14) was conjectured by Carl Friedrich Gauss in 1792 or 1793, when he wasaround 16 years old, by perusing tables of prime numbers. Gauss did not prove (14). P. L. Chebyshev madeprogress toward proving it in the 1850s, and it was finally proved in 1896, by Jacques Hadamard and Charlesde la Vallee-Poussin independently.

X162. Let (pn) be a strictly increasing sequence in (0,∞). For each x ∈ (0,∞), let

P (x) = n ∈ N : pn ≤ x

and let π(x) be the number of elements in P (x). Prove that the for the given sequence (pn), (14) holds ifand only if (15) holds. (Hints: For each x ∈ [pn, pn+1), we have π(x) = n . If (14) holds, then π(x) <∞ forlarge x , and hence each x ∈ (0,∞), and it follows that pn → ∞ .)

37


Remark. In view of problem X162, you may be wondering why the prime number theorem is usuallyexpressed in the form (14) rather than the more easily grasped form (15). The reason is that in the casewhere (pn) is the sequence of prime numbers listed in increasing order, although the error in each of thesetwo approximations tends to zero, the error in the approximation (14) tends to zero faster.

Remark. If the famous Riemann hypothesis holds, then an even better approximation is obtained in (14)if we replace x/ log x by L(x) =

∑xn=2(logn)

−1 . In this case, one has π(x) = L(x) + O(x1/2 log x) asx → ∞ . For a fascinating discussion of the prime number theorem and related matters, without proofs,but with heuristic explanations, see Serge Lang, The Beauty of Doing Mathematics, Three Public Dialogues,Springer-Verlag, 1985.

Remark. We know that as h→ 0, (1 + h)1/h → e . It is sometimes of interest to know that

(1 + h)1/h ↑ e as h ↓ 0

and that(1 + h)1/h ↓ e as h ↑ 0.

One way to show this would be to verify, by calculation, that

d

dh(1 + h)1/h < 0 for 0 < h <∞.

However, there is a more enlightening and elegant way which is based on things we have already checked.We know that

d

dulog u =

1

uis strictly decreasing on the interval where 0 < u <∞.

Thus the function u 7→ log u is strictly concave down on the interval (0,∞). In other words, the functionu 7→ − logu is strictly convex on (0,∞). Hence, if 0 < u < v < w < ∞ , then considering the slopes ofsuitable chords, we have

log v − log u

v − u>

logw − log u

w − u>

logw − log v

w − v. (16)

(You should draw a picture!) If 0 < h1 < h2 < ∞ , then taking u = 1, v = 1 + h1 , and w = 1 + h2 , andapplying the first inequality in (16), we get

log(1 + h1)

h1>

log(1 + h2)

h2,

or equivalently,(1 + h1)

1/h1 > (1 + h2)1/h2 .

Thus the map h 7→ (1 + h)1/h is strictly decreasing on the interval (0,∞). Similarly, if −1 < h1 < h2 < 0,then taking u = 1 + h1 , v = 1 + h2 , and w = 1, and applying the second inequality in (16), the we get

− log(1 + h1)

−h1>

− log(1 + h2)

−h2,

or again equivalently,(1 + h1)

1/h1 > (1 + h2)1/h2 .

Thus the map h 7→ (1 + h)1/h is also strictly decreasing on the interval (−1, 0). Putting all this together,we see that if we let

ψ(h) =

(1 + h)1/h if h ∈ (−1, 0) ∪ (1,∞),

e if h = 0,

then ψ is continuous and strictly decreasing on the interval (−1,∞).

38


X163. due 5Th

(a) The gamma function is defined for 0 < x <∞ by

Γ(x) =

∫ ∞

0

tx−1e−t dt.

For n = 1, 2, 3, . . . and for 0 < x <∞ , let

In(x) =

∫ n

0

tx−1(

1− t

n

)n

dt.

Prove that for each x ∈ (0,∞), we have limn→∞ In(x) = Γ(x). (Hint: Fix x ∈ (0,∞). Letf(t) = tx−1e−t for all t ∈ (0,∞). For n = 1, 2, 3, . . . , let

fn(t) =

tx−1(

1− tn

)n

if t ∈ (0, n),

0 if t ∈ [n,∞).

Use the preceding Remark to show that for each t ∈ (0,∞), fn(t) ↑ f(t) as n → ∞ . Thenuse Dini’s theorem25 to show that if 0 < a < b < ∞ , then fn → f uniformly on [a, b] . Thisuniform convergence could be established by calculation rather than by appealing to Dini’s theorem.However, by using Dini’s theorem, we save ourselves the trouble of calculating.)

(b) For −1 < z < ∞ , let Π(z) = Γ(z + 1). Using integration by parts, one can easily see that forz = 0, 1, 2, . . . , we have Π(z) = z! . For this reason, the function Π is called the factorial function.By explicitly evaluating In(x), deduce from part (a) that

Γ(z + 1) = limn→∞

n!(n+ 1)z

(z + 1)(z + 2) · · · (z + n)

= limn→∞

n∏

k=1

[

(

k + 1

k

)zk

z + k

]

=

[

(

2

1

)z1

z + 1

] [

(

3

2

)z2

z + 2

][

(

4

3

)z3

z + 3

]

· · ·

for each z ∈ (−1,∞). It was in the form of the latter infinite product that Euler (in 1729) inventedthe factorial function.26

X164. Let f ∈ R[0, 1] and suppose f is continuous at 1. Determine

limn→∞

n

∫ 1

0

xnf(x) dx.

Justify your answer. (Hint: If you understand thoroughly how to solve problem X131, then you should beable to solve this problem. Vice versa too.)

25 “Dini’s theorem” is the customary name for a theorem like Theorem 7.13 in Rudin, Principles of Mathematical Analysis,Third Edition. Of course, the sequence of functions in question can just as well be increasing, rather than decreasing as in thetheorem in Rudin.

26 If you know some complex analysis, then you may find the following remarks interesting. It not hard to show that theproduct whose factors are the reciprocals of the factors in the product for Γ(z+1) converges uniformly on each compact subsetof C . From this it follows that the reciprocal of the gamma function can be extended in a unique way to a function that isanalytic in the whole complex plane. The set of zeroes of this extension is the set of non-positive integers and each of theseis a simple zero. Thus the gamma function extends in a unique way to a function that is meromorphic in the whole complexplane. This extension has no zeroes and its set of poles is the set of non-positive integers and each of these is a simple pole. Ifyou didn’t understand some of the terms in this footnote, don’t worry. You will learn them when you study complex analysis.

39


Remark. Let u, v, w ∈ R with u < v < w . As we know, if f : [u,w] → R is Riemann integrable, then

∫ w

u

f(x) dx =

∫ v

u

f(x) dx +

∫ w

v

f(x) dx.

The same argument that establishes this actually proves that the upper and lower Riemann integrals satisfy

U(f, [u,w]) = U(f, [u, v]) + U(f, [v, w]) and L(f, [u,w]) = L(f, [u, v]) + L(f, [v, w]).

Now let E be a bounded subset of R . Suppose E ⊆ [a1, b1] and E ⊆ [a2, b2] , where a1, b1, a2, b2 ∈ R witha1 < b1 and a2 < b2 . Let a = max a1, a2 and b = min b1, b2 . Then a ≤ b and E ⊆ [a, b] . Notice that

U(1E , [a1, b1]) = U(1E , [a1, a]) + U(1E , [a, b]) + U(1E , [b, b1]) = 0 + U(1E , [a, b]) + 0 = U(1E , [a, b]).

Likewise, U(1E , [a2, b2]) = U(1E, [a, b]) . Hence U(1E , [a1, b1]) = U(1E, [a2, b2]) . Similarly, L(1E, [a1, b1]) =L(1E, [a2, b2]) . In other words, γ(E) and γ•(E), the inner and outer content of E , do not depend on whichcompact interval containing E they are computed with respect to.

Definition. Let E be a subset of R . To say that E is contented means that E is bounded and γ•(E) =γ(E). (This somewhat whimsical terminology is not standard.)

Remark. If E is a bounded interval in R , then E is contented and γ(E) is equal to the length of E .

Remark. If E ⊆ [a, b] , where a, b ∈ R with a < b , then E is contented iff 1E is Riemann integrable over[a, b] .

Remark. If E is a contented subset of R , then γ(E), the outer content of E , is also called simply thecontent of E .

X165. Let E1, . . . , En be contented subsets of R . Let E =⋃n

k=1 Ek and D =⋂n

k=1 Ek . Prove that Eand D are contented.

X166. Let A and B be contented subsets of R . Prove that A \B is contented.

X167. Let E1, . . . , En be contented subsets of R and let E =⋃n

k=1Ek . Suppose that E1, . . . , En aredisjoint. Prove that γ(E) =

∑nk=1 γ(Ek).

X168. Let E1, . . . , En be bounded subsets of R .(a) Let E ⊆ ⋃n

k=1 Ek . Prove that γ(E) ≤∑nk=1 γ(Ek).

(b) Now suppose in addition that E1, . . . , En are disjoint. Let F be a bounded subset of R such that⋃n

k=1 Ek ⊆ F . Prove that∑n

k=1 γ•(Ek) ≤ γ•(F ).

X169. Let E be a bounded subset of R .(a) Prove that γ(E) is the infimum of sums of the form

∑nk=1 γ(Ik), where n ∈ N , I1, . . . In are

bounded intervals in R , and E ⊆ ⋃nk=1 Ik .

(b) Prove that γ•(E) is the supremum of sums of the form∑n

k=1 γ(Ik), where n ∈ N and whereI1, . . . In are disjoint intervals in R which are contained in E .

X170. Let E be a bounded subset of R . Prove that

γ(E) = inf γ(F ) : E ⊆ F and F is contented

and that

γ•(E) = sup γ(D) : D ⊆ E and D is contented .

X171. Let E be a contented subset of R , let A ⊆ E , and let B = E \A . Prove that γ(E) = γ(A)+γ•(B).(Note that neither A nor B need be contented.)

40


Reminder. Let (Z, d) be a metric space. Let f : I → Z , where I is an interval in R . Let a, b ∈ I witha < b . Recall that a subdivision of [a, b] is an ordered n-tuple (x0, x1, . . . , xn) of real numbers such thata = x0 < x1 < · · · < xn = b . Recall that if P = (x0, x1, . . . , xn) is a subdivision of [a, b] , then the variationof f on P is

V (f, P ) =n∑

k=1

d(

f(xk−1), f(xk))

. (17)

Let’s write Π[a, b] for the set of subdivisions of [a, b] . Recall that the variation of f on [a, b] is

V (f, [a, b]) = sup V (f, P ) : P ∈ Π[a, b] .

Now let us specialize to the case where Z = R and d(z, z′) = |z′ − z| . Then (17) becomes

V (f, P ) =

n∑

k=1

∣

∣f(xk)− f(xk−1)∣

∣.

X172. Let f : [a, b] → R , where a, b ∈ R with a < b . Prove that V (f, [a, b]) <∞ iff there exist increasingfunctions ϕ, ψ : [a, b] → R such that f = ϕ − ψ . (Hint for the forward implication: Define ϕ and ψ byϕ(x) = V (f, [a, x]) and ψ(x) = ϕ(x) − f(x).)

Remark. Let f : [a, b] → R , where a, b ∈ R with a < b . In problem X172, we saw that V (f, [a, b]) <∞ ifff is a difference of increasing functions on [a, b] . This decomposition of f is clearly not unique. Furthermore,the decomposition suggested in problem X172 is not optimal. The next exercise will explain what we meanby optimal here and will reveal the optimal way of choosing such a decomposition. First we need somepreliminaries.

Reminder. Let y be a real number. Then y+ = max y, 0 and y− = max −y, 0 . We have

y = y+ − y−, |y| = y+ + y−, y+ =|y|+ y

2, y− =

|y| − y

2. (18)

Furthermore, if y = u− v , where u, v ∈ [0,∞), then y+ ≤ u and y− ≤ v .

Definitions. Let f : I → R , where I is an interval in R . Let a, b ∈ I with a < b . If P = (x0, x1, . . . , xn)is a subdivision of [a, b] , then by definition, the positive variation of f on P is

V +(f, P ) =

n∑

k−1

(

f(xk)− f(xk−1))+

and the negative variation of f on P is

V −(f, P ) =

n∑

k−1

(

f(xk)− f(xk−1))−.

In view of (18), it is obvious that

V +(f, P ) + V −(f, P ) = V (f, P )

and that

V +(f, P )− V −(f, P ) =

n∑

k=1

(

f(xk)− f(xk−1))

= f(xn)− f(x0) = f(b)− f(a),

so that

V +(f, P ) =V (f, P ) + f(b)− f(a)

2

41


and

V −(f, P ) =V (f, P ) + f(a)− f(b)

2.

Next, by definition, the positive variation of f on [a, b] is

V +(f, [a, b]) = sup

V +(f, P ) : P ∈ Π[a, b]

and the negative variation of f on [a, b] is

V −(f, [a, b]) = sup

V −(f, P ) : P ∈ Π[a, b]

.

X173. Let f : [a, b] → R , where a, b ∈ I with a < b .

(a) Prove that

V +(f, [a, b]) =V (f, [a, b]) + f(b)− f(a)

2,

V −(f, [a, b]) =V (f, [a, b]) + f(a)− f(b)

2,

and

V (f, [a, b]) = V +(f, [a, b]) + V −(f, [a, b]).

In particular, if one of the quantities V (f, [a, b]) , V +(f, [a, b]) , and V −(f, [a, b]) is finite, then allthree of them are finite.

(b) Prove that if c ∈ (a, b), then

V +(f, [a, c]) + V +(f, [c, b]) = V +(f, [a, b])

and

V −(f, [a, c]) + V −(f, [c, b]) = V −(f, [a, b])

Define F, g, h : [a, b] → [0,∞] by F (x) = V (f, [a, x]) , g(x) = V +(f, [a, x]) , and h(x) = V −(f, [a, x]) .Obviously F (a) = g(a) = h(a) = 0. Notice that by part (a), with b replaced by x , we have

g(x) =F (x) + f(x)− f(a)

2, h(x) =

F (x) + f(a)− f(x)

2,

and F (x) = g(x) + h(x) for each x ∈ [a, b] .

(c) We already know that F is increasing. Prove that g and h are increasing too.

(d) Prove that if V (f, [a, b]) <∞ , then f = f(a) + g − h .

(e) Suppose that f = f(a)+G−H , where G,H : [a, b] → R are increasing and satisfy G(a) = H(a) =0. Prove that V (f, [a, b]) <∞ , that g ≤ G and that h ≤ H . Hence F ≤ G+H , with equality iffG = g and H = h .

X174. Let f : [0, 1] → R be continuous. Suppose that do before MT

∫ 1

0

f(x)g′(x) dx = 0

for each continuously differentiable function g : [0, 1] → R satisfying g(0) = 0 = g(1). Prove that f must bea constant function. (Hint: One approach to solving this problem may be based on a thorough understandingof problem X131 and/or problem X164. I don’t mean that the specific form of the functions considered inthose problems is important. Rather, their qualitative behaviour is what is important.)

42


X175. Define complex-valued functions functions f and g by do before MT

f(t) =

∞∑

k=1

eikt

k2and g(t) =

∞∑

k=1

eikt

k.

Let each of these functions be defined for all t ∈ R for which the corresponding series converges.(a) Prove that the series for f converges uniformly on R .(b) Let’s write 2πZ for the set 2πn : n ∈ Z . Clearly for each t ∈ 2πZ , the series for g(t) reduces

to∑∞

k=1 1/k , which diverges. Let A = R \ 2πZ . (Notice that A is a dense open subset of R .)Prove that the series for g converges locally uniformly in A .27 (Hint: Recall that if z ∈ C with|z| = 1 but z 6= 1, then the series

∑∞k=1 z

k has bounded partial sums. Review the proof of thisfact and adapt it to prove a related uniform result. Then apply summation by parts.)

(c) Show that f is differentiable on A and that f ′ = ig on A .(d) Consider the functions defined by the series

∞∑

k=1

cos(kt)

k2and

∞∑

k=1

sin(kt)

k2.

What does part (c) tell us about the derivatives of these functions?

Let X be a topological space, let Y be a metric space, and let Φ be a collection of functions fromX into Y . If a ∈ X , then to say that Φ is equicontinuous at a means that for each ε > 0, there existsa nhd U of a in X such that for each f ∈ Φ, for each x ∈ U , d(f(x), f(a)) < ε . To say that Φ isequicontinuous on X means that for each a ∈ X , Φ is equicontinuous at a . And by the way, although wedon’t need this concept right now, let us mention that if X is also a metric space, then to say that Φ isuniformly equicontinuous on X means that for each ε > 0, there exists δ > 0 such that for each f ∈ Φ, forall x, x′ ∈ X , if dX(x, x′) < δ , then dY (f(x), f(x

′)) < ε . (Warning: In Rudin, Principles of MathematicalAnalysis, Third Edition, the term “equicontinuous on X ” is used instead of “uniformly equicontinuous onX .” When the metric space X is compact, which is the main case Rudin is interested in, one can showthat if Φ is equicontinuous on X , then Φ is uniformly equicontinuous on X . Perhaps this is why Rudindoes not bother to distinguish between these two notions. The proof of the fact just mentioned is almostword-for-word the same as the proof that a single continuous function from a compact metric space into ametric space is uniformly continuous.)

X176. Let X be a topological space, let Y be a complete metric space, let Φ be a collection of functions do before MT

from X into Y , let B be a filter base on Φ, and let D be a dense subset of X . Suppose Φ is equicontinouson X and limf→B f(x) exist in Y for each x ∈ D .

(a) Show that limf→B f(x) exists for each x ∈ X .(b) Define g : X → Y by g(x) = limf→B f(x) for each x ∈ X . Show that g is continuous.(c) State what parts (a) and (b) tell us in the case of a sequence (fn) of functions from X into Y .

27 To say that a series converges locally uniformly in A means that each point in A has a nhd in A in which the seriesconverges uniformly.

43


Reminder. Let E be a bounded subset of R . Then γ(E) denotes the outer content of E and γ•(E)denotes the inner content of E . To say that E is contented means that γ(E) = γ•(E). If E is contented,then γ(E) is also called simply the content of E .

X177.(a) Let E be a bounded subset of R . Prove that γ(E) is the infimum of the set of numbers of the

form γ(U), where U varies over all bounded open subsets of R containing E . (In fact, it isenough to consider sets U which are unions of finitely many bounded open intervals.)

(b) Let (Kn) be a decreasing sequence of compact subsets of R and let K =⋂∞

n=1Kn . Prove thatγ(Kn) ↓ γ(K).

Notation. The following notation is not standard but will be convenient to use in the next few exercises.For each n ∈ N , let Sn = 0, 1 n be the set of finite binary sequences of length n . Let S =

⋃

n∈N Sn , theset of non-empty finite binary sequences. Finally, let Σ = 0, 1 N be the set of infinite binary sequences.

Reminder. Let Sn , S , and Σ be as above. If (X, d) is a metric space, then by a Cantor scheme28 in(X, d), we shall mean a family (A(s))s∈S of non-empty subsets of X , indexed by S , such that:

(a) A(0) and A(1) are disjoint;(b) For each n ∈ N and each s ∈ Sn , A(s1, . . . , sn, 0) and A(s1, . . . , sn, 1) are disjoint and for i = 0, 1,

A(s1, . . . , sn) ⊇ A(s1, . . . , sn, i);

(c) For each σ = (s1, s2, s3, . . . ) ∈ Σ, diam(A(s1, . . . , sn)) → 0 as n→ ∞ .

X178. Let Sn , S , and Σ be as above. Let (X, d) be a complete metric space and let (A(s))s∈S bea Cantor scheme in (X, d). For each n ∈ N , let Bn =

⋃

s∈SnA(s). Let K =

⋂

n∈NBn . For eachσ = (s1, s2, s3, . . . ) ∈ Σ, let f(σ) be the unique point in

⋂

nA(s0, . . . , sn). (We can do this by problemX9.) We saw in problem X12 that h is a one-to-one map from Σ onto K and consequently that K isequinumerous to the interval [0, 1]. Give Σ the product topology.

(a) Prove that h is a homeomorphism from Σ onto K .(b) Deduce that K is homeomorphic to the ordinary Cantor set.

The ordinary Cantor set has zero total length, in the sense that its outer content is zero. In the extraproblems last quarter, we mentioned that there are so-called fat Cantor sets which resemble the ordinaryCantor set topologically but which have strictly positive total length. The next exercise treats these sets indetail.

X179. Let Sn , S , and Σ be as above. For all a, b ∈ R with a < b and for all u ∈ (0, 1), let M(a, b, u) be theopen interval of length (b−a)(1−u) centered at the midpoint of [a, b] and let J([a, b], u, 0) and J([a, b], u, 1)be the left and right portions of [a, b] that remain after we delete M(a, b, u) from [a, b] . Fix t ∈ (0, 1).Let (tn) be a strictly decreasing sequence in (t, 1) such that tn → t and let t0 = 1. For each n ∈ N , letun = tn/tn−1 . Notice that un ∈ (0, 1) and

∏nk=1 uk = tn for each n ∈ N . Let I(0) = J([0, 1], u1, 0) and let

I(1) = J([0, 1], u1, 1). If n ∈ N , s ∈ Sn , and I(s) has already been defined, let I(s, 0) = J(I(s), un+1, 0)and let I(s, 1) = J(I(s), un+1, 1). For each n ∈ N , let Fn =

⋃

s∈SnI(s). Let F =

⋂

n∈N Fn .(a) Prove that F is compact and has empty interior.(b) Prove that F is homeomorphic to the ordinary Cantor set. In particular, F has no isolated points.

In fact, each non-empty relatively open subset of F has the cardinality of the continuum.(c) Prove that for each n ∈ N , γ(Fn) = tn .(d) Prove that γ(F ) = t .(e) Prove that 1F is not Riemann integrable over [0, 1].

Reminder. Let X be a topological space and let f : X → [−∞,∞] . For each p ∈ X , let Up be thecollection of nhds of p in X . Of course Up is a filter base on X . The lower regularization of f is thefunction f∗ defined on X by

f∗(p) = lim infUp

f = supU∈Up

infx∈U

f(x).

28 This is not standard terminology.

44


The upper regularization of f is the function f∗ defined on X by

f∗(p) = lim supUp

f = infU∈Up

supx∈U

f(x).

We saw in problem X56 that f∗ is the largest lower semicontinuous function on X which is less than orequal to f , that f∗ is the smallest upper semicontinuous function on X which is greater than or equal tof , and that for each p ∈ X , f is continuous at p iff f∗(p) = f∗(p).X180. Let F be a fat Cantor set and let f = 1F . Prove that f∗ = 0 and f∗ = f .

X181. Let f : [a, b] → R . due 7Th

(a) Prove that the lower Riemann integral of f over [a, b] is the same as the lower Riemann integralof f∗ over [a, b] and that the upper Riemann integral of f over [a, b] is the same as the upperRiemann integral of f∗ over [a, b] .

(b) Prove that f is Riemann integrable over [a, b] iff f is bounded and f∗ − f∗ is Riemann null.

Remark. In a sense, problem X181 tells us that a Riemann integrable function must be nearly continuous.This is because the set where f is continuous is the set where f∗−f∗ = 0. When we study Lebesgue measure,we’ll make this more precise. In fact, once we have Lebesgue measure, it follows almost immediately fromproblem X181 that f is Riemann integrable iff f is bounded and the set where f is not continuous hasLebesgue measure zero.

Definition. Let Y be a set and let ϕ : [a, b] → Y , where a, b ∈ R with a < b . To say that ϕ is Riemannsimple means that the range of ϕ is a finite set and that for each y ∈ Y , the set

ϕ = y = x ∈ [a, b] : ϕ(x) = y

is contented.

X182. Let ϕ : [a, b] → R , where a, b ∈ R with a < b . Suppose that ϕ is Riemann simple. Prove that ϕ is

Riemann integrable over [a, b] and that∫ b

aϕ(x) dx =

∑

y∈Y yγ(ϕ = y), where Y denotes the range of ϕ .

X183. Let f : [a, b] → R , where a, b ∈ R with a < b . Prove that f is Riemann integrable iff there exists asequence (ϕn) of Riemann simple functions ϕn : [a, b] → R such that ϕn → f uniformly on [a, b] . (Hints:The reverse implication is the easy part, given the result of problem X182. To prove the forward implication,let m = inf f and M = sup f , use problem X156 with problem X166 and partition the interval [m,M ]appropriately.)

Reminder. If f : [a, b] → R , where a, b ∈ R with a < b , then U(f, [a, b]) denotes the upper Riemannintegral of f over [a, b] and L(f, [a, b]) denotes the lower Riemann integral of f over [a, b] . If no confusionis likely to result, we may write U(f) for U(f, [a, b]) and L(f) for L(f, [a, b]) .

X184. Let f : [a, b] → Rd , where a, b ∈ R with a < b . Prove that f is Riemann integrable over [a, b] iff

there exists a sequence of step functions (ϕn) such that U(|f−ϕn|) → 0, and that in this case,∫ b

aϕn(t) dt →

∫ b

af(t) dt .

Reminder. K denotes either R or C .

X185. Let ϕ : [a, b] → Y , where a, b ∈ R with a < b and where Y is a vector space over K . Suppose

ϕ is a step function. Then it is obvious what∫ b

a ϕ(x) dx should be. The only question is whether this iswell-defined (because there are many subdivisions of [a, b] such that ϕ is constant on the interior of each

subinterval associated with the subdivision). Show that∫ b

aϕ(x) dx is well-defined. (Hint: For any two

subdivisions of [a, b] , we can consider a common refinement.)

X186. Let ϕ, ψ : [a, b] → Y , where a, b ∈ R with a < b and where Y is a vector space over K . Supposethat ϕ and ψ are step functions.

(a) Prove that ϕ+ ψ is a step function and that

∫ b

a

ϕ(x) + ψ(x) dx =

∫ b

a

ϕ(x) dx +

∫ b

a

ψ(x) dx.

45


(b) Let c ∈ K . It is essentially obvious that cϕ is a step function. Check that

∫ b

a

cϕ(x) dx = c

∫ b

a

ϕ(x) dx.

Definition. Let Y be a normed linear space over K . Let f : [a, b] → Y , where a, b ∈ R with a < b . Tosay that y is a Riemann integral of f over [a, b] means that y ∈ Y and there exists a sequence (ϕn) of

step functions from [a, b] to Y such that U(|f −ϕn|) → 0 and∫ b

a ϕn(x) dx→ y . (Note that |f −ϕn| is thefunction gn on [a, b] defined by gn(x) = |f(x) − ϕn(x)| , where for each v ∈ Y , |v| denotes the norm of v .Since gn is real-valued, U(gn) makes sense.) To say that f is Riemann integrable over [a, b] means thatthere exists y ∈ Y such that y is a Riemann integral of f over [a, b] .

X187. Let Y be a normed linear space over K . Let f : [a, b] → Y , where a, b ∈ R with a < b .(a) Suppose that y and z are Riemann integrals of f over [a, b] . Prove that y = z . Thus it makes

sense to speak of the Riemann integral of f over [a, b] and to write∫ b

af(x) dx = y .

(b) Suppose that Y is a Banach space29 and there exists a sequence (ϕn) of step functions from [a, b]to Y such that U(|f − ϕn|) → 0. Prove that f is Riemann integrable over [a, b] . (Hint: For each

n , let yn =∫ b

aϕn(x) dx . Prove that (yn) is a Cauchy sequence in Y .)

Reminder. Let f : [a, b] → Y , where a, b ∈ R with a < b and where Y is a metric space. To say that fis regulated means that for each u ∈ (a, b] , the left hand limit f(u−) = limx→u− f(x) exists in Y , and foreach v ∈ [a, b), the right hand limit f(v+) = limx→v+ f(x) exists in Y . As we have seen, if f is regulated,then there is a sequence (ϕn) of step functions ϕn : [a, b] → Y such that ϕn → f uniformly on [a, b] . Theconverse is also true, provided Y is complete.

X188. Let Y be a Banach space and let a, b ∈ R with a < b . Let f : [a, b] → Y be regulated. Prove thatf is Riemann integrable over [a, b] .

X189. Let Y be a metric space. Let f : [a, b] → Y be regulated. For each ε > 0, let

L(ε) = u ∈ (a, b] : d(f(u−), f(u)) > ε and R(ε) = v ∈ [a, b) : d(f(v+), f(v)) > ε .

Show that for each ε > 0, R(ε) and L(ε) are finite sets. Conclude that the set

D = c ∈ [a, b] : f is not continuous at c

is countable.

X190. Let D = C \ 0 . Define g : D → C by g(z) = 1/z . Let z0 ∈ D . Let r = |z0| . Then r > 0 andB(z0, r) ⊆ D . Find a power series representation

∑∞k=0 ck(z−z0)k for g(z) that is valid for all z ∈ B(z0, r).

(You should not need calculus for this.) Deduce that g is analytic.

Remark. If f and g are analytic, we would like to know that their composition g f is analytic. The nextfew exercises deal with this.

X191. Consider a power series due 7Th

g(z) =∞∑

ℓ=0

bℓzℓ

whose radius of convergence R2 is strictly positive. Consider another power series

f(z) =

∞∑

k=1

akzk

29 Recall that a Banach space is a normed linear space which is complete with respect to the metric induced by its norm.For instance, R

d with its usual norm is a Banach space. Here is another example. Let X be a topological space and let Cb(X)be the vector space of all bounded continuous functions from X into K . For each f ∈ Cb(X) , let ‖f‖ = sup |f | . This definesa norm on Cb(X) . Under this norm, Cb(X) is a Banach space, because a uniformly Cauchy sequence of K-valued functions isuniformly convergent and because a uniform limit of continuous functions is continuous.

46


whose constant term is zero and whose radius of convergence R1 is strictly positive. For each k ∈ N , letαk = |ak| . The series for f converges absolutely for |z| < R1 . Hence we may define ϕ(z), for |z| < R1 , by

ϕ(z) =∞∑

k=1

αkzk.

By the Cauchy product formula, there are coefficients a(ℓ)k and α

(ℓ)k , where k and ℓ vary over ω =

0, 1, 2, . . . , such that for |z| < R1 , we have

f(z)ℓ =

∞∑

k=ℓ

a(ℓ)k zk

and

ϕ(z)ℓ =

∞∑

k=ℓ

α(ℓ)k zk.

(a) Prove that |a(ℓ)k | ≤ α(ℓ)k .

Let R0 = sup r ∈ [0, R1) : ϕ(r) < R2 .(b) Prove that R0 > 0 and that for |z| < R0 , we have

|f(z)| ≤ ϕ(|z|) < R2.

Hence g(f(z)) is defined for |z| < R0 .

(c) Prove that for |z| < R0 , we have

g(f(z)) =

∞∑

k=0

ckzk,

where

ck =

∞∑

ℓ=0

bℓa(ℓ)k .

(Informally, the proof of this is just a computation involving interchanging the order of summationin a certain double series. Of course you are expected to justify this interchange in order ofsummation.)

X192. Let D1 and D2 be open subsets of C . Let f : D1 → D2 and g : D2 → C be analytic. Let h = g f .Prove that h is analytic.

X193. Let D be an open subset of C and let f : D → C \ 0 be analytic. Prove that 1/f is analytic.(Hint: This is trivial if you just combine the results of a couple of the exercises above.)

47


Reminder. For each n ∈ N and each u ∈ C \ 1 , we know that

1 + u+ u2 + · · ·+ un−1 =1− un

1− u. (19)

X194. For each n ∈ N and each t ∈ R \ −1 , if we let u = −t in (19), we get

1

1 + t= 1− t+ t2 − · · ·+ (−1)n−1tn−1 +

(−1)ntn

1 + t.

Use this (but no theorems about series) to show that for each x ∈ (−1, 1],

log(1 + x) =

∞∑

k=1

(−1)k−1xk

k,

and that for each a ∈ (−1, 1), the series converges uniformly for x ∈ [a, 1]. In particular,

log 2 = 1− 1

2+

1

3− 1

4+− · · · .

X195. For each n ∈ N and each t ∈ R , if we let u = −t2 in (19), we get

1

1 + t2= 1− t2 + t4 − · · ·+ (−1)n−1t2n−2 +

(−1)nt2n

1 + t2.

Use this (but no theorems about series) to show that for each x ∈ [−1, 1],

arctanx =

∞∑

k=1

(−1)k−1 x2k−1

2k − 1,

and that the series converges uniformly for x ∈ [−1, 1]. In particular, since arctan 1 = π/4,

π

4= 1− 1

3+

1

5− 1

7+− · · · .

X196. due 8Th

(a) Let a ∈ R . Prove Newton’s binomial theorem: For each x ∈ (−1, 1),

(1 + x)a =

∞∑

k=0

(

a

k

)

xk, (20)

where(

a

k

)

=a(a− 1)(a− 2) · · · (a− k + 1)

k!.

(When k = 0,(

ak

)

= 1 because a product of no factors is 1.) Hint: Show that the series on theright in (20) converges. Let f(x) denote the sum of the series. Show that f satisfies the differentialequation

(1 + x)f ′(x) = af(x).

Use the method of integrating factors to solve this differential equation.30

(b) Now suppose a ∈ C . Then (20) still holds for each x ∈ (−1, 1), provided we define (1 + x)a asea log(1+x) . (It should be clear that the argument you gave in part (a) still works. You need notrepeat it.) Use the coincidence principal to show that for each z ∈ C with |z| < 1,

(1 + z)a =∞∑

k=0

(

a

k

)

zk,

where we take (1 + z)a to be eaLog(1+z) . (This is the principal value of (1 + z)a . Recall thatLog(1 + z) is the principal logarithm of 1 + z , namely the one whose imaginary part lies in theinterval (−π, π] .)

30 Alternatively, one could prove (20) by applying Taylor’s theorem, but this would involve more work, because it would betedious to estimate the remainder.

48


X197. For each n ∈ N , let kn ∈ N , let z(n, j) ∈ C for j = 1, . . . , kn , and let due 8Th

pn =kn∏

j=1

(1 + z(n, j)) and sn =

kn∑

j=1

z(n, j).

Suppose that∑kn

j=1 |z(n, j)|2 → 0 as n→ ∞ .

(a) Suppose z ∈ C with z 6= −1. Then Log(1 + z) = z + θz2 , where

θ =

Log(1+z)−zz2 if z 6= 0,

0 if z = 0.

Hence1 + z = ez+θz2

.

Show that if |z| ≤ 1/2, then |θ| ≤ 1. Hint: If |z| < 1, then

Log(1 + z) = z − z2

2+z3

3− z4

4+ − · · · .

(b) Show that pn ∼ esn as n→ ∞ .(c) Suppose in addition that sn → s ∈ C as n→ ∞ . Show that pn → es as n→ ∞ .

Remark. One important application of problem X197 occurs in the proof of the central limit theorem inprobability theory.

X198. Suppose that ak ∈ C \ −1 for each k ∈ N . Let pn =∏n

k=1(1 + ak) for each n ∈ N . Noticethat pn 6= 0 for each n ∈ N . Suppose that

∑∞k=1 ak converges and that

∑∞k=1 |ak|2 <∞ . Prove that (pn)

converges to a limit p in C \ 0 .Let E be a vector space over R . If v, w ∈ E , then the line segment from v to w is the set L(v, w) =

(1 − t)v + tw : t ∈ [0, 1] . Let S be a subset of E . To say that S is convex means that for all v, w ∈ S ,L(v, w) ⊆ S . To say that S is star-shaped with respect to v means that for each w ∈ S , L(v, w) ⊆ S . If Sis star-shaped with respect to v , then clearly v ∈ S . To say that S is star-shaped means that there exists vsuch that S is star-shaped with respect to v . Clearly S is convex iff S is star-shaped with respect to eachv ∈ S . Now let us mention some examples. Considering C as a vector space over R , each open or closeddisc is a convex subset of C , and so is each open or closed half-plane, while C \ (−∞, 0] is star-shaped withrespect to 1, but not convex.

Let X and Y be topological spaces. Let f0 and f1 be continuous maps from X into Y . To say thatf0 is homotopic to f1 in Y means that there exists a continuous map H from X × [0, 1] into Y such thatfor each x ∈ X , H(x, 0) = f0(x) and H(x, 1) = f1(x). Such a map H is called a homotopy from f0 tof1 in Y . Intuitively, such a homotopy describes a continuous deformation of the map f0 into the map f1 ,where each of the intermediate maps H(·, t), t ∈ [0, 1], is a continuous map from X into Y . It is not hardto show that the relation of homotopy between continuous maps from X into Y is an equivalence relationon C(X,Y ), the set of all continuous maps from X into Y .

Let X be a topological space. To say that X is contractible means that the identity map from X intoitself is homotopic in X to some constant map.

X199. Let X be a star-shaped subset of Rd . Show that X is contractible.

Let X , Y , and Z be topological spaces. Let f : X → Z be a continuous map. To say that f can befactored through Y means that there exist continuous maps g : X → Y and h : Y → Z such that f = h g .X200. Let C× = C \ 0 . Let X be a topological space and let f be a continuous map from X into C× .Consider the following seven statements:

(a) There exists a branch of log f .(b) f can be factored through C .(c) f can be factored through a convex subset of C .

49


(d) f can be factored through a star-shaped subset of C .(e) f can be factored through some contractible topological space.(f) f is homotopic in C× to some constant map.(g) f is homotopic in C× to the constant map 1: X → C× .

Prove that (a)⇒(b)⇒ (c)⇒(d)⇒(e)⇒(f)⇒ (g). (It is an important non-trivial fact that (g)⇒(a) too, butyou are not asked to prove this here.)

Remark. Let C× , X , and f be as in problem X200. Obviously f can always be factored through X ,because f = f idX , where idX denotes the identity function on X . Hence if X is contractible, then thereexists a branch of log f . In particular, if X is a star-shaped subset of C , then there exists a branch of log f .Still more particularly, if X is a star-shaped subset of C× , then there exists a branch of log in X . To seethis, take f to be idX . (If X ⊆ C× , then to say that g is a branch of log in X means that for each z ∈ X ,eg(z) = z . In other words, if X ⊆ C× , then a branch of log in X is a branch of log f , where f = idX .)

Definition. Let f : X → Y , where X is a topological space and Y is a metric space. Let p ∈ X and letUp be the collection of nhds of p in X . The oscillation of f at p is

osc(f, p) = infU∈Up

diam(f [U ]).

X201. Let f : X → Y , where X is a topological space and Y is a metric space. Define ϕ : X → [0,∞] by

ϕ(p) = osc(f, p)

for all p ∈ X .(a) Let p ∈ X . Prove that f is continuous at p iff osc(f, p) = 0.(b) Prove that ϕ is upper semicontinous.(c) Let C = p ∈ X : f is continuous at p . Prove that C is a Gδ subset of X .

X202. Let X be a topological space and let f : X → R be bounded. Let p ∈ X . Prove that

osc(f, p) = f∗(p)− f∗(p).

Reminder. Let X be a topological space and let A ⊆ X . To say that A is nowhere dense means that theclosure of A has empty interior. To say that A is meager means that A is a countable union of nowheredense sets. For instance, any countable subset of R is meager in R . So is any closed subset of R which hasempty interior. For instance, the Cantor set is meager in R . You should think of meager sets as sets whichare small in the sense of topology. Meager sets are also known as sets of the first category. Sets which arenot meager are also known as sets of the second category. To say that X is a Baire space means that nonon-empty open subset of X is meager. According to problem X74, X is a Baire space iff for each meagerset A ⊆ X , the set X \ A is dense in X iff for each sequence (Gn) of dense open subsets of X , we have⋂∞

n=1Gn is dense in X . The Baire category theorem states that each complete metric space is a Baire space.

Reminder. Clearly Q is an Fσ subset of R , so R \Q is a Gδ subset of R . By problem X77, Q is not aGδ subset of R . Hence R \Q is not an Fσ subset of R .

X203. Let (fn) be a sequence of continuous functions from a topological space X to a metric space (Y, d ),such that (fn) converges pointwise to a function f : X → Y . Prove that for each open set G ⊆ Y , f−1[G]is an Fσ subset of X .

X204. Prove that 1Q , the indicator function of the set Q of rational numbers, is not the limit of a sequenceof continuous functions from R to R .

X205. Let X be a topological space. Prove that X is a Baire space iff for each sequence (Fn) of closedsubsets of X , if X =

⋃

n Fn , then⋃

n Int(Fn) is dense in X .

X206. Let f be a function from a Baire space X to a separable metric space (Y, d ), such that for eachopen set G ⊆ Y , f−1[G] is an Fσ subset of X . Let C be the set of all x ∈ X such that f is continuous atx . By problem X201, we know that C is a Gδ subset of X . Prove that C is dense in X . (Hint: For eachn ∈ N , Y can be covered by a sequence of open sets of diameter at most 1/n .)

50


X207. Define f : R → R by

f(x) =

e−1/x if x > 0,

0 if x ≤ 0.

(a) Prove that for each n ∈ N , f (n)(x) → 0 as x→ 0+.(b) Prove that f is infinitely differentiable on R .

X208. Let a, b ∈ R with a < b . Find an infinitely differentiable function g : R → R such that g(x) = 1for each x ∈ (−∞, a] , 0 < g(x) < 1 for each x ∈ (a, b), and g(x) = 0 for each x ∈ [b,∞).

X209. Let D = C \ (−∞, 0]. Define f : D → C by f(z) = Log z . We have seen that f is continuous inD . Prove that f is analytic in D .

Reminder. In problem X187, we considered Riemann integration of functions taking values in a Banachspace.

X210. (The triangle inequality for a Riemann integrable function taking values in a Banach space.) Letf : [a, b] → Y , where a, b ∈ R with a < b and where Y is a Banach space. Suppose that f is Riemann

integrable over [a, b] . Let y =∫ b

a f(t) dt . Prove that |y| ≤∫ b

a |f(t)| dt . (Hint: This is easy if f is a stepfunction. For the general case, approximate f by step functions.)

X211. Let f : [a, b] → Y , where a, b ∈ R with a < b and where Y is a Banach space. Suppose f isRiemann integrable over [a, b] . From the discussion in problem X187, it is obvious that then f is integrableover [u, v] for all u, v ∈ [a, b] with u < v . Define F : [a, b] → Y by F (x) =

∫ x

a f(t) dt . Let c ∈ [a, b] suchthat f is continuous at c . Prove that F is differentiable at c and that F ′(c) = f(c). (By the way, in thecourse of your proof, you should notice that analogous statements hold for the one-sided derivatives of F atc and one-sided continuity of f at c .)

X212. (The straddle lemma.) Let Y be a normed linear space. Let f : E → Y , where E ⊆ R . Let E′ bethe set of limit points of E and let a ∈ E′ ∩E . Suppose that f is differentiable31 at a . Prove that for eachε > 0, there exists δ > 0 such that for all u, v ∈ E with u ≤ a ≤ v and 0 < v − u < δ , we have

∣

∣

∣

∣

f(v)− f(u)

v − u− f ′(a)

∣

∣

∣

∣

< ε.

Remark. The mean value theorem does not apply to vector-valued functions. Even for a function takingvalues in R2 , the points that the mean value theorem furnishes for the two components of the function areusually going to be different. The next result is an adequate substitute for the mean value theorem in manysituations involving vector-valued functions.

X213. (The mean value inequality.) Let f : [a, b] → Y , where a, b ∈ R with a < b and where Y is anormed linear space.

(a) Suppose that f is differentiable on [a, b] . Prove that there exists c ∈ [a, b] such that

|f ′(c)| ≥∣

∣

∣

∣

f(b)− f(a)

b− a

∣

∣

∣

∣

.

(Hint: First check that if u, v, w ∈ [a, b] with u < v < w , then at least one of the two quantities∣

∣

∣

∣

f(v)− f(u)

v − u

∣

∣

∣

∣

and

∣

∣

∣

∣

f(w) − f(v)

w − v

∣

∣

∣

∣

is greater than or equal to∣

∣

∣

∣

f(w)− f(u)

w − u

∣

∣

∣

∣

.

In particular, this holds if v is the midpoint of [u,w] . Then start with the whole interval [a, b] ,apply a bisection argument, and finish by applying the straddle lemma.)

(b) Suppose that f is continuous on [a, b] and differentiable on (a, b). Let

M = sup |f ′(t)| : t ∈ (a, b) .Prove that |f(b)− f(a)| ≤M · (b− a).

31 Differentiability for a function taking values in a normed linear space is defined just like for a function taking values inR .

51


X214. Let Y be a normed linear space and let a, b ∈ R with a < b .(a) Let H : [a, b] → Y be continuous on [a, b] and differentiable on (a, b). Suppose also that H ′(x) = 0

for each x ∈ (a, b). Prove that H is constant on [a, b] .(b) Let F,G : [a, b] → Y be continuous on [a, b] and differentiable on (a, b). Suppose also that

F ′(x) = G′(x) for each x ∈ (a, b). Prove that there is a constant C ∈ Y such that G = F + C .

X215. (The fundamental theorem of calculus for a continuous function taking values in a Banach space.)Let f : [a, b] → Y be continuous, where a, b ∈ R with a < b and where Y is a Banach space. Suppose thatG : [a, b] → Y is continuous on [a, b] and differentiable on (a, b). Suppose also that G′(x) = f(x) for eachx ∈ (a, b). Prove that

∫ b

a

f(t) dt = G(b)−G(a).

X216. Give an example of a differentiable function f : R → R and sequences (un) and (vn) in R suchthat un 6= vn for all n , un → 0 and vn → 0, but

lim supn→∞

f(vn)− f(un)

vn − un= ∞ and lim inf

n→∞

f(vn)− f(un)

vn − un= −∞.

Explain why your example does not contradict the straddle lemma.

X217. Define functions u and v by due 9Th

u(t) =

∞∑

k=1

cos(kt)

k(21)

and

v(t) =∞∑

k=1

sin(kt)

k. (22)

Let each of these functions be defined for all t ∈ R for which the sum of the corresponding series is defined.We know from problem X175 that these series converge locally uniformly in R\ 2πZ . Clearly the series (21)diverges to +∞ on 2πZ while the series (22) converges on 2πZ , because each term is zero there.

(a) Find simple closed form expressions for u(t) and v(t). Hint: As in problem X175, define a functiong by

g(t) =∞∑

k=1

eikt

k

for each t ∈ R for which the series converges in C . Notice that u(t) is the real part of g(t) andv(t) is the imaginary part of g(t). Now for |z| ≤ 1 with z 6= −1, the principal logarithm of 1 + zis given by

Log(1 + z) = z − z2

2+z3

3−+ · · · .

Hence for |z| ≤ 1 with z 6= 1,

−Log(1 − z) = z +z2

2+z3

3+ · · · .

Substituting z = eit in the preceding equation, we find that

g(t) = −Log(1− eit)

for each t ∈ R \ 2πZ . To get simple expressions from this for u and v , check that

1− eit = 2ei(t−π)/2 sint

2.

The expressions you find for u and v should involve only real numbers.

52


(b) Sketch the graph of v . If you found as simple an expression for v as you should have in (a), thenthis should be a piece of cake.

(c) Even though the series (22) does not converge uniformly,32 we can still hope that each of theintegrals,

∫ 2π

0

v(t) cos(kt) dt, k = 0, 1, 2, . . .

and∫ 2π

0

v(t) sin(kt) dt, k = 1, 2, 3, . . .

can be calculated by expressing v(t) as in (22) and integrating term-by-term. We shall soon seethat this is indeed the case. Verify that the values so obtained for these integrals are consistentwith your answer to (a).

X218. Let a ∈ C . We saw in problem X196 that for |z| < 1, the principal value of (1 + z)a is given by due 9Th

(1 + z)a =∞∑

k=0

(

a

k

)

zk. (23)

If a ∈ ω = 0, 1, 2, . . . , then of course(

ak

)

= 0 for each k ∈ a+ 1, a+ 2, . . . , so (23) reduces to the

ordinary binomial expansion (1 + z)a =∑a

k=0

(

ak

)

zk , which is valid for all z ∈ C . For the remainder of thisexercise, let us assume that a /∈ ω , so that the right side of (23) really is an infinite series. Then by theratio test, it is easy to see that the radius of convergence of the series is exactly 1.

(a) Suppose that Re(a) ≤ −1. Show that∣

∣

(

ak

)∣

∣ ≥ 1 for each k ∈ ω . Deduce that the series in (23)diverges for all z with |z| = 1.

(b) Consider general a ∈ C again. As we know,(

a0

)

= 1. Verify that for each k ∈ N ,(

ak

)

= (−1)kbk ,where

bk =

k∏

ℓ=1

(

1− a+ 1

ℓ

)

. (24)

Then use problem X197(a), together with problem X158, to show that for each k ∈ N ,

bk =ckka+1

, (25)

where (ck) is a sequence in C\ 0 which converges to a limit c belonging to C\ 0 , and wherewe mean the principal value of ka+1 .

(c) Suppose that Re(a) > 0. Show that the series in (23) converges absolutely for each z with |z| = 1and that for each such z , the equation (23) holds.33

(d) Suppose that −1 < Re(a) ≤ 0. Use (25) to show that bk → 0. Use (24) to show that

bk − bk+1 = bkdk, (26)

where dk is a suitable simple factor that you should determine explicitly. Then use (25) and (26)to show that the sequence (bk) is of bounded variation. Finally, deduce that if |z| = 1 but z 6= −1,then the series in (23) converges, although not absolutely, and that the equation (23) holds foreach such z .

(e) Suppose again that −1 < Re(a) ≤ 0. Suppose also that z = −1. Prove that the series in (23)diverges. (Hint: This is easy if you use a theorem that we proved in class recently. If the seriesdid converge, what would that tell us about limx↓−1(1 + x)a ? Remember that a 6= 0 because weare assuming that a /∈ ω .)

32 It cannot converge uniformly, because v is discontinous at each integer multiple of 2π , as should be clear from your answerto (a).

33 For w ∈ C \ 0 , if ζ is the principal value of wa , then |ζ| = eRe(a) log |w|−Im(a)Arg(w) . Hence if Re(a) > 0, then asw → 0, the principal value of wa tends to 0, because Re(a) log |w| tends to negative infinity while Arg(w) stays between −πand π . Hence when Re(a) > 0, we consider the principal value of 0a to be 0. In particular, when Re(a) > 0 and z = −1,the left side in (23) has the value 0.

53


Remark. If (ak)k∈N is a sequence in [0,∞] , then it is easy to check that

∑∞k=1 ak = sup

∑

k∈B ak : B is a finite subset of N

.

This motivates the following definition.

Definition. Let (ak)k∈A be a family of elements of [0,∞] , where A is an arbitrary index set. (A couldeven be uncountably infinite.) Then by definition,

∑

k∈A ak = sup∑

k∈B ak : B is a finite subset of A

X219. Let V be an inner product space and let (uk)k∈A be an orthonormal family in V . Let f ∈ V . Foreach k ∈ A , let ck = 〈uk |f〉 .

(a) Let B be a finite subset of A and let h =∑

k∈B ckuk . Show that

‖f − h‖2 +∑

k∈B

|ck|2 = ‖f‖2.

(b) Use part (a) to prove Bessel’s inequality in its abstract form:

∑

k∈A

|ck|2 ≤ ‖f‖2. (27)

54


Reminder. R(T) denotes the (almost) inner product vector space of complex-valued 1-periodic functionson R which are Riemann integrable over [0, 1]. For each k ∈ Z , we define ek : R → C by ek(t) = e2πikt .We have seen that (ek)k ∈ Z is an orthonormal family in R(T).

Remark. Let g ∈ R(T). Since (ek)k∈Z is an orthonormal family in R(T), Bessel’s inequality (27) tells usthat

∑

k∈Z

|g(k)|2 ≤ ‖g‖22. (28)

Actually, according to Parseval’s equation, we have equality in (28), though we don’t need that fact for theapplication of (28) in the next exercise.

X220. Let f : R → C be 1-periodic and differentiable. Suppose f ′ is Riemann integrable over [0, 1]. due 10Th

(a) Prove that∑

k∈Z |f(k)| < ∞ . (Hint: Let g = f ′ . Find an expression for g(k) in terms of f(k).Then use (28).)

(b) For each n ∈ ω = 0, 1, 2, . . . , let Sn =∑n

k=−n f(k)ek . Prove that the sequence of functions(Sn) converges uniformly to f . (Hint: Prove first that (Sn) is uniformly Cauchy and henceconverges uniformly to some continuous function S : R → C . Then show that ‖f − S‖2 = 0 anduse the continuity of f and of S to conclude that for each t ∈ R , S(t) = f(t).)

(c) Show that in fact,∑

k∈Z f(k)ek = f , where the series converges uniformly to f in the sense ofunordered sums. In other words, show that for each ε > 0, there exists a finite subset B ⊆ Z suchthat for each finite subset C ⊆ Z with B ⊆ C , and for each t ∈ R , we have

∣

∣

∣f(t)−

∑

k∈Cf(k)ek(t)

∣

∣

∣≤ ε.

X221. Let P1(t) = t− 1/2 for 0 < t < 1. Let P1(0) = 0 and extend P1 to R by periodicity.34 Show thatthe Fourier series for P1 may be written as

−∞∑

k=1

2 sin 2πkt

2πk. (29)

Then deduce from Parseval’s equation that∑∞

k=1 k−2 = π2/6.

Remark. Let P1 be as in problem X221. Let

P2(t) =

∞∑

k=1

2 cos 2πkt

(2πk)2,

let

P3(t) =

∞∑

k=1

2 sin 2πkt

(2πk)3,

and define P4 , P5 , P6 , P7 , and so on, by

P2r(t) = (−1)r−1∞∑

k=1

2 cos 2πkt

(2πk)2r

and

P2r+1(t) = (−1)r−1∞∑

k=1

2 sin 2πkt

(2πk)2r+1.

34 By the way, for each t ∈ R \ Z , P1(t) = t − ⌊t⌋ − 1/2, where ⌊t⌋ is the greatest integer that is less than or equal to t .Of course for each t ∈ Z , P1(t) = 0.

55


From problem X175 and problem X217, after some elementary modifications, we know that the series (29)converges to P1(t) for each t ∈ R and that P ′

2(t) = P1(t) for each t ∈ R \ Z . The series for P2 , P3 , andso on, all converge uniformly on R and so it is elementary that for each r ∈ N with r ≥ 2, and for eacht ∈ R , P ′

r+1(t) = Pr(t). It follows that for each r ∈ N , the restriction of the 1-periodic function Pr tothe open interval (0, 1) agrees with a certain polynomial pr of degree r , where the constant term in pr is

determined by the fact that∫ 1

0pr(t) dt = 0. These polynomials are known as the Bernoulli polynomials, after

James Bernoulli, who introduced essentially these polynomials35 in 1713, in connection with the problemof determining the sums

∑nk=1 k

r . Note that it is straightforward to determine the Bernoulli polynomialsexplicitly, by successive integration, beginning from p1 . Then, just as we applied Parseval’s equation to P1

to evaluate∑∞

k=1 k−2 , we may apply Parseval’s equation to Pr to evaluate

∑∞k=1 k

−2r .

Definition. Let (xj) be a sequence of real numbers. To say that (xj) is equidistributed modulo 1 meansthat for each f ∈ C(T),

limn→∞

1

n

n−1∑

j=0

f(xj) =

∫ 1

0

f(x) dx. (30)

X222. Let α ∈ R . Let x0 ∈ R and let xj = x0 + jα for j ∈ N . Prove that the sequence (xj) due 10Th

is equidistributed modulo 1 iff α is irrational. (Hint: The forward implication is trivial. For the reverse

implication, first prove it for f of the form f(x) = e2πikx where k ∈ Z , by explicit calculation of∑n−1

j=0 f(xj).Then use the fact that the set of trigonometric polynomials is dense in C(T).)

X223. Let (xj) be a sequence of real numbers which is equidistributed modulo 1. By definition, (30) holdsfor each f ∈ C(T). Prove that in fact, (30) holds for each f ∈ R(T). (Hint: It suffices to consider the casewhere f is real-valued. Verify that then for each ε > 0, there exist functions g, h ∈ C(T) with g ≤ f ≤ h

and∫ 1

0h(x)− g(x) dx < ε .)

Terminology. Let x be a real number. Then the fractional part of x is the number x− ⌊x⌋ , where ⌊x⌋ isthe greatest integer less than or equal to x .

X224. Let (xj) be a sequence of real numbers. Prove that (xj) is is equidistributed modulo 1 iff foreach interval I ⊆ [0, 1), we have n−1#n(I) → |I| , where |I| is the length of I and #n(I) is the numberof j ∈ 0, . . . , n− 1 such that the fractional part of xj is in I . (Hint: The forward implication is aparticular case of the result of problem X223. To prove the reverse implication, remember that a continuouscomplex-valued function on [0, 1] is a uniform limit of step functions.)

Reminder. Let f, g ∈ R(T). Then the convolution of f and g is the function f ∗ g defined on R by

(f ∗ g)(t) =∫ 1

0

f(t− s)g(s) ds.

X225. Let f, g ∈ R(T). Prove that sup |f ∗ g| ≤ ‖f‖2‖g‖2 .X226. Let f, g ∈ R(T) and let (fn) and (gn) be sequences in R(T) such that ‖f − fn‖2 → 0 and‖g − gn‖2 → 0. Prove that fn ∗ gn → f ∗ g uniformly.

Reminder. Let f ∈ R(T) and let k ∈ Z . By definition, f(k) = 〈ek |f〉 =∫ 1

0 e−2πiktf(t) dt . As we saw in

class, it is easy to check that f ∗ ek = f(k)ek .

X227. Let j, k ∈ Z . Find ej(k) and ej ∗ ek .X228. Let j, k, ℓ ∈ Z . Check that ej ∗ (ek ∗ eℓ) = (ej ∗ ek) ∗ eℓ .X229. Let f, g, h ∈ R(T). Prove that f ∗ (g ∗ h) = (f ∗ g) ∗ h . (Hint: For now, we can prove this by usingthe fact that the set of trigonometric polynomials is dense in R(T) with respect to the 2-norm. If we hadthe basic theory of double integrals at our disposal, then we could prove it by a routine calculation involvingan interchange in order of integration.)

X230. Let f, g ∈ R(T). Prove that f ∗ g is continuous.

35 Some authors adopt related but slightly different definitions of the Bernoulli polynomials.

56

Math 653 Extra Problems Spring, 2006

These problems are numbered consecutively with the ones from last quarter. (That is why the first of themis not numbered X1.)

Remark. Our goal at present is to study notions such as length, area, volume, mass, charge, and so on,with a view to developing a theory of integration that will be superior to the Riemann theory of integration.Such notions are all special cases of the general notion of measure. In due course, we shall define preciselywhat we mean by measure. For now, let us just say in order to study the measure of fairly general sets, wemust first study the measure of very special sets. The collection of these special sets usually forms what isknown as a pre-ring.

Reminder. To say that H is a pre-ring (of sets) means that H is a collection of sets such that for allA,B ∈ H , we have A ∩B ∈ H and A \B is a finite disjoint union of elements of H . If X is a set, thento say that H is a pre-ring of subsets of X or, more briefly, that H is a pre-ring on X , means that His a pre-ring of sets and each element of H is a subset of X .

Example. Let H = (a, b] : −∞ < a ≤ b <∞ . Then H is a pre-ring on R . Notice that H is notclosed under relative complements. For instance (1, 4] \ (2, 3] is not an element of H , although it is thedisjoint union of two elements of H , namely (1, 2] and (3, 4]

Example. The collection of all bounded intervals in R is a pre-ring on R . However, this pre-ring is lessconvenient to take as a collection of basic sets than the one in the preceding example, because in our proofs,we would have more cases to consider.

X231. Let H1 and H2 be pre-rings on a set X . Let

H = A ∩B : A ∈ H1 and B ∈ H2 .

Prove that H is a pre-ring.

Reminder. Let A and B be collections of sets. Then by definition,

A ⊙ B = A×B : A ∈ A and B ∈ B .

Warning. You should not write A × B for A ⊙ B . The notation A × B denotes a different collection,namely the set of ordered pairs (A,B) : A ∈ A and B ∈ B .Reminder. As we saw in class, it follows from the result of problem X231 that if H1 and H2 are pre-ringson sets X1 and X2 respectively, and if H = H1 ⊙ H2 , then H is a pre-ring on X1 ×X2 . For instance,the collection

(a, b]× (c, d] : −∞ < a ≤ b <∞ and −∞ < c ≤ d <∞

is a pre-ring on R2 . Of course the result of problem X231 can be generalized by induction to any finitenumber of pre-rings instead of just two. From this it follows that if Hk is a pre-ring on a set Xk fork = 1, . . . , n , and if H is the collection of all Cartesian products of the form

∏nk=1 Ak , where Ak ∈ Hk for

k = 1, . . . , n , then H is a pre-ring on∏n

k=1Xk . For instance, the collection of Cartesian products of theform

∏nk=1(ak, bk] , where −∞ < ak ≤ bk <∞ for k = 1, . . . , n , is a pre-ring on Rn .

X232. Let Φ be a collection of pre-rings. Suppose that Φ is upwards directed by set-inclusion.36 LetH =

⋃

Φ. Prove that H is a pre-ring.

Definition. Let X be a set. To say that H is a pre-field of subsets of X or, more briefly, that H isa pre-field on X , means that H is a pre-ring on X and X ∈ H . (Another name for a pre-field is apre-algebra.)

Example. The collection of all intervals in R , including unbounded intervals, is a pre-field on R .

Example. Let H = (a, b] : 0 ≤ a ≤ b ≤ 1 . Then H is a pre-field on (0, 1].

36 To say that Φ is upwards directed by set-inclusion means that for each H1 ∈ Φ and each H2 ∈ Φ, there exists H3 ∈ Φsuch that H1 ⊆ H3 and H2 ⊆ H3 .

57


Remark. The next exercise is relevant to the study of measure in infinite dimensional spaces. (The theoryof measure in infinite dimensional spaces is important in probability theory, but we shall not pursue thistopic in this course.)

X233. Let (Xα)α∈A be a family of sets and let X =∏

α∈AXα . For each α ∈ A , let Hα be a pre-fieldon Xα . Let H be the collection of all sets of the form

∏

α∈AHα , where Hα ∈ Hα for each α ∈ A andHα = Xα for all but finitely many α ∈ A . Prove that H is a pre-field on X . (Hint: For each finite setB ⊆ A , let HB be the collection of all sets of the form

∏

α∈AHα , where Hα ∈ Hα for each α ∈ A andHα = Xα for each α ∈ A \B . Use earlier work to show that each HB is a pre-field on X . Let

Φ = HB : B is a finite subset of A .

Notice that H =⋃

Φ. To finish, use problem X232.)

Remark. After the basic sets that we start with in our study of measure, the next simplest kind of sets arethe ones which are finite disjoint unions of the basic sets. If the collection of basic sets that we start with isa pre-ring, then the collection of finite disjoint unions of such basic sets will be a ring.

Reminder. To say that R is a ring (of sets) means that R is a collection of sets such that Ø ∈ R andfor all A,B ∈ R , we have A∪B ∈ R and A \B ∈ R . If X is a set, then to say that R is a ring of subsetsof X or, more briefly, that R is a ring on X , means that R is a ring of sets and each element of R is asubset of X .

X234. Let R be a ring of sets. Prove that R is a pre-ring. (Obviously, to prove this, it remains onlyto show that R is closed under pairwise intersections. In other words, you just need to prove that for allA,B ∈ R , we have A ∩B ∈ R .)

Example. Let X be a set with at least two elements. Let S =

x : x ∈ X

∪Ø . Then Ø ∈ S andS is closed under pairwise intersections and relative complements, but S is not a ring of sets.

X235. (If you attended recitation on Tuesday, March 27, then you are excused from turning in this exercise.) due 1F

Let H be a pre-ring. Let R be the collection of finite disjoint unions of elements of H . The object of thisexercise is to lead you through a proof of the fact that R is a ring. Notice that Ø ∈ R because Ø is theunion of the empty collection of elements of H . Notice also that R is obviously closed under finite disjointunions.

(a) Let A,B ∈ R . Prove that A ∩B ∈ R .(b) Let A ∈ H and let B ∈ R . Prove that A \B ∈ R .(c) Let A,B ∈ R . Prove that A \B ∈ R .(d) Let A,B ∈ R . Prove that A ∪B ∈ R .(e) Prove that R is the smallest ring of sets that contains H .

Remark. Let A be an interval, let (Bk) be a finite sequence of intervals, and let C = A \⋃kBk . Sincethe collection of all intervals is a pre-ring, it follows from problem X235 that C is a finite disjoint union ofintervals. It is important to realize that the analog of this statement when we replace “finite” by “countable”is false!

X236. Give two examples of an interval A and an infinite sequence of intervals (Bk) such that the setC = A \⋃k Bk is not a countable disjoint union of intervals. (Hint: For one example, think about the setof irrational numbers. For another example, think about the Cantor set.)

Definition. Let X be a set. To say that F is a field of subsets of X or, more briefly, that F is a fieldon X means that F is a ring on X and X ∈ F . (Another name for a field in this context is an algebra.)

Remark. Let X be a set, let H be a pre-field on X , and let F be the set of finite disjoint unions ofelements of H . Then F is a field on X . This obviously follows from problem X235.

58


Reminder. Let H = (u, v] : −∞ < u ≤ v <∞ . As we know, H is a pre-ring on R . Let V be acommutative group. Consider any function F : R → V . Define τ : H → V by

τ((u, v]) = F (v) − F (u) (31)

for all u, v ∈ R with u ≤ v . (Notice that τ is well-defined because if (u, v] 6= Ø, then u and v are uniquelydetermined by (u, v] , while if (u, v] = Ø, then u = v , so F (v)−F (u) = 0.) As we saw in class, τ is additive.The most important special case is the one where V = R and F (x) ≡ x . In this case, τ((u, v]) = v− u , thelength of the interval (u, v] .

Remark. In the special case where V = R and where F : R → R is C1 , for all u, v ∈ R with u ≤ v , wehave

F (v)− F (u) =

∫ v

u

dF

dxdx,

so in this case (31) may rewritten as

τ((u, v]) =

∫ v

u

dF

dxdx. (32)

We shall not make mathematical use of (32), which is just as well since we are in the process of redevel-oping the theory of integration “from scratch.” We only mention (32) because its generalizations to higherdimensions will be useful to help us remember the generalizations of (31) to higher dimensions.

X237. Let H = (u, v] : −∞ < u ≤ v <∞ . Let V be a commutative group. Consider any additivefunction τ : H → V . Prove that there exists a function F : R → V such that for all u, v ∈ R with u ≤ v ,we have

τ((u, v]) = F (v)− F (u).

Although F is not unique, because we can add any constant to it, show that one suitable choice for F isgiven by

F (x) =

τ((0, x]) if x ≥ 0,

−τ((x, 0]) if x < 0.

Remark. In problem X237, F is called a distribution function for τ .

X238. Let H be a pre-ring and let E be a finite collection of elements H . Prove that there is a finitedisjoint collection A ⊆ H such that

⋃

A =⋃

E and for each E ∈ E , we have E =⋃

AE , where

AE = A ∈ A : A ⊆ E .

X239. Prove that the analog of problem X238 with “finite” replaced by “countable” is false. (Hint: Considerthe countable collection of sets E = (−∞, x] : x ∈ Q . You can take H to be the pre-ring of all subsetsof R .)

X240. Let H1 and H2 be pre-rings. Let H = A ∩B : A ∈ H1 and B ∈ H2 . As we know from problemX231, H is a pre-ring. Let V be a commutative semigroup with 0 and let τ : H → V . Suppose that foreach B ∈ H2 , the function A 7→ τ(A ∩ B) is additive on H1 , and that for each A ∈ H1 , the functionB 7→ τ(A ∩B) is additive on H2 . Prove that τ is additive on H . (Hint: problem X238 should help.)

X241. Let H1 and H2 be pre-rings on sets X1 and X2 respectively. Let H = H1 ⊙ H2 . As we saw inclass, H is is a pre-ring on X1 × X2 . Let V be a commutative semigroup with 0 and let τ : H → V .Suppose that for each B ∈ H2 , the function A 7→ τ(A× B) is additive on H1 , and that for each A ∈ H1 ,the function B 7→ τ(A×B) is additive on H2 . Prove that τ is additive on H . (Hint: This can be deducedeasily from problem X240.)

Remark. Of course the results of problem X240 and problem X241 can be generalized from two pre-ringsto any finite number of pre-rings.

59


X242. Let H1 = (u, v] : −∞ < u ≤ v <∞ . As we know, H1 is a pre-ring on R . Let H2 be anypre-ring and let H = H1 ⊙ H2 . Let V be a commutative group. Let W be the set of all finitely additivefunctions from H2 to V . If σ, τ ∈W , then of course σ+ τ denotes the function from H2 to V defined by(σ+τ)(B) = σ(B)+τ(B) for all B ∈ H2 , and it is obvious that σ+τ ∈ W . Notice that with this operationof addition, W is itself a commutative group. Consider any function G : R → W . Define τ : H → V by

τ((u, v]×B) = G(v)(B) −G(u)(B)

for all u, v ∈ R with u ≤ v and all B ∈ H2 . (You should explain why τ is well-defined.) Prove that τ isadditive on H . (Hint: Use problem X241.)

X243. Let H1 = H2 = (u, v] : −∞ < u ≤ v <∞ and let H = H1 ⊙ H2 . As we know, H1 and H2

are pre-rings on R and H is a pre-ring on R2 . Let V be a commutative group. Consider any functionF : R2 → V . Define τ : H → V by

τ((u1, v1]× (u2, v2]) = F (v1, v2)− F (u1, v2)− F (v1, u2) + F (u1, u2) (33)

for all u1, v1, u2, v2 ∈ R with u1 ≤ v1 and u2 ≤ v2 . (You should explain why τ is well-defined.) Prove thatτ is additive on H . (Hint: Use problem X242.)

Remark. Notice that even if V = R in problem X243, the W that we consider in applying problem X242to solve problem X243 is an infinite dimensional vector space over R , namely the space of all additivereal-valued functions on the pre-ring (u, v] : −∞ < u ≤ v <∞ . Thus we finally see a significant payoffto our having taken the trouble from the outset to consider additive functions taking values in semigroupsmore general than R .

Remark. In the special case where V = R and where F : R2 → R is C2 , for all u1, v1, u2, v2 ∈ R withu1 ≤ v1 and u2 ≤ v2 , we have

F (v1, v2)− F (u1, v2)− F (v1, u2) + F (u1, u2) =

∫ v1

u1

∫ v2

u2

∂2F

∂x2∂x1dx2 dx1,


τ((u1, v1]× (u2, v2]) =

∫ v1

u1

∫ v2

u2

∂2F

∂x2∂x1dx2 dx1.

Remark. Let us consider a particular case of the situation treated in problem X243. Let V = R and letF : R2 → R satisfy

F (x1, x2) = F1(x1)F2(x2)

for all x1, x2 ∈ R , where F1, F2 : R → R . Then for all u1, v1, u2, v2 ∈ R with u1 ≤ v1 and u2 ≤ v2 , wehave

τ((u1, v1]× (u2, v2]) = F1(v1)F2(v2)− F1(u1)F2(v2)− F1(v1)F2(u2) + F1(u1)F2(u2)

=(

F1(v1)− F1(u1))

F2(v2)−(

F1(v1)− F1(u1))

F2(u2)

=(

F1(v1)− F1(u1))(

F2(v2)− F2(u2))

.

The most important special case is the one where

F (x1, x2) = x1x2

for all x1, x2 ∈ R . In this case, τ((u1, v1] × (u2, v2]) = (v1 − u1)(v2 − u2), the area of the rectangle(u1, v1]× (u2, v2] .

Remark. The next exercise is a converse to problem X243.

60


X244. Let H1 = H2 = (u, v] : −∞ < u ≤ v <∞ and let H = H1⊙H2 . As we know, H1 and H2 arepre-rings on R and H is a pre-ring on R2 . Let V be a commutative group. Let τ : H → V be additive.Prove that there exists a function F : R2 → V such that for all u1, v1, u2, v2 ∈ R with u1 ≤ v1 and u2 ≤ v2 ,we have

τ((u1, v1]× (u2, v2]) = F (v1, v2)− F (u1, v2)− F (v1, u2) + F (u1, u2)

(Hint: Let W be the group of all finitely additive functions from H2 to V . Define σ : H1 → W byσ(A)(B) = τ(A ×B) for all A ∈ H1 and all B ∈ H2 . Check that σ is additive. Hence, by problem X237,there is a function G : R → W such that for all u1, v1 ∈ R with u1 ≤ v1 , σ((u1, v1]) = G(v1)−G(u1). Tofinish, for each x1 ∈ R , apply problem X237 to the additive function B 7→ G(x1)(B) on H2 .)

Remark. In problem X244, F is called a distribution function for τ . Of course F is not unique. It is noteven unique up to a constant. Rather, one may add to it any function C from R2 to V which satisfies

C(v1, v2)− C(v1, u2)− C(u1, v2) + C(u1, v1) = 0

for all u1, v1, u2, v2 ∈ R with u1 ≤ v1 and u2 ≤ v2 . When V = R and C : R2 → R is C2 , it can be shownthat this condition on C is equivalent to the condition

∂2C

∂x2∂x1≡ 0.

Remark. Now we wish to generalize problem X243 and problem X244 from R2 to Rn . To formulate thegeneralization succinctly, we introduce some notation. But first, to prepare for interpreting a particular caseof this generalization, we consider a generalization of the binomial theorem.

X245. Prove that for each n ∈ N , for all b1, . . . , bn ∈ R , we have

n∏

i=1

(1 + bi) =∑

I

∏

i∈I

bi,

where the index I of summation ranges over all subsets of the set 1, . . . , n . (Note: If I = Ø, then∏

i∈I bi = 1 by convention. This makes sense because for each real number c , the product of c and∏

i∈Ø bi)should be c .)

Remark. By a slight refinement of the method of proof for problem X245, one can prove the generalizedbinomial theorem, which states that for each n ∈ N , for all real numbers a1, b1, . . . , an, bn , we have

n∏

i=1

(ai + bi) =∑

I

[

(

∏

i/∈I

ai

)(

∏

i∈I

bi

)

]

,

where again the index I of summation ranges over all subsets of the set 1, . . . , n and where for each suchI , the i in

∏

i/∈I ai ranges over the set 1, . . . , n \ I .Notation. Let u = (u1, . . . , un) and v = (v1, . . . , vn) be elements of Rn . To write u ≤ v means that fork = 1, . . . , n , we have uk ≤ vk . If u ≤ v , then by definition,

(u, v] =n∏

k=1

(uk, vk].

If I ⊆ 1, . . . , n , then we’ll write u : I : v for the n-tuple w = (w1, . . . , wn) defined by

wk =

uk if k ∈ I,

vk if k /∈ I.

(This is not standard notation.) For instance, if I = Ø, then u : I : v = v , while if I = 1, . . . , n , thenu : I : v = u .

61


X246. Let H = (u, v] : u, v ∈ Rn and u ≤ v . As we know, H is a pre-ring on Rn . Let V be acommutative group.

(a) Consider any function F : Rn → V . Define τ : H → V by

τ((u, v]) =∑

I

(−1)|I|F (u : I : v) (34)

for all u, v ∈ Rn with u ≤ v , where the index I of summation ranges over all subsets of the set 1, . . . , n and where |I| denotes the number of elements in I . (You should explain why τ iswell-defined.) Prove that τ is additive on H .

(b) Conversely, consider any additive function τ : H → V . Prove that there exists a functionF : Rn → V such that for all u, v ∈ Rn with u ≤ v , (34) holds.

Remark. In the special case where V = R and where F : Rn → R is Cn , it is easy to show by inductionon n that for all u = (u1, . . . , un) ∈ Rn and all v = (v1, . . . , vn) ∈ Rn with u ≤ v , we have

∑

I

(−1)|I|F (u : I : v) =

∫ v1

u1

· · ·∫ vn

un

∂nF

∂xn · · · ∂x1dxn · · · dx1,


τ((u, v]) =

∫ v1

u1

· · ·∫ vn

un

∂nF

∂xn · · · ∂x1dxn · · · dx1.

Remark. Let us consider a particular case of the situation treated in problem X246(a). Let V = R andlet F : Rn → R satisfy

F (x1, . . . , xn) =n∏

k=1

Fk(xk)

for all x1, . . . , xn ∈ R , where F1, . . . , Fn : R → R . Then for all u = (u1, . . . , un) ∈ Rn and all v =(v1, . . . , vn) ∈ Rn with u ≤ v , we have

τ((u, v]) =∑

I

(−1)|I|F (u : I : v) =∑

I

n∏

k=1

(−1)1I (k)F ((u : I : v)k) =n∏

k=1

(

Fk(vk)− Fk(uk))

,

where we used the generalized binomial theorem in the last step. The most important special case is the onewhere

F (x1, . . . , x2) =n∏

k=1

xk

for all x1, . . . , xn ∈ R . In this case, τ((u, v]) =∏n

k=1(vk−uk), the n-dimensional measure of (u, v] . (Lengthis 1-dimensional measure, area is 2-dimensional measure, volume is 3-dimensional measure, and so on.)

Remark. In problem X246(b), F is called a distribution function for τ . Of course F is not unique. It isnot even unique up to a constant. Rather, one may add to it any function C from Rn to V which satisfies

∑

I

(−1)|I|C(u : I : v) = 0

for all u ∈ Rn with u ≤ v . When V = R and C : Rn → R is Cn , it can be shown that this condition onC is equivalent to the condition

∂nC

∂xn · · · ∂x1≡ 0.

X247. Let R be a ring of sets and let τ : R → [−∞,∞] . Suppose that for all disjoint A,B ∈ R , we haveτ(A ∪ B) = τ(A) + τ(B). (It is implicit in this assumption that for all such A and B , τ(A) + τ(B) isdefined, or in other words, τ(A) + τ(B) is not of the form ∞+ (−∞), nor is it of the form (−∞) +∞ .)

(a) Prove that for all A,B ∈ R , if A ⊆ B and τ(A) is finite, then τ(B \A) = τ(B) − τ(A).(b) Prove that for all A,B ∈ R , if A ⊆ B and τ(A) = ∞ , then τ(B) = ∞ .(c) Prove that for all A,B ∈ R , if A ⊆ B and τ(A) = −∞ , then τ(B) = −∞ .(d) Prove that τ takes on at most one of the values ∞ and −∞ .(e) For definiteness, assume that τ does not take on the value −∞ . Thus τ takes values in (−∞,∞] .

Of course under addition, (−∞,∞] is a commutative semigroup with 0. Suppose in addition thatτ(C) 6= ∞ for some C ∈ R . Prove that τ is additive.

62


Terminology. In the light of problem X247, we shall slightly extend the definition of additivity. Let Rbe a ring of sets and let τ : R → [−∞,∞] . To say that τ is additive will mean that τ takes values in(−∞,∞] or in [−∞,∞) and that τ is additive in the sense already defined for a set function taking valuesin a commutative semigroup with 0.

X248. Give an example of a pre-ring H and a function τ : H → [−∞,∞] such that for all disjointA,B ∈ H for which A ∪ B happens to belong to H , we have τ(A ∪ B) = τ(A) + τ(B), and yet τ takeson both of the values −∞ and ∞ . (Hint: There is an example where H is a finite set.)

X249. Let R be a ring of sets, let V be a commutative semigroup with 0, and let τ : R → V be additive.Prove that for all A,B ∈ R , we have

τ(A ∪B) + τ(A ∩B) = τ(A) + τ(B).

(Warning: Your proof must not involve subtraction, such as τ(C)− τ(D), where C,D ∈ R . Since V is onlya semigroup, τ(C)−τ(D) need not be defined. For instance, if V = [0,∞] under addition, then τ(C)−τ(D)could be ∞−∞ . You should be able to write your proof so that it only involves addition.)

X250. Let R be a ring of sets, let V be a commutative group, and let µ : R → V be additive. Then byproblem X249, for all A,B ∈ R , we have

µ(A ∪B) + µ(A ∩B) = µ(A) + µ(B). (35)

Now since we are assuming that V is a commutative group, it follows from (35) that for all A,B ∈ R , wehave

µ(A ∪B) = µ(A) + µ(B) − µ(A ∩B).

From this, it is a simple calculation to show that if A,B,C ∈ R , then

µ(A ∪B ∪ C) = µ(A) + µ(B) + µ(C)

− µ(A ∩B)− µ(A ∩ C)− µ(B ∩ C)+ µ(A ∩B ∩ C).

The corresponding formula when we consider n sets A1, . . . , An ∈ R is

µ(⋃n

i=1Ai

)

=∑

i

µ(Ai)−∑

i<j

µ(AiAj) +∑

i<j<k

µ(AiAjAk)−+ · · ·+ (−1)n+1µ(A1 · · ·An), (36)

where to save space, we have written AiAj for Ai ∩ Aj and so on. Equation (36) is called the inclusion-exclusion formula. It can be proved by induction on n . Do so.

Remark. Let us note that (36) can be written more succinctly as follows.

µ(⋃n

i=1 Ai

)

=∑

I 6=Ø

(−1)|I|+1 µ(⋂

i∈I

Ai

)

,

where the index I of summation ranges over all non-empty subsets of the set 1, . . . , n and where |I|denotes the number of elements in I . To see this, write Jm for the set of m-element subsets of 1, . . . , n ,and notice that the sum on the right in (36) is

n∑

m=1

(−1)m+1∑

I∈Jm

µ(⋂

i∈I

Ai

)

=

n∑

m=1

∑

I∈Jm

(−1)m+1 µ(⋂

i∈I

Ai

)

=

n∑

m=1

∑

I∈Jm

(−1)|I|+1 µ(⋂

i∈I

Ai

)

=∑

I 6=Ø

(−1)|I|+1 µ(⋂

i∈I

Ai

)

,

where the last step follows from the fact that each non-empty subset I ⊆ 1, . . . , n belongs to exactly oneof the sets J1, . . . ,Jn .

63


X251. Suppose n men put their hats in a sack and then each man draws a hat from the sack at random. due 2Th

(a) Determine the probability that at least one man gets his own hat back. (Hint: Let E = 1, . . . , n and let Ω be the set of one-to-one maps from E into E . Let F be the set of all subsets of Ω.Then F is a ring of sets on Ω. For each A ∈ F , let |A| denote the number of elements in A .Define P : F → [0, 1] by

P (A) =|A||Ω|

for each A ∈ F . Then P is additive on F . We seek P (B), where B =⋃n

i=1 Ai and Ai =ω ∈ Ω : ω(i) = i .)

(b) Find the limit that the probability in (a) approaches as n→ ∞ . (The number e will play a role.)

Definition. A topological semigroup is a semigroup V endowed with a topology which makes the semigroupoperation, which is a function from V × V into V , continuous.

Examples. [0,∞] under addition, and endowed with its usual topology, is a topological semigroup. So are(−∞,∞] , R , and Rn . All of these examples are commutative and have 0. The last two are groups.

Unordered sums. Let V be a Hausdorff topological commutative semigroup with 0. Let (yk)k∈K be afamily of elements of V . (The index set K is arbitrary. It could even be uncountable.) Let Λ be thecollection of finite subsets of K . For each λ ∈ K , let

Sλ =∑

k∈λ

yk.

Notice that Λ becomes a directed set when ordered by inclusion. This makes (Sλ)λ∈Λ into a net in V . Tosay that y is a sum for (yk)k∈K in V means that the net (Sλ) converges to y in V . (Explicitly, this meansthat for each nhd U of y in V , there exists a finite set λ0 ⊆ K such that for each finite set λ ⊆ K withλ0 ⊆ λ , we have

∑

k∈λ yk ∈ U .) Since V is Hausdorff, limits in V are unique, so there can be at most onesuch y . Hence if y is a sum for (yk)k∈K in V , then it makes sense to write

∑

k∈K yk = y and to call y thesum of the family (yk)k∈K in V . To say that (yk)k∈K is summable in V means that there exists y in Vsuch that y is a sum for (yk)k∈K in V .

X252. Let (yk)k∈K be a family in [0,∞] . As we’ve mentioned, under addition, [0,∞] is a Hausdorfftopological semigroup with 0. Let Λ be the collection of finite subsets of K . For each λ ∈ Λ, let Sλ =∑

k∈λ yk . Let y = sup Sλ : λ ∈ Λ . Prove that y =∑

k∈K yk in the sense discussed immediately above.

Remark. The next exercise is a sort of generalized associative law for unordered sums. While it is not thebest result of its type, it will be sufficient for the application we have in mind, which will be the extensionof a countably additive set function from a pre-ring to a ring.

X253. Let V be a Hausdorff topological commutative semigroup with 0. Let (yk)k∈K be a family in V .Suppose this family is summable in V and let y =

∑

k∈K yk . Let (Kℓ)ℓ∈L be a disjoint family of finitesubsets of K whose union is all of K . For each ℓ ∈ L , let zℓ =

∑

k∈Kℓyk . Prove that

∑

ℓ∈L zℓ = y too.

Reminder. If S is a set, then S<N denotes the set of finite sequences of elements of S , including the emptysequence, SN denotes the set of infinite sequences of elements of S (in other words, the set of functionsfrom N into S ), and S≤N denotes the set of finite or infinite sequences of elements of S (in other words,S<N ∪ SN ).

Reminder. Let H be a collection of sets, let V be a Hausdorff topological commutative semigroup with0, and let τ : H → V . To say that τ is countably additive, or that τ is σ-additive, means that for eachdisjoint (Hk) ∈ H ≤N , if

⋃

kHk ∈ H , then τ(⋃

kHk) =∑

k τ(Hk). If τ is countably additive, then τ isadditive, because H <N ⊆ H ≤N . In particular, if τ is countably additive and Ø ∈ H , then τ(Ø) = 0.

X254. Let H be a pre-ring and let R be the ring of sets generated by H . Let τ be an additive setfunction on H . As we’ve seen in class, τ has a unique extension to an additive set function σ on R . Nowsuppose in addition that τ takes values in a Hausdorff topological commutative semigroup with 0 and thatτ is countably additive. Prove that σ is countably additive.

64


X255. Let R be a ring of sets, let V be a Hausdorff commutative topological semigroup with 0, Letτ : R → V . Prove that the following are equivalent.

(a) τ is countably additive on R .(b) τ is additive on R and for each increasing sequence (Rn) in R , if R =

⋃

nRn ∈ R , thenτ(Rn) → τ(R) as n→ ∞ .

X256. Let H be a pre-ring and let τ : H → [0,∞] . Prove that if τ is countably additive, then τ iscountably subadditive. (Hint: Use problem X254 and mimic the proof, which we did in class, that if τ isadditive, then τ is subadditive.)

X257. Let R be a ring of sets, let V be a Hausdorff topological commutative group, and let τ : R → V .(a) Prove that if τ is countably additive, if (Rn) is a decreasing sequence in R , and if R =

⋂

Rn ∈ R ,then τ(Rn) → τ(R) as n→ ∞ .

(b) Prove that if τ is additive and if for each decreasing sequence (Rn) in R , with⋂

nRn = Ø, wehave τ(Rn) → 0 as n→ ∞ , then τ is countably additive.

Example. Let R be the collection of all subsets of N . Of course R is a ring of subsets of N . Defineτ : R → [0,∞] as follows: For each A ⊆ N , let τ(A) be the number of elements in A , if A is a finite set,and let τ(A) = ∞ if A is an infinite set. (By the way, τ is called counting measure on N .) It is easyto check that τ is countably additive. But τ does not satisfy the conclusion of problem X257(a). For letRn = n, n+ 1, n+ 2, . . . for each n ∈ N and let R =

⋂

nRn . Then (Rn) is a decreasing sequence ofelements of R but τ(Rn) does not tend to τ(R) as n → ∞ , because τ(Rn) = ∞ for each n ∈ N butR = Ø so τ(R) = 0. This does not contradict problem X257(a), because [0,∞] is just a semigroup underaddition, not a group.

Remark. Let R be a ring of sets and let τ : R → [−∞,∞] . Suppose that τ is additive. Then by problemX247, τ takes on at most one of the values ∞ and −∞ , so either τ takes values in (−∞,∞] or τ takesvalues in [−∞,∞). Hence it makes sense to talk about whether or not τ is countably additive. Suppose τis countably additive. Let (Rn) be a decreasing sequence in R such that −∞ < τ(R1) <∞ . Let R =

⋂

Rn

and suppose R ∈ R . Then τ(Rn) → τ(R) as n→ ∞ . To see this, consider S = S ∈ R : S ⊆ R1 . ThenS itself is a ring of sets. Also, R ∈ S and each Rn ∈ S . Let σ be the restriction of τ to S . By parts(b) and (c) of problem X247, σ : S → R . Of course σ is countably additive. Since R is a group underaddition, it follows from problem X257(a) that σ(Rn) → σ(R) as n→ ∞ . But this says that τ(Rn) → τ(R)as n→ ∞ .

Remark. Length is countably additive on the collection of bounded intervals in R . The example in thenext exercise shows that the analogous statement with R replaced by Q is false. This example also showsthat for problem X257(b), it matters that R is a ring. A pre-ring would not do.

X258. Let X be a totally ordered set.37 For all u, v ∈ X with u ≤ v , let (u, v] = x ∈ X : u < x ≤ v . due 2Th

Let H = (u, v] : u, v ∈ X, u ≤ v .(a) Prove that H is a pre-ring on X .(b) Let A,B ∈ H . By the definition of H , A = (s, t] and B = (u, v] for some s, t, u, v ∈ X with

s ≤ t and u ≤ v . Suppose that Ø 6= A ⊆ B . Prove that u ≤ s < t ≤ v . In particular, ifØ 6= A = B , then s = u and t = v .

Now let Γ be a commutative group, let F : X → Γ, and define τ : H → Γ by τ((u, v]) = F (v) − F (u) forall u, v ∈ X with u ≤ v .

(c) Prove that τ is additive on H . (Remember to check that τ is well-defined.)For the remainder of this exercise, assume that X = Q , the set of rational numbers, that Γ is R underaddition, and that F (x) = x for all x ∈ X .

(d) Prove that τ is not countably subadditive on H .(e) Prove that τ is not countably additive on H .(f) Prove that for each decreasing sequence (Hn) in H , if

⋂

nHn = Ø, then τ(Hn) → 0 as n→ ∞ .

37 To say that X is a totally ordered set means that X is a partially ordered set in which any two elements are comparable.If X is a partially ordered set and u, v ∈ X , then to say that u and v are comparable means that u ≤ v or v ≤ u .

65


About Measurability.

If X is a set and µ is the outer measure on X generated by a function τ : H → [0,∞] , where H is a setof subsets of X , we would like to identify a collection E of subsets of X such that H ⊆ E , µ is countablyadditive on E , and E is as large as possible subject to the condition that it have reasonable properties as aset of sets. The next exercise shows that in a quite satisfying sense, Mµ is a good choice for E .

X259. Let H be a set of subsets of a set X , let τ : H → [0,∞] , and let µ be the outer measure on Xgenerated by τ . Let A be a set of subsets of X . Since µ is countably subadditive on P(X), µ will becountably additive on A iff µ is superadditive on A . Suppose A is a pre-ring on X such that H ⊆ Aand µ is countably additive on A . Prove that A ⊆ Mµ . Thus Mµ is the largest pre-ring on X containingH on which µ is countably additive.

X260. Let X be a set and let µ : P(X) → [0,∞] . For each T ⊆ X , define µT : P(X) → [0,∞] by

µT (E) = µ(E ∩ T )

for all E ⊆ X .(a) Let E, S, T ⊆ X and let ν = µT . Prove that E ν-splits S iff E µ-splits S ∩ T .(b) Let T ⊆ X and let ν = µT . Prove that Mµ ⊆ Mν .(c) Suppose µ is subadditive. Let N = µT : T ⊆ X and µ(T ) <∞ . Prove that

Mµ =⋂ Mν : ν ∈ N .

X261. Let X be a set and let µ : P(X) → [0,∞] . due 3Th

(a) Suppose µ is subadditive, E ⊆ X , and µ(E) + µ(X \E) = µ(X) <∞ . Let F ∈ Mµ . Prove thatE µ-splits F . (Suggestion: Draw a Venn diagram.)

(b) Suppose µ is subadditive, E ⊆ H ∈ Mµ , and µ(E)+µ(H \E) = µ(H) <∞ . Let F ∈ Mµ . Provethat E µ-splits F . (Hint: For part of the proof, let ν = µH , apply part (a) with µ replaced byν , and apply parts (a) and (b) of problem X260.)

(c) Let H be a pre-ring on X and let τ : H → [0,∞] be countably additive. Suppose µ is the outermeasure on X generated by τ . Let H ∈ H with τ(H) <∞ and let E ⊆ H . Prove that E ∈ Mµ

iff µ(E) + µ(H \ E) = τ(H).

Remark. Intuitively, in problem X261(c), µ(E) is an overestimate for what τ(E) ought to be and µ(H \E)is an overestimate for what τ(H \ E) ought to be. (We say “ought to be” because E and H \ E need notbelong to the domain of τ , namely H , and because our goal is to extend τ in as natural a way aspossible to a countably additive function on a collection of sets larger than the starting collection H .) Ifµ(E) +µ(H \E) = τ(H) <∞ , then intuitively, these two “overestimates” are not really overestimates afterall, but are actually exactly right.

Outer Content.

Paralleling the theory of measure is the theory of content. The theory of measure came later than the theoryof content and is superior to it. In this sense, the theory of content is obsolete. However, by a little study ofthe theory of content, and by contrasting it with the theory of measure, you can improve your understandingof the theory of measure. For now, let’s compare and contrast outer measure and outer content. Soon we’llalso introduce, and compare and contrast, inner measure and inner content.

Let X be a set, let τ : H ⊆ P(X) → [0,∞] , and define γ : P(X) → [0,∞] by

γ(A) = inf

∑

n

τ(Hn) : A ⊆ ⋃nHn and (Hn) ∈ H <N

.

We call γ the outer content generated by τ . Note that the definition of outer content involves only finitecoverings of A , whereas the definition of outer measure involves countable coverings of A . Clearly if µ isthe outer measure on X generated by τ , then µ ≤ γ .

X262. Let H = (a, b] : −∞ < a ≤ b <∞ and define τ : H → [0,∞) by τ(

(a, b])

= b−a for −∞ < a ≤b < ∞ . Let γ be the outer content on R generated by τ and let µ be the outer measure on R generatedby τ . (Of course, µ is Lebesgue outer measure on R .) It is obvious from the definitions that µ ≤ γ .

(a) Let S = x ∈ [0, 1] : x is rational . Prove that γ(S) = 1 and µ(S) = 0.(b) Let K be a compact subset of R . Prove that γ(K) = µ(K).

66


In the next few exercises, you are asked to establish some properties of outer content which are analogousto properties of outer measure that we proved in class. You may find this to be a good way to review theproofs of those properties of outer measure.

X263. Let X be a set, let τ : H ⊆ P(X) → [0,∞] , and let γ be the outer content generated by τ . Provethe following statements.

(a) γ(H) ≤ τ(H) for all H ∈ H .

(b) γ is subadditive on P(X).

(c) γ is an extension of τ iff τ is subadditive on H .

Let X be a set. To say that γ is an outer content on X means that γ : P(X) → [0,∞] and γ issubadditive on P(X). By problem X263, if τ : H ⊆ P(X) → [0,∞] and γ is the outer content on Xgenerated by τ , then γ is indeed an outer content on X .

Let X be a set. Recall that for any γ : P(X) → [0,∞] , if γ(Ø) = 0, then Mγ is a field of sets on Xand γ is additive on Mγ . In particular, if γ is an outer content on X , then Mγ is a field of sets on X andγ is additive on Mγ . Also, if γ is an outer content on X , and Nγ = A ⊆ X : γ(A) = 0 , then, with thehelp of the finite subadditivity of γ , it is easy to see that Nγ ⊆ Mγ .

X264. Let X be a set, let τ : H ⊆ P(X) → [0,∞] , and let γ be the outer content on X generated by τ .Let E ⊆ X . Prove that the following are equivalent.

(a) E ∈ Mγ .

(b) For each H ∈ H , E γ-splits H .

(c) For each H ∈ H , γ(E ∩H) + γ(Ec ∩H) ≤ τ(H).

X265. Let X be a set, let H be a pre-ring on X , let τ : H → [0,∞] be superadditive, and let γ be theouter content on X generated by τ . Prove that H ⊆ Mγ .

This completes the group of exercises on the properties of outer content that are analogous to theproperties of outer measure that we proved in class. The next result is analogous to problem X259.

X266. Let H be a set of subsets of a set X , let τ : H → [0,∞] , and let γ be the outer content on Xgenerated by τ . Let A be a set of subsets of X . Since γ is subadditive on P(X), γ will be additive onA iff γ is superadditive on A . Suppose A is a pre-ring on X such that H ⊆ A and γ is additive on A .Prove that A ⊆ Mγ . Thus Mγ is the largest pre-ring on X containing H on which γ is additive.

Note that parts (a) and (b) of problem X261 apply to any function µ : P(X) → [0,∞] , so in particularthey apply if µ is an outer content. Part (c) of problem X261 applies to outer measures. Now here is theanalog of that for outer contents.

X267. Let X be a set, let H be a pre-ring on X , and let τ : H → [0,∞] be additive. Suppose µ is theouter content on X generated by τ . Let H ∈ H with τ(H) <∞ and let E ⊆ H . Prove that E ∈ Mµ iffµ(E) + µ(H \ E) = τ(H).

Adapted Outer Measures and Adapted Outer Contents.

Let X be a set and let µ : P(X) → [0,∞] . To say that µ is adapted38 means that µ(Ø) = 0 and for eachA ⊆ X , µ(A) = inf µ(B) : A ⊆ B ∈ Mµ .

X268. Let H be a pre-ring on a set X , let τ : H → [0,∞] be superadditive, let γ be the outer content onX generated by τ , and let µ be the outer measure on X generated by τ . Prove that γ and µ are adapted.

Let X be a set and let µ : P(X) → [0,∞] . Let us write µ• for the function from P(X) → [0,∞]defined by

µ•(A) = inf µ(B) : A ⊆ B ∈ Mµ

for all A ⊆ X . Clearly µ is adapted iff µ(Ø) = 0 and µ = µ• .

38 This is not a standard use of the term adapted , so if you employ it outside this course, you should explain what you meanby it. Some authors do use the word regular in this context to mean the same thing that we mean by adapted here. However,the adjective regular is probably the most overworked word in mathematics. Even in measure theory, it is used in severaldifferent ways. You should always check its meaning when you see it used, and you should always explain what you mean byit when you use it yourself.

67


X269. Let X be a set, let µ : P(X) → [0,∞] with µ(Ø) = 0, and let τ be the restriction of µ to Mµ . Aswe know, Mµ is a field of sets on X and τ is additive.

(a) Prove that µ• is the outer content on X generated by τ .(b) Deduce that in particular, if µ is adapted, then µ is subadditive on P(X).(c) Prove that Mµ ⊆ Mµ• .(d) Suppose in addition that µ is subadditive and µ(X) <∞ . Prove that Mµ = Mµ• .

X270. Let X be a set and let µ : P(X) → [0,∞] be adapted. Let E ⊆ X and suppose that E µ-splitseach element of Mµ . Prove that E ∈ Mµ .

X271. Let X be a set, let µ be an outer measure on X , and let τ be the restriction of µ to Mµ . Provethat µ is adapted iff µ is the outer measure on X generated by τ .

X272. Let X be a set and let µ be an adapted outer measure on X . Let (An) be an increasing sequence due 4M

of subsets of X and let A =⋃∞

n=1An . Prove that µ(An) ↑ µ(A) as n→ ∞ . (Neither A nor the An’s needbe measurable.)

Inner Measure and Inner Content.

Let X be a set and let µ : P(X) → [0,∞] . Let us write µ• for the function from P(X) to [0,∞] definedby

µ•(A) = sup µ(B) : B ∈ Mµ and B ⊆ A .

If µ is the outer measure (respectively, the outer content) generated by a function τ : H ⊆ P(X) → [0,∞] ,then let us call µ• the inner measure (respectively, the inner content) on X generated by τ .

X273. Let X be a set, let H be a pre-ring on X , let τ : H → [0,∞] be additive, let γ be the outercontent generated by τ , and let γ• be the inner content generated by τ . Let A ⊆ X such that γ(A) <∞ .Prove that γ•(A) is the supremum of sums of the form

∑

n τ(Hn) where (Hn) varies over finite disjointsequences of elements of H with

⋃

nHn ⊆ A .

Note that problem X273 does not have an analog for inner measure. This is not a defect of innermeasure. Rather, the result of problem X273 is a defect of inner content. For instance, if A is the set ofirrational numbers in [0, 1], then the Lebesgue inner measure of A is 1, whereas if (Hi) is any disjointfamily of intervals which are contained in A , then

∑

i length(Hi) = 0 because each Hi is either empty or asingleton.

By the way, originally, the result of problem X273 was taken as the definition of inner content. We havechosen instead to define inner content in a way that permits a unified treatment of inner measure and innercontent.

X274. Let X be a set and let µ : P(X) → [0,∞] be adapted.(a) Let A ⊆ H where H ∈ Mµ . Prove that µ•(A) + µ(H \A) = µ(H).(b) Let A ⊆ X with µ(A) <∞ . Prove that A ∈ Mµ iff µ(A) = µ•(A).

X275. Let H be a pre-ring on a set X and let τ : H → [0,∞] . Suppose either that τ is countablyadditive and µ is the outer measure on X generated by τ , or that τ is additive and µ is the outer contenton X generated by τ .

(a) Let A ⊆ X with µ(A) <∞ . Prove that A ∈ Mµ iff µ•(A) = µ(A).(b) Let B ⊆ X . Prove that B ∈ Mµ iff for each A ∈ Mµ with µ(A) <∞ , we have B ∩A ∈ Mµ .

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Reminder. Let X be a set, let Γ be a non-empty collection of σ-fields on X , and let F =⋂

Γ. In otherwords, let

F = F : F ∈ G for each G ∈ Γ .

Then F is a σ-field on X .

Reminder. Let X be a set and let H be a set of subsets of X . Then there is a smallest σ-field F onX containing H . In fact, F =

⋂

Γ where Γ is the set of all σ-fields A on X such that H ⊆ A . (Notethat Γ is non-empty, because P(X) ∈ Γ.) By the way, it is clear that there can be at most one such F ,because if F1 and F2 are two such “smallest” elements of Γ, then F1 ⊆ F2 and F2 ⊆ F1 .) We call Fthe σ-field on X generated by H . We denote F by σ(H ). (Warning: σ(H ) depends on X as well as onH . If H is a set of subsets of X and X is a proper subset of X ′ , then the σ-field on X ′ generated by His different from the σ-field on X generated by H . For instance, X ′ belongs to the former σ-field but notto the latter. Thus the notation σ(H ) is context-dependent. Normally this does not cause confusion.)

Reminder. Let X be a set and let G and H be sets of subsets of X . Then σ(G ) ⊆ σ(H ) iff G ⊆ σ(H ).It follows that σ(G ) = σ(H ) iff G ⊆ σ(H ) and H ⊆ σ(G ).

Reminder. Let X be a topological space, let G be the set of open subsets of X , and let B = σ(G ). ThenB is called the Borel σ-field on X . To say that E is a Borel subset of X means that E ∈ B .

Reminder. Let B be the Borel σ-field on R . Then each interval belongs to B . For instance, if a is a realnumber, then although [a,∞) is not open, it is a countable intersection of open sets, because it is equal to⋂∞

n=1(a− n−1,∞), so it is a Borel set.

X276. Let B be the Borel σ-field on R .(a) Let H = (a,∞) : a ∈ R . Prove that σ(H ) = B . (Hint: Apply one of the reminders above.)

(b) Let I be a set of intervals. Suppose that for each a ∈ R , there exists a sequence (In) in Isuch that (a,∞) =

⋃∞n=1 In . Prove that σ(I ) = B . (Similarly, if for each b ∈ R , there exists a

sequence (In) in I such that (−∞, b) =⋃∞

n=1 In , then σ(I ) = B , although you are not askedto prove this.)

Example. Let X be a dense subset of R and let B be the Borel σ-field on R . If

I = (a, b] : a, b ∈ X and a < b ,

then by problem X276(b), σ(I ) = B . Similarly, if

I = [a, a+ 2006) : a ∈ X ,

then σ(I ) = B .

Reminder. Let B be the Borel σ-field on R . Let F : R → R be increasing and right-continuous. Let µbe the Lebesgue-Stieltjes outer measure on R with distribution function F . Then B ⊆ Mµ . In particular,if we take F (x) ≡ x , we see that each Borel subset of R is Lebesgue measurable.

Remark. Let M be the σ-field of Lebesgue measurable subsets of R and let B be the Borel σ-field onR . As we have just observed, B ⊆ M . So it is natural to ask whether B = M . The answer is no. In fact,it can be shown that the cardinality of B is the same as the cardinality of R and it is easy to show thatthe cardinality of M is the same as the cardinality of the power set of R , so that the cardinality of B isstrictly less than the cardinality of M .

Now here is why the cardinality of M is the same as the cardinality of the power set of R . Let C bethe ordinary Cantor set (not a fat Cantor set). Then C has Lebesgue measure zero. Hence each subset ofC is Lebesgue measurable. In other words, P(C) ⊆ M . Of course, M ⊆ P(R). So to show that M hasthe same cardinality as P(R), it suffices39 to show that P(C) has the same cardinality as P(R). Butthis follows immediately from the fact that the cardinality of C is the same as the cardinality of R .

39 by the Schroeder-Bernstein theorem, which states that if each of two given sets has the same cardinality as a subset ofthe other, then the two given sets have equal cardinality.

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Remark. Let M be the σ-field of Lebesgue measurable subsets of R . Since the cardinality of M is thesame as the cardinality of the set of all subsets of R , it is natural to ask whether M is actually equal to theset of all subsets of R . Using the axiom of choice, one can show that the answer is no. There are subsets ofR that are not Lebesgue measurable. We now turn to a proof of this.

Notation and Terminology. For each A ⊆ R and each x ∈ R , let us write A + x for a+ x : a ∈ A .We call A + x the translate of A by x . Let µ be an outer measure on R . To say that µ is translation-invariant means that for each A ⊆ R and each x ∈ R , µ(A+x) = µ(A). Clearly Lebesgue outer measure istranslation-invariant. The next exercise tells in particular that not every subset of R is Lebesgue measurable.

X277. Let µ be a translation-invariant outer measure on R . Suppose in addition that µ((0, 1]) = 1. Provethat there exists E ⊆ R such that E /∈ Mµ . (Hint: If x, y ∈ R and x− y is rational, then Q+ x = Q+ y .If x, y ∈ R and x− y is irrational, then Q+ x and Q+ y are disjoint. Let Π = (Q+ x) ∩ (0, 1] : x ∈ R .Then Π is a set of non-empty subsets of R . By the axiom of choice, there exists a family (xP )P∈Π of realnumbers such that for each P ∈ Π, xP ∈ P . Let E = xP : P ∈ Π . Check that R =

⋃

E + r : r ∈ Q .Deduce that µ(E) > 0. Check that if r, s ∈ Q with r 6= s , then E + r and E + s are disjoint. Deduce that(0, 2] contains infinitely many disjoint translates of E . From this, deduce that if E were µ-measurable, thenµ(E) would have to be zero.)

Remark. One might have hoped that the problem of defining the length of an arbitrary subset of R couldbe solved by producing a function µ : P(R) → [0,∞] such that µ is countably additive, µ is translation-invariant, and for each bounded interval I , µ(I) is the length of I . But from problem X277, it follows thatno such function exists.

Remark. The construction in problem X277 is due to Vitaly (1905) and was one of the early, strikingapplications of the full axiom of choice. The first application of the full axiom of choice was Zermelo’stheorem (1904) that every set can be well-ordered. The full axiom of choice was introduced by Zermelo forthe express purpose of proving this well-ordering theorem.

Remark. Let R be the ring of all bounded subsets of Rd . It can be shown that there does exist anadditive (not countably additive), translation-invariant function µ : R → [0,∞) such that µ((0, 1]d) = 1.When d ≥ 2, it is natural to ask whether we can also require that µ be invariant under rotations. It canbe shown that this is possible when d = 2. But when d ≥ 3, no such function that is also invariant underrotations exists. The reason is that Banach and Tarski (1924), building on work of Hausdorff (1914), showedthat if A and B are bounded subsets of R3 having non-empty interior, then it is possible to partitionA and B into finitely many pieces A1, . . . , An and B1, . . . , Bn respectively, in such a way that for eachk ∈ 1, . . . , n , there is a proper rigid motion of R3 that maps Ak onto Bk . In colloquial terms, A can bedisassembled into finitely many pieces which can be reassembled to form B . For instance, a ball of radiusone can be cut up into finitely many pieces which can be reassembled to form two disjoint balls of radiusone. This stunning result is known as the Banach-Tarski paradox. The reason why it is not contradictory isthat the pieces are so bizarre that it is impossible to define their volume. The axiom of choice is used in theproof of the Banach-Tarski paradox. It is interesting to note that the way it is used is to show that one maychoose one point from each coset of a suitable subgroup of the group of rotations of R3 , just as Vitaly’sconstruction of a subset of R that is not Lebesgue-measurable depended on choosing one point from eachcoset of Q , considered as a subgroup of the group (R,+).

Some Counterexamples to Uniqueness of Measures.

Reminder. A measurable space is an ordered pair (X,A ) such that X is a set and A is a σ-field on X .

Reminder. Let (X,A ) be a measurable space. To say that µ is a measure on A means that µ : A →[0,∞] and µ is countably additive on A .

Reminder. A measure space is an ordered triple (X,A , µ) such that X is a set, A is a σ-field on X , andµ is a measure on A . (In French, this is called “une espace mesuree,” which may be literally translatedas “a measured space.” Thus, from the French point of view, when we put a measure µ on a measurablespace (X,A ), then we have “measured” it and it becomes a “measured space.” Unfortunately, in Englishthe terminology is slightly less logical.)

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Reminder. Let H be a pre-ring on a set X , let τ : H → [0,∞] be countably additive, let τ∗ be the outermeasure on X generated by τ , let A be the σ-field on X generated by H , and let µ be the restriction ofτ∗ to A . Suppose X can be covered by a sequence (Hk) of elements of H with τ(Hk) < ∞ for each k .Then µ is the unique extension of τ to a measure on A .

Definition. Let (X,A ) be a measurable space. To say that µ is a probability measure on A means thatµ is a measure on A and µ(X) = 1.

X278.(a) Give an example of a measurable space (X,A ), a subset H ⊆ A with A = σ(H ), and proba-

bility measures µ and ν on A , such that µ(H) = ν(H) for all H ∈ H but µ 6= ν . (Hint: Thereis an example where X is a small finite set.)

(b) Explain why the example you found in part (a) does not conflict with what we learned in classand recalled above about uniqueness of measures.

X279. Let X be an uncountable set, let H =

x : x ∈ X

∪Ø , let A be the σ-field on X generatedby H , and let τ be the function on H which is identically zero. Clearly H is a pre-ring on X and τ iscountably additive on H .

(a) Explain exactly which sets belong to A .(b) Let α ∈ [0,∞] . Prove that there is a unique measure µα on A such that µα is an extension of

τ and µα(X) = α .(c) If α, β ∈ [0,∞] with α 6= β , then µα and µβ are different measures on A which agree on

H . Explain why this does not conflict with what we learned in class and recalled above aboutuniqueness of measures.

(d) Let µ be the restriction to A of the outer measure on X generated by τ . Find α ∈ [0,∞] suchthat µ = µα . Justify your answer.

Definition. Let (X,A , µ) be a measure space. To say that (X,A , µ) is σ-finite, or that µ is σ-finite,means that there exists a sequence (Ak) of elements of A such that X =

⋃∞k=1 Ak and for each k ,

µ(Ak) <∞ .

X280. Let H = (a, b] : −∞ < a ≤ b <∞ and let A be the σ-field on R generated by H . As weknow, H is a pre-ring on R and A is the Borel σ-field on R . Define µ : A → [0,∞] by µ(Ø) = 0 andµ(A) = ∞ if A 6= Ø. Let τ be the restriction of µ to H . It is easy to see that µ is a measure on A whichis not σ-finite, and obviously µ is an extension of τ .

(a) Define ν : A → [0,∞] by letting ν(A) be the number of elements in A∩Q , if this is finite, and byletting ν(A) = ∞ otherwise. Prove that ν is a σ-finite measure on A and that ν is an extensionof τ .

(b) The measures µ and ν are different extensions of τ . Explain why this does not conflict with whatwe learned in class and recalled above about uniqueness of measures.

More About Measurability.

We have seen that there are far more Lebesgue measurable subsets of R than Borel subsets of R . However,it turns out that given any Lebesgue measurable set E ⊆ R , there exist Borel sets A,B ⊆ R such thatA ⊆ E ⊆ B and B \ A has Lebesgue measure zero. (This follows from the next exercise.) Thus eachLebesgue measurable subset of R differs from some Borel subset of R only by a set of Lebesgue measurezero.

X281. Let X be a set, let H be a pre-ring on X , let τ : H → [0,∞] be countably additive, and let µ due 4Th

be the outer measure on X generated by τ . Then we know that µ is an extension of τ , that H ⊆ Mµ ,that µ is countably additive on Mµ , and that Mµ is a σ-field on X . Let B = σ(H ). Since H ⊆ Mµ

and Mµ is a σ-field on X , B ⊆ Mµ .(a) Let E ⊆ X . Prove that there exists C ∈ B such that E ⊆ C and µ(E) = µ(C). (Comment: Of

course, this is only interesting when µ(E) <∞ . When µ(E) = ∞ , we can just take C = X .)(b) Let E ⊆ C ∈ Mµ with µ(E) = µ(C) < ∞ . Prove that E ∈ Mµ iff µ(C \ E) = 0. (Hint: Recall

that since µ is subadditive, Nµ ⊆ Mµ , where Nµ is the set of all N ⊆ X such that µ(N) = 0.)(c) Suppose in addition that µ is σ-finite. Let E ⊆ X . Prove that E ∈ Mµ iff there exist A,B ∈ B

such that A ⊆ E ⊆ B and µ(B \A) = 0. (Hint: The reverse implication is essentially trivial. Toprove the forward implication, first consider the case where µ(E) <∞ .)

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Approximating Measurable Sets by Simpler Sets.

Recall that if A and B are sets, then the symmetric difference of A and B is

A∆B = (A \B) ∪ (B \A).

It is worth noticing that A∆ B = 1A 6= 1B , where 1A is the function that is 1 on A and 0 elsewhereand 1B is the function that is 1 on B and 0 elsewhere.

X282. Let X be a set and let µ : P(X) → [0,∞] be subadditive. Let A ⊆ X and suppose that for eachε > 0, there exists E ∈ Mµ such that µ(A∆ E) < ε . Prove that A ∈ Mµ .

X283. Let X be a set, let H be a pre-ring on X , let τ : H → [0,∞) be countably additive (note that τdoes not assume the value ∞), and let µ be the outer measure on X generated by τ . Let R be the set offinite disjoint unions of elements of H . (As we know, R is the smallest ring of sets on X containing H .)Let A ⊆ X . Prove that the following are equivalent:

(a) A ∈ Mµ and µ(A) <∞ .(b) For each ε > 0, there exists E ∈ R such that µ(A∆E) < ε .

Example. Let F : R → R be increasing and right continuous. Let µ be the Lebesgue-Stieltjes outermeasure on R with distribution function F . Let R be the set of finite disjoint unions of intervals of theform (a, b] where −∞ < a ≤ b <∞ . Let A ⊆ R . Then the following are equivalent:

(a) A ∈ Mµ and µ(A) <∞ .(b) For each ε > 0, there exists R ∈ R such that µ(A∆R) < ε .

To see this, apply problem X283 with H = (a, b] : −∞ < a ≤ b <∞ and τ((a, b]) = F (b)− F (a) for alla, b ∈ R with a < b .

X284. Let X be a set and let µ : P(X) → [0,∞] be subadditive. Let F = A ⊆ X : µ(A) <∞ . Defineρ : F ×F → [0,∞) by ρ(A,B) = µ(A∆B). Prove that ρ is a pseudometric on F . (In other words, provethat ρ has all the properties of a metric on F except that if ρ(A,B) = 0, it need not follow that A = B .)

Remark. Let X , H , τ , µ , and R be as in problem X283. Let F = A ⊆ X : µ(A) <∞ and letE = F ∩ Mµ . Let ρ be as in problem X284. In terms of the result of problem X284, problem X283 hasthe following interpretation: E is the closure of R with respect to the topology induced on F by thepseudometric ρ .

Subadditivity and Superadditivity Revisited.

With the help of the integral for non-negative simple functions, we can give an enlightening alternative proofthat for a non-negative set function on a field of sets, additivity implies subadditivity and superadditivity.

X285. Let A be a field of subsets of a set X and let µ : A → [0,∞] be additive.(a) Let A ∈ A and let (Bk) be a finite sequence of elements of A such that A ⊆

⋃

kBk . Use thetheory of integrals of non-negative simple functions to prove that µ(A) ≤ ∑k µ(Bk). (Hint: LetN =

∑

k 1Bk. Notice that for each x ∈ X , N(x) is equal to the number of values of k for which

x ∈ Bk . In particular, 1A ≤ N .)(b) Let (Ak) be a finite disjoint sequence of elements of A and let B ∈ A such that

⋃

k Ak ⊆ B .Use the theory of integrals of non-negative simple functions to prove that

∑

k µ(Ak) ≤ µ(B).

The Inclusion-Exclusion Formula Revisited.

Recall from problem X250 that if X is a set, R is a ring of sets on X , V is a commutative group, µ : R → Vis additive, and A1, . . . , An ∈ R , then the inclusion-exclusion formula states that

µ(⋃n

i=1Ai

)

=∑

i

µ(Ai)−∑

i<j

µ(AiAj) +∑

i<j<k

µ(AiAjAk)−+ · · ·+ (−1)n+1µ(A1 · · ·An),

where to save space, we have written AiAj for Ai ∩ Aj and so on. By the remark which follows problemX250, the inclusion-exclusion formula may be written more succinctly as follows:

µ(⋃n

i=1 Ai

)

=∑

I 6=Ø

(−1)|I|+1 µ(⋂

i∈I

Ai

)

, (37)

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where the index I of summation ranges over all non-empty subsets of the set 1, . . . , n and where |I|denotes the number of elements in I . My aim here is to lead you to a slick proof of the inclusion-exclusionformula in the form (37). As a small first step, note that it suffices to treat the case where R is a field ofsets on X , because we are free to replace X by X ′ =

⋃nk=1 Ak , R by A = R ∈ R : R ⊆ X ′ (which is a

field of sets on X ′ ), and µ by the restriction of µ to A .In class, we have treated integrals of [0,∞)-valued simple functions with respect to an additive [0,∞]-

valued function µ on a field A . In the same way, we may treat integrals of integer-valued simple functionswith respect to an additive function µ on a field A taking values in a commutative group, because just aswe can multiply elements of [0,∞] by elements of [0,∞), we can multiply elements of a commutative groupby integers. You may take this for granted in the next exercise.

X286. Let X be a set, let A be a field of sets on X , let V be a commutative group, and let µ : A → V beadditive. Use the result of problem X245 to give a slick proof of the generalized inclusion-exclusion formula.(Hint: Given n ∈ N and sets A1, . . . , An ∈ A , let B =

⋃ni=1 Ai . Verify that

1B = 1− 1Bc = 1− 1∩ni=1

Aci= 1−

n∏

i=1

1Aci= 1−

n∏

i=1

(

1− 1Ai

)

.

Use the result of problem X245 to expand the last product. In the expression that you thereby obtain for1B , integrate over X with respect to µ . You should get the inclusion-exclusion formula in the form (37).(Notice that the functions you are integrating are all integer-valued, so it makes sense to integrate them withrespect to µ .)

Remark. Perhaps this is a good place to mention a further generalization of the theory of integrals of simplefunctions. Let U , V , and W be commutative semigroups with 0. Suppose we are given an operation of“multiplication” (u, v) → uv from U × V to W , such that for all u, u′ ∈ U and all v, v′ ∈ V , we have(u+u′)v = uv+u′v , u ·0 = 0, u(v+v′) = uv+uv′ , and 0 ·v = 0. Let X be a set, let A be a field of subsetsof X , and let µ : A → V be additive. Then we may develop a theory of integrals of U -valued A -simplefunctions with respect to µ . If ϕ : X → U is A -simple, then by definition, its integral with respect to µ is

I(ϕ) =∑

u

uµ(ϕ = u).

The proofs of the basic properties of this integral that we gave in the special case where U = [0,∞) andV = [0,∞] all go through in this more general situation with essentially no change. We have alreadymentioned the case where U = Z under addition, V is a commutative group, and W = V . Here are someother cases:

(a) U = ω , the set of non-negative integers under addition, V is any commutative semigroup with 0,and W = V .

(b) U = V =W = R under addition.(c) U is a field, V is a vector space over U , and W = V . (Integral of a scalar-valued simple function

with respect to a vector-valued µ .)(d) U is a vector space over a field V and W = V . (Integral of a vector-valued simple function with

respect to a scalar-valued µ .)(e) U is an inner product space over K , V = U , W = K , and uv = 〈u|v〉 , the inner product of u

and v , for all u, v ∈ U .(f) U = V =W = R3 and uv = u× v , the cross product of v and w , for all u, v ∈ R3 .

More About Measurable Functions.

X287. Let (X,A ) be a measurable space. Let f, g : X → R be measurable. Prove that f < g ∈ A .(Hint: For each x ∈ X , we have f(x) < g(x) iff there exists a rational number r such that f(x) < r andr < g(x).)

X288. Let (X,A ) be a measurable space. Let f, g : X → R be measurable. Prove that f = g ∈ A .

X289. Let (X,A ), (Y,B), and (Z,C ) be measurable spaces. Let f : X → Y and g : Y → Z be measur-able. Let h = g f . Prove that h : X → Z is measurable.

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X290. Let (X,A ) be a measurable space and let Y be a metric space. Let f : X → Y . Prove that f ismeasurable iff for each continuous function g : Y → R , gf is measurable. (Hint for the reverse implication:For each non-empty set C ⊆ Y , the function y 7→ d(y, C) = inf d(y, y′) : y′ ∈ C is continuous and if C isclosed, then d(y, C) = 0 iff y ∈ C .)

X291. Let (X,A ) be a measurable space and let Y be a metric space. Let (fn) be a sequence of measurablefunctions from X to Y . Let f : X → Y and suppose that for each x ∈ X , we have fn(x) → f(x) as n→ ∞ .Prove that f is measurable.

Pullbacks of σ-Fields.

X292. Let (Y,B) be a measurable space, let X be a set, and let g : X → Y . Let A =

g−1[B] : B ∈ B

. due 5Th(a) Prove that A is the smallest σ-field on X which makes g measurable. (We call A the σ-field on

X generated by g (from (Y,B)) and we denote A by σ(g).)(b) Let (W, E ) be a measurable space and let f : W → X . Prove that f is E /A -measurable iff g f

is E /B-measurable.(c) Prove that A is the unique σ-field on X that has the property described in (b) for all measurable

spaces (W, E ) and all maps f : W → X .(d) Suppose D is a set of subsets of Y such that B = σ(D) on Y . Let C =

g−1[D] : D ∈ D

.Prove that A = σ(C ) on X .

Subspaces of Measurable Spaces.

Example. Let (Y,B) be a measurable space. Let X be a subset of Y . (We do not assume that X ∈ B .)Let A = B ∩X : B ∈ B . Notice that for each T ⊆ Y , we have T ∩X = g−1[T ] , where g is the inclusionmap40 from X to Y . Therefore from problem X292, it is clear that:

(a) A is the smallest σ-field on X which makes the inclusion map from X to Y measurable. (Wecall A the subspace σ-field that X inherits from (Y,B).)

(b) If (W, E ) be a measurable space and f : W → X , then f is E /A -measurable from W to X iff fis E /B-measurable from W to Y .

(c) A is the unique σ-field on X that has the property described in (c) for all measurable spaces(W, E ) and all maps f : W → X .

(d) If D is a set of subsets of Y such that B = σ(D) on Y and if C = D ∩X : D ∈ D , thenA = σ(C ) on X .

Example. Let Y be a topological space and let B be the Borel σ-field on Y . Thus B = σ(G (Y )) onY , where G (Y ) is the collection of open subsets of Y . Let X ⊆ Y . Then the subspace topology that Xinherits from Y is G (X) = G ∩X : G ∈ G (Y ) . Let A be the subspace σ-field that X inherits from(Y,B). Then by the preceding example, A = σ(G (X)) on X . In other words, A is equal to the Borelσ-field on X .

X293. Let (Y,B) be a measurable space, let X ⊆ Y , and let A be the subspace σ-field that X inheritsfrom (Y,B). Let (Z,C ) be a measurable space and let h : Y → Z be B/C -measurable. Prove that therestriction of h to X is A /C -measurable.

X294. Let (Y,B) be a measurable space and let g : Y → R be B-measurable. Let X = g 6= 0 and letA be the subspace σ-field that X inherits from (Y,B). Prove that 1/g is A -measurable from X to R .(Take 1/∞ = 0 = 1/(−∞).)

X295. Let (Y,B) be a measurable space and let fn : Y → R be B-measurable for each n ∈ N . Let

X =

x ∈ Y : limn→∞

fn(x) exists in R

.

(a) Prove that X ∈ B .(b) Define f : X → R by f(x) = limn→∞ fn(x) for all x ∈ X . Let A be the subspace σ-field that X

inherits from (Y,B). Prove that f is A -measurable.

40 Given sets X and Y with X ⊆ Y , to say that g is the inclusion map from X to Y means that g is the function fromX to Y defined by g(x) = x for all x ∈ X .

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X296. Let (Y,B) be a measurable space, let X ⊆ Y , let A be the subspace σ-field that X inherits fromY , and let E = B ∈ B : B ⊆ X .

(a) Prove that E ⊆ A .(b) Prove that E = A iff X ∈ E iff X ∈ B .

Cartesian Products of Measurable Spaces.

For simplicity, we shall consider just Cartesian products of two measurable spaces. By induction, our resultsextend to Cartesian products of finitely many measurable spaces. (It is also possible to consider Cartesianproducts of infinitely many measurable spaces.)

Notation. Let (X,A ) and (Y,B) be measurable spaces and let Z = X × Y . Recall that

A ⊙ B = A×B : A ∈ A and B ∈ B .

We call the elements of A ⊙ B measurable rectangles. Except in trivial cases, A ⊙ B is not a σ-field onZ . The σ-field on Z generated by A ⊙B is called the product of the σ-fields A and B and is denoted byA ⊗ B .

X297. Let (X,A ) and (Y,B) be measurable spaces. Let Z = X × Y . Make Z into a measurable spaceby equipping it with the σ-field C = A ⊗ B .

(a) Let π1 : Z → X and π2 : Z → Y be the usual projection maps. Prove that π1 and π2 aremeasurable.

(b) Let (W, E ) be a measurable space, let f : W → Z , and let g = π1 f and h = π2 f , so thatf(w) = (g(w), h(w)) for all w ∈ W . Prove that f is measurable iff g and h are measurable.

(c) Prove that the product σ-field on Z is the unique σ-field on Z having the property described inpart (b) for all measurable spaces (W, E ) and all functions f : W → Z .

X298. Let X and Y be topological spaces and let Z = X × Y . Let A and B be the Borel σ-fields on due 5Th

X and Y respectively. Let C be the Borel σ-field on Z .(a) Prove that A ⊗ B ⊆ C . (By the way, this inclusion can be strict, although you are not asked to

give an example where it is. Warning: No hand waving allowed! Hint: If A ⊆ X and B ⊆ Y ,then A×B = π−1

1 [A] ∩ π−12 [B] .)

(b) Suppose that X is second countable. Prove that A ⊗ B = C . (Hint: Let Un : n ∈ N be acountable base for the topology of X . Let G be open in Z . Prove that there exist open setsVn in Y such that G =

⋃∞n=1(Un × Vn). Remark: Similarly, if Y is second countable and X is

arbitrary, then A ⊗ B = C , although you are not asked to prove this.)

Example. Let B be the Borel σ-field on R and let C be the Borel σ-field on R ×R . Then by problemX298(b), C = B ⊗ B , because R is a second countable topological space. Hence by problem X297(b), if(W, E ) is any measurable space, f : W → R×R , and g = π1f and h = π2f , so that f(w) = (g(w), h(w))for all w ∈ W , then f is measurable as a map into the topological space R×R iff g and h are measurableas maps into the topological space R . To say the same thing in different words, if f : W → C , then f ismeasurable iff Re(f) and Im(f) are both measurable.

X299. Let X and Y be sets and let Z = X×Y . Let D and E be sets of subsets of X and Y respectively.Let A = σ(D) on X and let B = σ(E ) on Y .

(a) Let F = D × Y : D ∈ D ∪ X × E : E ∈ E . Prove that A ⊗ B = σ(F ) on Z .(b) Suppose that X =

⋃∞k=1Dk for some (Dk) ∈ DN and Y =

⋃∞k=1 Ek for some (Ek) ∈ DN . Prove

that A ⊗ B = σ(D ⊙ E ) on Z .

75


Reminder. Let m∗ be Lebesgue outer measure on R . Let M = Mm∗ . If A ⊆ R , then to say that Ais Lebesgue measurable means that A ∈ M . Let m be the restriction of m∗ to M . The measure m iscalled Lebesgue measure on R . Let f : R → Y , where Y is a topological space. To say that f is Lebesguemeasurable means that f is measurable with respect to the σ-field M . Now let f : R → R . To say that f isLebesgue integrable means that f is integrable with respect to the measure m . If f is Lebesgue integrable,then of course

∫

f dm is called the Lebesgue integral of f .

The Riemann Integral versus the Lebesgue Integral.

We wish to characterize the functions f which are properly Riemann integrable and to show that for eachsuch function f , the Lebesgue integral of f is the same as the Riemann integral of f . We may as wellconsider only functions f : R → R , since if f is defined only on a subinterval of R , we may extend it toall of R by defining it to be zero outside that subinterval. Now let us formulate a convenient definitionof the Riemann integral. (It should be obvious that it is equivalent to the usual definition.) If Y is a setand ϕ : R → Y , then to say that ϕ is a step function means that R can be partitioned into finitely manyintervals on each of which ϕ is constant. Just as for simple functions, any way of combining finitely many stepfunctions yields another step function. For instance, if ϕ1, ϕ2 : R → R are step functions, then so are ϕ1+ϕ2

and ϕ1 ∨ ϕ2 = max ϕ1, ϕ2 . Notice that if ϕ : R → R is a step function, then ϕ is Lebesgue integrableiff ϕ 6= 0 is bounded. Let us write Φ for the set of Lebesgue integrable step functions ϕ : R → R . Foreach ϕ ∈ Φ, let I(ϕ) =

∫

ϕdm . (Of course if ϕ ∈ Φ, then since R can be partitioned into finitely manyintervals on each of which ϕ is constant, I(ϕ) is really just a simple finite sum.) Now let f : R → R . Thenthe lower Riemann integral of f is

R∗(f) = sup I(ϕ) : ϕ ∈ Φ and ϕ ≤ f

and the upper Riemann integral of f is

R∗(f) = inf I(ϕ) : ϕ ∈ Φ and f ≤ ϕ .

To say that f is properly Riemann integrable means that −∞ < R∗(f) = R∗(f) < ∞ , and in this case,the (proper) Riemann integral of f is R(f) = R∗(f) = R∗(f). Notice that if f is not bounded below orif f < 0 is not bounded, then there is no ϕ ∈ Φ satisfying ϕ ≤ f , so R∗(f) = −∞ . Similarly, if f isnot bounded above or if f > 0 is not bounded, then there is no ϕ ∈ Φ satisfying f ≤ ϕ , so R∗(f) = ∞ .Thus if f is properly Riemann integrable, then f must be a bounded function and f 6= 0 must be abounded set.

X300. Let f : R → R be bounded below and suppose f < 0 is bounded. Let f∗ be the lower regular-ization of f . The object of this exercise is to lead you to a proof that the lower Riemann integral of f isequal to the Lebesgue integral of f∗ .

(a) Prove that∫

f∗ dm is defined and∫

f∗ dm > −∞ .(b) Prove that R∗(f) ≤

∫

f∗ dm . (Hint: If ϕ ∈ Φ and ϕ ≤ f and if ϕ∗ is the lower regularization ofϕ , then ϕ∗ < ϕ is a finite set and ϕ∗ ≤ f∗ .)

(c) Prove that∫

f∗ dm ≤ R∗(f). (Hint: If x ∈ R , y ∈ Q , and y < f∗(x), then there exist a, b ∈ Qwith a < x < b such that for each x′ ∈ (a, b), we have y < f∗(x′). Deduce that there is a sequence(ψn) in Φ such that f∗ ≤ supn ψn ≤ f . Then let ϕn = maxk≤n ψk to get an increasing sequence(ϕn) in Φ such that f∗ ≤ limn ϕn ≤ f .)

X301. Let f : R → R be bounded and suppose f 6= 0 is bounded. Let f∗ and f∗ be the lower andupper regularizations of f respectively and let

D = x0 ∈ R : f is not continuous at x0 .

From problem X300, we see that∫

f∗ dm is defined,∫

f∗ dm > −∞ , and R∗(f) =∫

f∗ dm . Similarly,∫

f∗ dm is defined,∫

f∗ dm <∞ , and R∗(f) =∫

f∗ dm .(a) Prove that f is (properly) Riemann integrable iff m(D) = 0.(b) Suppose that f is (properly) Riemann integrable. Prove that f is Lebesgue measurable, that f

is Lebesgue integrable, and that R(f) =∫

f dm .

76


Riemann Null Functions versus Lebesgue Null Functions.

Definition. Let f : R → R . To say that f is Lebesgue null means that f = 0 m-almost everywhere.

X302. Let f : R → R . Prove that f is Riemann null iff f is Riemann integrable and Lebesgue null.

Reminder. Let f = 1Q . Then f is Lebesgue null. However, f is not Riemann null. Indeed, f is not evenRiemann integrable.

X303. Let m be Lebesgue measure on R . Let f : R → C be Lebesgue measurable.(a) Let C be the set of p ∈ R such that f is continuous at p . Prove that if f = 0 m-a.e., then for

each p ∈ C , f(p) = 0. In particular, if f is continuous on R and f = 0 m-a.e., then for eachp ∈ R , f(p) = 0.

(b) Suppose f is regulated. Prove that f = 0 m-a.e. iff f(p) = 0 for all but countably many p ∈ R .(c) Give an example where f = 0 m-a.e., but f 6= 0 is uncountable.

Absolutely Convergent Improper Integrals.

As we have seen, a properly Riemann integrable function on R is Lebesgue integrable and its Riemannintegral is equal to its Lebesgue integral. The next exercise treats the relation between absolutely convergentimproper integrals over [0,∞) and Lebesgue integrals over [0,∞). Absolutely convergent improper integralsover other intervals may be treated similarly.

X304. Let f : [0,∞) → R be Lebesgue measurable. The integrals in this problem are to be understood asLebesgue integrals.

(a) Prove that limb→∞

∫ b

0|f(x)| dx exists in [0,∞] and is equal to

∫∞

0|f(x)| dx .

(b) Prove that f is Lebesgue integrable over [0,∞) iff limb→∞

∫ b

0|f(x)| dx <∞ .

(c) Prove that if f is Lebesgue integrable over [0,∞), then limb→∞

∫ b

0 f(x) dx exists and is equal to∫∞

0 f(x) dx .

Warning. If f : (a, b) → R is improperly Riemann-integrable over (a, b), but the improper Riemann integral

of f over (a, b) is only conditionally convergent, then∫ b

a f+(x) dx = ∞ =

∫ b

a f−(x) dx and consequently the

Lebesgue integral of f over (a, b) is undefined. The Lebesgue theory of integration is a theory of absolutelyconvergent integrals. In the Lebesgue approach to integration, conditionally convergent integrals are viewedas limits of integrals, not as integrals per se.

Some Applications of the Integral Limit Theorems.

From now on, unless otherwise stated, we shall interpret integrals such as∫ b

af(x) dx as Lebesgue integrals,

or in other words, as integrals with respect to Lebesgue measure.

X305. In problem X163, we saw that for each x ∈ (0,∞), we have

limn→∞

∫ n

0

tx−1(

1− t

n

)n

dt = Γ(x) =

∫ ∞

0

tx−1e−t dt.

Use one of the Corollaries of Fatou’s lemma to give a simpler proof of this fact. (The inequality 1 + u ≤ eu

should be almost the only other ingredient that you will need for the proof.)

X306. Determine the following limits. Justify your answers.

(a) limn→∞

∫ 1

0 (1 + nx2)(1 + x2)−n dx

(b) limn→∞

∫∞

0 n sin(x/n)[x(1 + x2)]−1 dx

X307.(a) Let a ∈ (−1,∞). Define f : (0,∞) → (0,∞) by f(x) =

∫∞

0tae−tx dt . Use an appropri-

ate integral limit theorem, together with the definition of the derivative, to show that f ′(x) =−∫∞

0ta+1e−tx dt .

(b) Notice that∫∞

0e−tx dt = x−1 . Use this, together with part (a), to show that

∫∞

0tne−t dt = n! .

77


A Nasty Lebesgue Integrable Function.

Reminder. Let X be a topological space and let A ⊆ X . To say that A is nowhere dense means that theclosure of A has empty interior. To say that A is meager means that A is a countable union of nowheredense sets.

For instance, any countable subset of R is meager in R . So is any closed subset of R that has emptyinterior. For instance, the ordinary Cantor set is meager in R and so is any fat Cantor set. Thus theordinary Cantor set, which is small in the sense of Lebesgue measure (we mean of Lebesgue measure zero)is also small in the sense of topology (we mean meager), whereas a fat Cantor set is not small in the senseof Lebesgue measure but is small in the sense of topology.

Meager sets are also known as sets of the first category. Sets that are not meager are also known assets of the second category. To say that X is a Baire space means that no non-empty open subset of X ismeager.

X308. Define g : R → (0,∞] by

g(x) =

|x|−1/2e−|x| if x ∈ R \ 0 ,∞ if x = 0.

Let r1, r2, r3, . . . be an enumeration of the rationals and define f : R → (0,∞] by f(x) =∑∞

n=1 2−ng(x−rn).

Let m denote Lebesgue measure on R .(a) Prove that f is Lebesgue integrable over R . Hence f < ∞ m-a.e. on R . (Thus f = ∞ is

small in the sense of Lebesgue measure.)(b) Prove that if x0 ∈ R such that f(x0) <∞ , then f is discontinuous at x0 . Thus f is discontinuous

m-a.e. on R .(c) Prove that f is lower semicontinuous.(d) Prove that if x0 ∈ R such that f(x0) = ∞ , then f is continuous at x0 .(e) Prove that f <∞ is meager in R . (Thus f <∞ is small in the sense of topology. You

should contrast this with part (a).)(f) Prove that f = ∞ is uncountable. Thus, although it is clear that f = ∞ contains Q ,

f = ∞ is not equal to Q . (In fact, f = ∞ has the cardinality of the continuum, althoughyou are not asked to prove this.)

Integration with Respect to Counting Measure.

X309. Let I be a set. Define γ : P(I) → [0,∞] by

γ(A) =

|A| if A is a finite set,

∞ otherwise,

for all A ⊆ I , where |A| denotes the number of elements in A .(a) Prove that γ is a measure on the σ-field P(I). (Reminder: γ is called counting measure on I .)(b) Let f : I → [0,∞] . Prove that

∫

If dγ =

∑

i∈I f(i), where by definition,

∑

i∈I

f(i) = sup

∑

i∈I0

f(i) : I0 is a finite subset of I

.

78


Notation. Let (X,A , µ) be a measure space. Let f be a measurable function on X taking values in Ror Rd and let A ∈ A . Then we write

∫

A f dµ for∫

f1A dµ .

X310. Let (X,A , µ) be a measure space, let f : X → R , and suppose∫

f dµ is defined. Prove that foreach A ∈ A ,

∫

A f dµ is defined.

Semifinite Measures.

Let (X,A , µ) be a measure space. To say that µ is semifinite means that for each A ∈ A , if µ(A) > 0,then there exists B ∈ A such that B ⊆ A and 0 < µ(B) <∞ .

X311. Let (X,A , µ) be a measure space. Suppose that µ is σ-finite. Prove that µ is semifinite.

X312. Let X be a set and let γ be counting measure on X .(a) Verify that γ is semifinite.(b) Prove that γ is σ-finite iff X is countable. So for instance, if X = R , then γ is semifinite but

not σ-finite.

X313. Let (X,A , µ) be a measure space. Let f, g : X → R such that∫

f dµ and∫

g dµ are both defined.Assume that µ is semifinite.

(a) Suppose that for each A ∈ A , we have∫

Af dµ ≤

∫

Ag dµ . Prove that f ≤ g µ-almost everywhere.

(Hint: f > g is the union of the sets of the form f > r and s > g with r and s rational andr > s .)

(b) Suppose that for each A ∈ A , we have∫

Af dµ =

∫

Ag dµ . Prove that f = g µ-almost everywhere.

X314. Give an example of a measure space (X,A , µ) and two measurable functions f, g : X → [0,∞) suchthat

∫

A f dµ =∫

A g dµ for each A ∈ A , but µ(f 6= g) > 0. Of course µ cannot be semifinite. (Hint: Thereis an example where X has just one point.)

More Applications of the Integral Limit Theorems.

Let f : R → C be Lebesgue integrable. The Fourier transform of f is the function f : R → C defined by

f(y) =

∫

R

f(t)e−2πity dt

for all y ∈ R . It is an important fact that if f(t) = e−πt2 for all t ∈ R , then f = f . The following exercise41

outlines a particularly elegant proof of this.

X315. due 7Th

(a) Let f : R → C be Lebesgue measurable and suppose∫

R|f(t)|ext dt < ∞ for all x ∈ R . Let

z ∈ C . Justify the following calculation:

∫

R

f(t)ezt dt =

∫

R

f(t)

∞∑

n=0

(zt)n

n!dt =

∞∑

n=0

cnzn,

where

cn =

∫

R

f(t)tn

n!dt.

(b) Use the fact that∫

Re−πt2 dt = 1 (which you are not asked to prove here) to check that

∫

R

e−πt2e2πxt dt = eπx2

41 This exercise relies on a fact that we have not yet proved, namely that∫

Re−πt2 dt = 1. The most common way to prove

this is to write the square of this integral as an integral over R2 and switch to polar coordinates to evaluate the double integral.

Since we have not discussed double integrals, it is worth mentioning that another proof of this, in the form∫

Re−s2 ds =

√π ,

can be found on page 194 in Rudin, Principles of Mathematical Analysis, Third Edition.

79


for each x ∈ R . Then use part (a) and some complex analysis to explain why we are justified inreplacing x by −iy in the last equation to get

∫

R

e−πt2e−2πiyt dt = e−πy2

for each y ∈ R .

The next exercise generalizes part (a) of problem X315.

X316. Let f : R → C be Lebesgue measurable. Let S =

s ∈ R :∫

Resx|f(x)| dx <∞

. Let a = inf Sand b = supS . Assume that a < b . Let u0 = s0+it0 , where s0, t0 ∈ R and a < s0 < b (and where of coursei =

√−1). Let r = min b− s0, s0 − a and let D = u ∈ C : |u− u0| < r . (Thus D is the largest open

disk centered at u0 and contained in the strip u ∈ C : a < Reu < b .) Prove that for all u ∈ D ,

∫

R

euxf(x) dx =∞∑

n=0

cn(u− u0)n,

where for each n ,

cn =1

n!

∫

R

xneu0xf(x) dx.

Don’t forget to explain why the integrand in the expression for cn is integrable.

Definitions. A probability measure on a measurable space (X,A ) is a measure µ on A such that µ(X) = 1.A Borel measure on a topological space is simply a measure on the Borel σ-field on that space. If µ is aBorel probability measure on R , then the characteristic function of µ is the function ϕ : R → C defined byϕ(t) =

∫

Reitx dµ(x) for all t ∈ R .

X317. Let µ be a Borel measure on R and let f : R → C be µ-integrable. Define g : R → C by

g(t) =

∫

R

eitxf(x) dµ(x)

for all t ∈ R .(a) Prove that g is uniformly continuous on R .(b) Suppose in addition that

∫

R|xf(x)| dµ(x) < ∞ . Prove that g is continuously differentiable and

that for each t ∈ R ,

g′(t) =

∫

R

ixeitxf(x) dµ(x).

Remark. Let µ be a Borel probability measure on R with characteristic function ϕ(t) =∫

Reitx dµ(x).

Suppose that∫

Rx2 dµ(x) < ∞ . Then by applying problem X317(b) twice (first with f(t) = 1 and then

with f(t) = ix), we find that ϕ is C2 and that for each t ∈ R ,

ϕ′′(t) =

∫

R

−x2eitx dµ(x).

In particular,

−ϕ′′(0) =

∫

R

x2 dµ(x).

Since ϕ(0) = 1, l’Hospital’s rule plus the definition of ϕ′′(0) shows that as t→ 0,

2− ϕ(t)− ϕ(−t)t2

→ −ϕ′′(0),

so2− ϕ(t)− ϕ(−t)

t2→∫

R

x2 dµ(x).

This last fact still holds even without the assumption that∫

Rx2 dµ(x) <∞ , as part (a) of the next exercise

shows.

80


X318. due 7Th

(a) Let µ be a Borel probability measure on R with characteristic function ϕ . Prove that as t→ 0,

2− ϕ(t)− ϕ(−t)

t2→

∫R

x2dµ(x).

(This holds whether or not the integral on the right is finite. There is a simple proof that doesnot require considering cases and that does not depend on problem X317(b). Hint: Recall that1− cosu = 2 sin2(u/2).)

(b) For 0 < α <∞ , define ϕα on R by ϕα = exp(−|t|α). Prove that if α > 2, then there is no Borelprobability measure on R with characteristic function ϕα .

Remark. What makes problem X318(b) interesting is that in contrast, if 0 < α ≤ 2, then ϕα is thecharacteristic function of a Borel probability measure on R , a so-called symmetric stable distribution ofindex α . (You are not asked to prove this.) To those who know probability theory, it will be a familiar factthat if α = 2, then ϕα is the characteristic function of the normal distribution with mean 0 and variance2.

Reminder. If (an) and (bn) are sequences of non-zero real or complex numbers, then to say that an isasymptotic to bn as n tends to infinity (written an ∼ bn as n→ ∞) means that an/bn → 1 as n→ ∞ .

X319. Let a > 1. The question of the asymptotic behaviour of the sequence of numbers

In =

∫ ∞

na

xn−1e−x dx, n = 1, 2, 3, . . . ,

arises in probability theory.42 Prove that

In ∼ Cnn−1an−1e−na as n→ ∞,

where C ∈ (0,∞) is a suitable constant, and determine the value of C . (Hint: In = nn−1an−1e−naJnwhere Jn is a certain integral over the interval (na,∞). The constant C will be limn→∞ Jn . Perform asubstitution in the integral Jn to get an integral over the interval (0,∞) to which you can apply a suitablelimit theorem.)

Definitions. Let (X,A ) be a measurable space and let µ and ν be measures on A . To say that ν isabsolutely continuous with respect to µ (denoted by ν ≪ µ) means that for each A ∈ A , if µ(A) = 0, thenν(A) = 0. To say that µ and ν are equivalent means that µ≪ ν and ν ≪ µ .

X320. Let (X,A , µ) be a measure space. Let ρ : X → [0,∞] be measurable. Define ν : A → [0,∞] by

ν(A) =

∫

A

ρ dµ

for all A ∈ A .(a) Prove that ν is a measure on A . (We call ν the measure with density ρ with respect to µ .)(b) Prove that ν ≪ µ .(c) Prove that ν ≈ µ iff µ(ρ = 0) = 0.

(d) Prove that for each measurable function f : X → [0,∞] , we have∫

f dν =∫

fρ dµ . (Hint: First,check this if f = 1A where A ∈ A . Then check it if f : Y → [0,∞) is simple. Then check it inthe general case. By the way, this sort of reasoning arises often in the theory of integration.)

(e) Let f : X → R be measurable. Prove that if either of the integrals∫

f dν and∫

fρ dµ exists,then so does the other and they are equal.

42 For those who know probability theory, we mention that In is the probability that (X1 + · · ·+Xn)/n exceeds a , whereX1,X2, X3, . . . are independent exponentially distributed random variables each of which has expected value equal to 1.

81


The Radon-Nikodym Theorem. Let (X,A , µ) be a measure space and let ν be another measure onA . In the light of problem X320, it is natural to ask when ν will have a density with respect to µ . In otherwords, when will there exist a measurable function ρ : X → [0,∞] such that for each A ∈ A , we have

ν(A) =

∫

A

ρ dµ.

The obvious necessary condition is that ν must be absolutely continuous with respect to µ . The Radon-Nikodym theorem says that this condition is also sufficient, provided that µ is σ-finite. It is also natural toask to what extent such a density of ν with respect to µ is unique. The next exercise deals with this.

X321. Let (X,A , λ) be a semifinite measure space. Let g, h : X → [0,∞] be measurable. Let µ be themeasure with density g with respect to λ and let ν be the measure with density h with respect to λ .

(a) Suppose that µ ≤ ν . Prove that g ≤ h λ-almost everywhere. (Hint: This follows immediatelyfrom an earlier exercise.)

(b) Suppose that µ = ν . Prove that g = h λ-almost everywhere.

X322. (The abstract change of variables formula.) Let (X,A ) and (Y,B) be measurable spaces and letT : X → Y be measurable. Let λ be a measure on X . Define µ : B → [0,∞] by µ(B) = λ(T−1[B]) for allB ∈ B .

(a) Prove that µ is a measure on B .(b) Prove that for each measurable function f : Y → [0,∞] , we have

∫

Yf dµ =

∫

Xf T dλ . (Hint:

You can use the an approach similar to the one that was suggested for problem X320(d).)(c) Let f : X → R be measurable. Prove that if either of the integrals

∫

Y f dµ and∫

X f dλ exists,then so does the other, and they are equal.

Notation and Terminology. Let (X,A ) be a measure space and let p ∈ X . Define δp on A by

δp(A) =

1 if p ∈ A,

0 if p /∈ A,

for all A ∈ A . It is easy to check that δp is a probability measure on A . We call δp the unit point massat p . Another name for δp is the Dirac measure at p . (The reason for this is that when X = R and A isthe Borel σ-field on R , δp is the mathematically rigorous realization of the notorious Dirac delta functioncentered at p .) By the way, the definition of δp may be written more succinctly as follows: δp(A) = 1A(p)for all A ∈ A .

Remark. Let (X,A ), (Y,B), T , λ , and µ be as in problem X322. The measure µ is called the imageof λ under the map T . It is easy to check that if p ∈ X and λ is δp , the unit point mass at p , then µis the unit point mass at f(p). In this sense, the notion of the image of a measure under the map T is ageneralization of the notion of the image of a point under T .

X323. Let a, b ∈ R with a 6= 0. Define T : R → R by T (x) = ax+ b . Let m be Lebesgue measure on R ,let M be the σ-field of Lebesgue measurable subsets of R , and let B be the Borel σ-field on R .

(a) Check that for each interval I ⊆ R , we have m(T−1[I]) = |a−1|m(I).(b) Prove that for each B ∈ B , we have m(T−1[B]) = |a−1|m(B). (Hint: Apply the uniqueness part

of the Hahn extension theorem.)(c) Prove that for each E ∈ M , we have T−1[E] ∈ M and m(T−1[E]) = |a−1|m(E).(d) Prove that for each Lebesgue measurable function f : R → [0,∞] , f T is Lebesgue measurable

and∫

Rf(ax+ b) dx = |a−1|

∫

Rf(x) dx . (Hint: Apply problem X322.)

(e) Let f : R → R be Lebesgue measurable. Prove that if either of the integrals∫

Rf(ax+ b) dx and

∫

R|a−1|f(x) dx exists, then so does the other and they are equal.

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X324. Let (X,A , µ) be a probability space. Let p, q ∈ [1,∞] with p ≤ q and let f ∈ L0(µ). Provethat ‖f‖p ≤ ‖f‖q . (Hint: This can be deduced from Holder’s inequality. Take one of the functions to beidentically 1.)

Remark. Let (X,A , µ) be a finite measure space. (In other words, let (X,A , µ) be a measure space forwhich µ(X) < ∞ .) Let p, q ∈ [1,∞] with p ≤ q . Then there is a constant C ∈ [0,∞), depending only onp , q , and µ(X), such that for each f ∈ L0(µ), ‖f‖p ≤ C‖f‖q . This can be proved in the same way as theresult of problem X324. In particular, Lq(µ) ⊆ Lp(µ).

X325. Let X be a set and let γ be counting measure on X . For each p ∈ [1,∞] and for each f : X → K ,let ‖f‖p denote the p-norm of f relative to the measure γ . In other words, for each p ∈ [1,∞] and for eachf : X → K , let

‖f‖p =

(∑

x∈X |f(x)|p)1/p

if p <∞,

sup |f(x)| : x ∈ X if p = ∞.

Let p, q ∈ [1,∞] with p ≤ q and let f : X → K . Prove that ‖f‖p ≥ ‖f‖q . (Hint: This is almost trivial.Reduce to the case where ‖f‖p = 1.)

Remark. Let X be a set. For each p ∈ [1,∞] , ℓp(X) denotes Lp(γ), where γ is counting measure on X .It follows immediately from problem X325 that if p, q ∈ [1,∞] with p ≤ q , then ℓp(X) ⊆ ℓq(X).

Remark. Note that the inequalities in the conclusions of problem X324 and problem X325 are opposites ofone another.

X326. Suppose µ is Lebesgue measure on R . Fix p, q ∈ [1,∞] with p < q .(a) Show that Lq(µ) is not a subset of Lp(µ).(b) Show that Lp(µ) is not a subset of Lq(µ).

Reminder. L0(T) denotes the set of Lebesgue measurable 1-periodic functions from R to C .

X327. Let f ∈ L0(T). Let a ∈ R . Prove that if either of the integrals∫ a+1

af(t) dt and

∫ 1

0f(t) dt exists,

then so does the other and they are equal.

Definition. Let f, g ∈ L0(T). The convolution of f and g is the function f ∗ g defined by

(f ∗ g)(t) =∫ 1

0

f(t− s)g(s) ds

for all t ∈ R for which the integral exists.

X328. Let f, g ∈ L0(T). Prove that f ∗g = g∗f . (In other words, for each t ∈ R , if either of the quantities(f ∗ g)(t) and (g ∗ f)(t) is defined, then so is the other and they are equal.)

X329. Let p, q ∈ [1,∞] with p−1 + q−1 = 1. Let f ∈ Lp(T) and let g ∈ Lq(T).(a) Prove that for each t ∈ R , (f ∗ g)(t) is defined and |(f ∗ g)(t)| ≤ ‖f‖p‖g‖q .(b) If in addition f ∈ C(T) or g ∈ C(T), prove that f ∗ g ∈ C(T).(c) Without assuming that f ∈ C(T) or g ∈ C(T), prove that f ∗ g ∈ C(T). (Hint: Either

p ∈ [1,∞) or q ∈ [1,∞). Say p ∈ [1,∞). Then there is a sequence fn in C(T)∩Lp(T) such that‖f − fn‖p → 0. Prove that fn ∗ g → f ∗ g uniformly on R .)

Notation. For each f ∈ L1(T) and for each k ∈ Z , f(k) =∫ 1

0 e−2πikxf(x) dx , by definition. (Earlier, for

f ∈ L1(R), we defined f differently, namely by f(y) =∫

Rf(t)e−2πity dt for all y ∈ R . Which meaning is

intended for f should be clear from the context.)

X330. Let f, g ∈ C(T). Suppose f = g . Prove that f(t) = g(t) for all t ∈ R .

X331. Let c ∈ ℓ1(Z).(a) Prove that the series

∑

k∈Z c(k)ek converges in C(T), in the sense of unordered sums, to somef ∈ C(T).

(b) Prove that f(k) = c(k) for each k ∈ Z .

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X332. Let g ∈ C(T). Suppose that g ∈ ℓ1(Z). Let f =∑

k∈Z g(k)ek . By problem X331, this makes senseand f ∈ C(T). Prove that f(t) = g(t) for all t ∈ R .

Reminder. From problem X315, we know that∫

Re−2πikxe−πx2

dx = e−πk2

for all k ∈ R .

X333. Prove that due 8Th

(4πt)1/2∑

k∈Z

e−4π2k2te2πikx =∑

n∈Z

e−(x−n)2/(4t) for all x ∈ R and all t > 0 . (38)

(Hint: Fix t > 0. Let ϕ be the function of x on the right side of the equation. Show that ϕ ∈ L2(T)

and∫ 1

0e−2πikxϕ(x) dx = (4πt)1/2e−4π2k2t for all k ∈ Z . Then explain how we can bridge the gap between

convergence in L2(T) and convergence at x for all x ∈ R in this case. This is partly a special case of someof the preceding few exercises, but please make your proof reasonably self-contained.)

The identity (38) in problem X333 is one of Jacobi’s theta function identities. Note that when t is closeto 0, then a substantial number of terms are needed to calculate the left side of (38) accurately, but theright side is well-approximated by the few terms with n near x . Note also that the right side of the (38)is obviously positive, whereas it is not at all obvious that the left side is positive. The method suggestedabove to prove (38) can also be used to establish other interesting identities. In fact, there is a generalresult of which all such identities are special cases. It is called the Poisson summation formula. The Poissonsummation formula is discussed in Folland, Real Analysis: Modern Techniques and Their Applications. Butof course I do not want you to solve problem X333 merely by pointing out that it follows from the Poissonsummation formula! I want you to give a reasonably self-contained proof.

Notation for Partial Derivatives. Let G be an open subset of Rd and let f : G→ R . A multi-index isa d-tuple of non-negative integers. If α = (α1, . . . , αd) is a multi-index and if f is sufficiently differentiable,then Dαf denotes the partial derivative

∂α1+···+αdf

∂α1x1 · · · ∂αdxd.

The identity (38) is the key to the next exercise, which is closely related to Weierstrass’s own proof thatthe trigonometric polynomials are dense in C(T).

X334. Let f ∈ C(T). If we use separation of variables to find a solution u of the heat equation

∂u

∂t=∂2u

∂x2for x ∈ R and t > 0 , (39)

subject to the initial conditionlimt↓0

u(x, t) = f(x) for x ∈ R , (40)

then we are led to the function u defined by

u =∑

k∈Z

uk, where uk(x, t) = e−4π2k2tf(k)e2πikx for k ∈ Z , x ∈ R and t > 0 . (41)

(a) Prove that for each t > 0, u(·, t) = f ∗ gt where gt is a certain function in C(T) which you shouldbe able to figure out from problem X333. Also, calculate ‖gt‖1 for each t > 0.

(b) For each t > 0, define Gt : C(T) → C(T) by Gt(h) = h ∗ gt . Prove that each Gt is a boundedlinear operator on C(T) and that in fact, supt>0 ‖Gt‖ <∞ , where ‖Gt‖ denotes the norm of Gt

as a linear operator on C(T).(c) Prove that Gt(f) → f uniformly as t ↓ 0. (Hint: Show that Gt(h) → h uniformly in the special

case where h is a trigonometric polynomial and then show that Gt(f) → f uniformly by usingthis special case in conjunction with part (b) and the fact that the trigonometric polynomials aredense in C(T).)

(d) Part (c) shows that u does satisfy the initial condition (40), and does so in as strong a sense aswe could hope for given the fact that the initial temperature distribution f is merely continuous.

84


It remains to show that u does satisfy the heat equation. Let Ω = R × (0,∞). Prove thatu ∈ C∞(Ω), and that the series (41) converges to u in the following very strong sense: Foreach multi-index α , for each t0 > 0, and for each ε > 0, there exists a finite subset A0 ⊆ Zsuch that for each finite subset A1 ⊆ Z with A0 ⊆ A1 , and for each (x, t) ∈ R × [t0,∞),|(Dαu)(x, t)−

∑

k∈A1(Dαuk)(x, t)| ≤ ε . Deduce in particular that u does satisfy (39).

The method outlined in the hint for part (c) of problem X334 is the model for a large variety of differentresults about convergence of Fourier series and related expressions. For instance, one can use it to show thatif f is merely in Lp(T), where p ∈ [1,∞), then the conclusion of part (d) still holds and u(·, t) → f inLp(T), so that the initial condition (40) is still satisfied when suitably interpreted. The proof is not muchdifferent than for the case where f ∈ C(T). One only needs to observe that ‖f ∗ gt‖p ≤ ‖f‖p‖gt‖1 andthat the trigonometric polynomials are dense in Lp(T). Note that the sense in which the initial condition(40) holds is adapted to the function space in which f is assumed to lie. Note also that the Fourier series∑∞

k=−∞ f(k)e2πikx need not itself converge for the initial condition (40) to be satisfied. It is sufficient that

the “regularized” series∑

k∈Z e−4π2k2tf(k)e2πikx behave appropriately as t ↓ 0, and this holds in much

greater generality.43 This is a pleasant feature of the heat equation. Laplace’s equation is nicely behaved ina similar way, but the wave equation is not.

Notation. Let X be a set. Then c0(X) denotes the set of functions g : X → K which tend to zero atinfinity in the sense that for each ε > 0, the set |g| ≥ ε is finite. For instance, c0(Z) is the set of functionsg : Z → K such that g(k) → 0 as |k| → ∞ . When c0(Z) is mentioned in the context of Fourier series, it isunderstood that K = C .

X335. (The Riemann-Lebesgue lemma for Fourier series.) Let f ∈ L1(T). It is obvious that f ∈ ℓ∞(Z)

and ‖f‖∞ ≤ ‖f‖1 . Prove that f ∈ co(Z). (Hint: Let E be the set of all g in L1(T) such that g ∈ c0(Z).Verify that E is closed in L1(T). This is easy. Then prove that E is dense in L1(T).)

Notation. Suppose f ∈ L1(T). For all integers m,n ≥ 0, define Sm,nf by

Sm,nf =

n∑

k=−m

f(k)ek.

We call Sm,nf the (m,n)-th asymmetric partial sum of the Fourier series for f . For each integer n ≥ 0,define Snf by

Snf =

n∑

k=−n

f(k)ek.

We call Snf the n-th symmetric partial sum of the Fourier series for f . For each integer n ≥ 0, define Dn

by

Dn =

n∑

k=−n

ek.

The function Dn is called the Dirichlet kernel of order n . As we know, for each k ∈ Z , we have ek ∗ f =f(k)ek . It follows immediately that Snf = Dn ∗ f .

The elegant treatment outlined in the next couple of exercises is taken from an article by Paul R.Chernoff.

43 Du Bois-Reymond (1873) gave an example of a function f ∈ C(T) whose Fourier series fails to converge at a certainpoint. In fact, there is a function f ∈ C(T) whose Fourier series converges only on a meager set. A meager set can be of fullLebesgue measure, but Kolmogorov showed that there is a function f ∈ L1(T) whose Fourier series diverges almost everywhere(1923) or even everywhere (1926). Marcel Riesz (1927) showed that if f ∈ Lp(T) , where 1 < p < ∞ , then the Fourier seriesof f converges to f in Lp(T) ; however, unless p = 2, the convergence is in general not in the sense of unordered sums. Lusin(1912) had conjectured that if f ∈ L2(T) , then the Fourier series of f converges almost everywhere. The evidence of the nextfew decades seemed to point in the opposite direction however and Zygmund (1959), in the preface of the second edition ofhis famous treatise on trigonometric series, wrote “The problem of the existence of a continuous function with an everywheredivergent Fourier series is still open.” Carleson (1966) astonished the experts by proving Lusin’s conjecture. R. A. Hunt (1968)simplified Carleson’s proof and confirmed Carleson’s claim that the conclusion of Lusin’s conjecture holds for f ∈ Lp(T) wherep > 1, not just for f ∈ L2(T) .

85


X336. Let f ∈ L1(T), let a ∈ R , and suppose due 8Th

∫ 1/2

−1/2

|f(a+ h)− f(a)||h| dh <∞. (42)

Prove that (Sm,nf)(a) → f(a) as m,n → ∞ . (Hint: To simplify the calculations, shift the origin andsubtract a suitable constant from f to reduce to the case where a = 0 and f(0) = 0. Let g(x) =

f(x)/(e2πix − 1). Use (42) to show that g ∈ L1(T). Use the fact that f(x) = (e2πix − 1)g(x) to relate fand g . Conclude that (Sm,nf)(0) is a telescoping sum and apply the Riemann-Lebesgue lemma to g toshow that (Sm,nf)(0) → 0 as m,n→ ∞ .)

Remark. Let f ∈ L1(T) and let a ∈ R . It is clear that if f satisfies any one of the following successivelymore general conditions, then (42) holds.

(a) f is differentiable at a .(b) f has finite one-sided derivatives at a .(c) f satisfies a Lipschitz condition at a ; in other words, there exists a constant C ∈ [0,∞) for that

for each x in some nhd of a , |f(x) − f(a)| ≤ C|x − a| .(d) f satisfies a Holder condition at a ; in other words, there exist constants C ∈ [0,∞) and α ∈ (0, 1]

such that for each x in some nhd of a , |f(x)− f(a)| ≤ C|x− a|α .X337. Let f ∈ L1(T) and let a ∈ R , and suppose

∫ 1/2

−1/2

|f(a+ h) + f(a− h)− 2f(a)||h| dh <∞. (43)

Prove that (Snf)(a) → f(a) as n → ∞ . (Hint: To simplify the calculations, shift the origin and subtracta suitable constant from f to reduce to the case where a = 0 and f(0) = 0. Then apply problem X336 tothe function (f(x) + f(−x))/2.)Remark. Let f ∈ L1(T), let a ∈ R , and suppose f(a+) and f(a−) exist and are finite. Suppose also thatf satisfies any one of the following successively more general conditions:

(a) limh→0+(f(a+ h)− f(a+))/h and limh→0−(f(a+ h)− f(a−))/h both exist and are finite.(b) f satisfies one-sided Lipschitz conditions at a ; in other words, there exists a constant C ∈ [0,∞)

and there exists δ > 0 such that for each x ∈ (a, a+ δ), we have |f(x) − f(a+)| ≤ C|x− a| , andfor each x ∈ (a− δ, a), we have |f(x)− f(a−)| ≤ C|x− a| .

(c) f satisfies one-sided Holder conditions at a ; in other words, there exist constants C ∈ [0,∞) andα ∈ (0, 1], and there exists δ > 0 such that for each x ∈ (a, a + δ), we have |f(x) − f(a+)| ≤C|x− a|α , and for each x ∈ (a− δ, a), we have |f(x)− f(a−)| ≤ C|x− a|α .

Then Snf(a) → (f(a+) + f(a−))/2 as n → ∞ . I leave it to you to to use problem X337 to verify this.Note that the restriction to symmetric partial sums is essential here, as one sees by considering the “squarewave” function f defined by f(x) = sgn(sin 2πx), where sgn(u) = u/|u| for u 6= 0 and sgn(0) = 0.

86


X338. Let X , Y , and Z be normed linear spaces. Define f : L (X,Y )× L (Y, Z) → L (X,Z) by

f(S, T ) = T S.

Prove that f is continuous.

X339. Define f : R2 → R byf(x, y) = (x2 − y)(y − 2x2).

Let Z = f = 0 and let G = f 6= 0 .(a) Sketch the set Z .(b) Find all the components of G and determine the sign of f in each of them.(c) Show that for each straight line L through the origin in R2 , the restriction of f to L has a strict

local maximum at the origin. (Do not use calculus for this.)(d) Show that f does not have a local maximum at the origin. (Do not use calculus for this.)

X340. Let f : R → R be differentiable. Suppose that f achieves a local minimum at the origin. Supposealso that the origin is the only critical point for f . Show that f has a global minimum at the origin.

X341. Define f : R2 → R by due 9Th

f(x, y) = x2(1 + y)3 + y2.

(a) Show that f achieves a strict local minimum at the origin. (Do not use calculus for this.)(b) Show that the origin is the only critical point for f .(c) Show that f does not achieve a global minimum at the origin.

X342. Let X and Y be Banach spaces. Define f : G L (X,Y ) → G L (Y,X) by f(S) = S−1 . Let S0 ∈G L (X,Y ). Show that f is differentiable at S0 and that f ′(S0) is the linear map from L (X,Y ) to L (Y,X)such that for each T ∈ L (X,Y ),

f ′(S0)(T ) = −S−10 TS−1

0 . (44)

Remark. When X = Y = K , then under the usual identification of L (X,Y ) and L (Y,X) with K , (44)reduces to the familiar formula

d

dss−1

∣

∣

∣

∣

s=s0

= −s−20 .

However, when X = Y but dim(X) > 1, such a reduction is not possible, because S−10 and T need not

commute.

X343. Let X and Y be Banach spaces, let Ω be an open subset of X , let f : Ω → Y , and let a ∈ Ω. Showthat changing the norms on X and Y to other equivalent norms does not affect whether f is differentiableat a , nor does it affect f ′(a).

X344. Let X and Y be normed linear spaces. Let ν be any norm on R2 . Let

‖(x, y)‖ = ν(‖x‖, ‖y‖) (45)

for all (x, y) ∈ X × Y .(a) Show that (45) defines a norm on X × Y which induces the product topology on X × Y .(b) Show that if X and Y are Banach spaces, then X × Y is a Banach space under this norm.

Remark. Different choices of the norm ν in (45) yield different but equivalent norms on X × Y . Thusconfusion should not arise if we simply speak of X×Y as a normed linear space, without specifying the norm.Any norm that induces the product topology will do. And problem X344 allows us easily to give examplesof such norms. For instance, ‖(x, y)‖ = ‖x‖+ ‖y‖ defines such a norm. So does ‖(x, y)‖ = (‖x‖2+ ‖x‖2)1/2 .(A different but equivalent one, of course.)

X345. (Coordinatewise differentiation.) Let X , Y , and Z be Banach spaces. Let Ω be an open subset ofX , let f : Ω → Y , let g : Ω → Z , and define h : Ω → Y × Z by h(x) =

(

f(x), g(x))

. Let p ∈ Ω. Show thath is differentiable at p iff f and g are both differentiable at p , and that in this case, for each u ∈ X , wehave h′(p)(u) =

(

f ′(p)(u), g′(p)(u))

.

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X346. (The product rule.) Let X , Y , and Z be Banach spaces. Let ϕ : X × Y → Z be continuous andbilinear.

(a) Let (a, b) ∈ X × Y . Prove that ϕ is differentiable at (a, b) and that for each (x, y) ∈ X × Y ,

ϕ′(a, b)(x, y) = ϕ(a, y) + ϕ(x, b).

(b) Let E be another Banach space, let Ω be an open subset of E , let f : Ω → X , let g : Ω → Y , letp ∈ Ω, and suppose f and g are both differentiable at p . Define h : Ω → Z by

h(u) = ϕ(

f(u), g(u))

.

Prove that h is differentiable at p and that for each u ∈ E ,

h′(p)(u) = ϕ(

f(p), g′(p)(u))

+ ϕ(

f ′(p)(u), g(p))

.

(Hint: Combine part (a), the chain rule, and problem X345.)

Example. For all y, z ∈ R3 , let y × z ∈ R3 be the usual cross product of the vectors y and z . Let E bea Banach space, let Ω be an open subset of E , let f, g : E → R3 , and define h : Ω → R3 by

h(u) = f(u)× g(u).

Let p ∈ Ω and suppose that f and g are differentiable at p . Then h is differentiable at p and for eachu ∈ E ,

h′(p)(u) = f(p)× g′(p)(u) + f ′(p)(u)× g(p).

To see this, just apply problem X346(b) with ϕ(y, z) = y × z .

Remark. Similar considerations apply to the product of a scalar-valued function and a vector-valued func-tion. In a Hilbert space, similar considerations apply to the inner product of two vector-valued functions.

X347. Let X be a real Hilbert space. Define f, g : X → R by f(x) = ‖x‖2 and g(x) = ‖x‖ .(a) Let a ∈ X . Show that f is differentiable at a and that for each x ∈ X , f ′(a)(x) = 2 〈a|x〉 .(b) Let a ∈ X \ 0 . Show g is differentiable at a and that for each x ∈ X , g′(a)(x) = 〈a|x〉 /‖a‖ .(c) Show that g is not differentiable at 0, except in the trivial case where X = 0 .

X348. Define f, g : R2 → R by f(0) = 0, g(0) = 0, and for each x = (x1, x2) ∈ R2 \ 0 , due 9Th

f(x) =x1x

22

x21 + x22and g(x) =

x31x2x41 + x22

.

Clearly f and g are continuously differentiable on R2 \ 0 . What about at 0?(a) Show that for each v ∈ R2 , the directional derivative

limt→0

f(tv)− f(0)

t= A(v)

exists in R2 . However, show that the function v 7→ A(v) is not linear. Deduce that f is notdifferentiable at 0.

(b) Show that for each v ∈ R2 , the directional derivative

limt→0

g(tv)− g(0)

t= B(v)

exists in R2 . Furthermore, show that the function v 7→ B(v) is linear. But show that nevertheless,g is not differentiable at 0.

88


Definition. Let X and Y be metric spaces, let f, g : X → Y , and let a ∈ X . To say that f and g aretangent at a means that for each ε > 0, there exists δ > 0 such that for each x ∈ BX(a, δ), we have

dY (f(x), g(x)) ≤ εdX(x, a).

Notice that if a is not an isolated point of X , then f and g are tangent at a iff f(a) = g(a) and

limx→a

dY (f(x), g(x))

dX(x, a)= 0.

X349. Let X and Y be Banach spaces, let Ω be an open subset of X , let f : Ω → Y , let a ∈ Ω, letT ∈ L (X,Y ), and define g : Ω → Y by g(x) = f(a) + T (x− a). Show that the following are equivalent:

(a) The function f is differentiable at a and f ′(a) = T .

(b) The functions f and g are tangent at a .

X350. Let X and Y be metric spaces and let a ∈ X . Show that the relation of being tangent at a is anequivalence relation on the set of all functions from X into Y .

X351. Let X and Y be metric spaces, let f, g : X → Y , and let a ∈ X . Suppose that f and g are tangentat a . Prove that f is continuous at a iff g is continuous at a .

Definition. Let X and Y be metric spaces, let f : X → Y , and let a ∈ X . To say that f satisfiesStepanoff’s condition at a means that there exist C, δ ∈ (0,∞) such that for each x ∈ BX(a, δ), we havedY (f(x), f(a)) ≤ CdX(x, a).

Evidently, if f is a Lipschitz function, then f satisfies Stepanoff’s condition at each point. It is alsoobvious that if f satisfies Stepanoff’s condition at a , then f is continuous at a .

X352. Let X and Y be metric spaces, let f, g : X → Y , and let a ∈ X . Suppose that f and g are tangentat a . Prove that f satisfies Stepanoff’s condition at a iff g satisfies Stepanoff’s condition at a .

X353. Let X , Y , and Z be metric spaces and let a ∈ X . Let f, F : X → Y be tangent at a . Let b = f(a).Note that b = F (a) too. Let g,G : Y → Z be tangent at b . In addition, suppose that F satisfies Stepanoff’scondition at a and that G is a Lipschitz function. Let h = g f and H = G F . Prove that h and H aretangent at a .

X354. Use problem X353 to give a proof of the chain rule.

Remark. Let X be a Banach space over C . Then we may view X as a Banach space over R simply byforgetting that we can multiply vectors in X by scalars that are not real.

Remark. Let X and Y be Banach spaces over C , let Ω be an open subset of X , let f : Ω → Y , andlet a ∈ Ω. Then differentiability of f at a means differentiability in the complex sense, but we may alsospeak of differentiability of f at a in the real sense, just by viewing X and Y as Banach spaces over R ,as described in the previous remark. Suppose f is differentiable in the complex sense at a . Let T be thederivative of f at a in the complex sense. Then T is a continuous complex-linear map from X to Y . Butthen T is also real-linear and clearly f is differentiable in the real sense at a and the derivative of f at ain the real sense is also T .

X355. Let X and Y be Banach spaces over C , let Ω be an open subset of X , let f : Ω → Y , and leta ∈ Ω. Suppose that f is differentiable at a in the real sense. Let T be the derivative of f at a in the realsense. Note that T is a continuous real-linear map from X to Y . Prove that the following are equivalent:

(a) The map f is differentiable at a in the complex sense.

(b) For each w ∈ X , T (iw) = iT (w), where i =√−1.

(c) The map T is complex-linear.

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X356. View C as a two-dimensional vector space over R . Let T be a real-linear map from C to C . Provethat the following are equivalent:

(a) The map T is complex-linear.

(b) There is a complex number γ such that for each w ∈ C , T (w) = γw .

(c) The matrix of T relative to the basis

1, i

is of the form

[T ] =

(

α −ββ α

)

for suitable α, β ∈ R . (Hint: You should find that α = Re(γ) and β = Im(γ), where γ is as in(b).)

X357. Let Ω be an open subset of C , let f : Ω → C , let u = Re(f), let v = Im(f), and let p ∈ Ω. Provethat the following are equivalent:

(a) f is differentiable at p in the complex sense.

(b) The limit

limz→p

f(z)− f(p)

z − p

exists in C .

(c) f is differentiable at p in the real sense and f satisfies the Cauchy-Riemann equations at p :

∂u

∂x(p) =

∂v

∂y(p) and

∂u

∂y(p) = −∂v

∂x(p).

Show furthermore that if f is differentiable at p in the complex-sense, and if γ is the value of the limit in(b), then for each w ∈ C , we have f ′(p)(w) = γw .

X358. (Uniqueness in the implicit function theorem.) Let X , Y , and Z be Banach spaces, let Ω be anShould f becontinuous?

Should f becontinuous?

open subset of X × Y , let f : Ω → Z , and suppose D2f exists and is continuous throughout Ω. Let

Ω1 = (a, b) ∈ Ω : D2f(a, b) ∈ G L (Y, Z) .

Note that Ω1 is open in X×Y . Suppose U is a connected open subset of X , h1, h2 : U → Y are continuous,h1(a) = h2(a) for some a ∈ U , and for each x ∈ U , we have (x, h1(x)) ∈ Ω1 , (x, h2(x)) ∈ Ω1 ,

f(x, h1(x)) = 0 and f(x, h2(x)) = 0.

Prove that h1 = h2 .

X359. In problem X358, it is natural to ask what happens if instead of requiring that (x, hk(x)) ∈ Ω1

for k = 1, 2, we require only that (x, hk(x)) ∈ Ω for k = 1, 2. Investigate this question for the functionf : R2 → R defined by

f(x, y) =(

x2 + (y − 1)2 − 1)(

x2 + (y − 2)2 − 4)

.

You should find that there is more than one C1 solution of the equation f(x, y) = 0 passing through thepoint (x, y) = (0, 0).

Remark. In Chapter 9 of Rudin, Principles of Mathematical Analysis, Third Edition, the inverse functiontheorem (Theorem 9.24) is used to prove the implicit function theorem (Theorem 9.28). This approachrequires stronger differentiability assumptions in the implicit function theorem than the approach we tookin class. For this reason, it is better to reverse Rudin’s order and use the implicit function theorem to provethe inverse function theorem. The next two exercises outline this approach to the inverse function theorem.

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X360. (The Inverse Function Theorem, Part 1 — Local Existence and Uniqueness; Continuity). Let Xand Y be Banach spaces, let E be an open subset of X , let f : E → Y be differentiable. Let a ∈ E suchthat f ′(a) ∈ G L (X,Y ) and let b = f(a). Suppose f ′ : E → L (X,Y ) is continuous at a . Prove that thereexist open sets U ⊆ X and V ⊆ Y such that a ∈ U ⊆ E , b ∈ V , the restriction of f to U is a one-to-onemap from U onto V , and the inverse g of this restriction is continuous. (Hint: Let Ω = E × Y . Then Ω isan open subset of X × Y and (a, b) ∈ Ω. Let Z = Y . Define ϕ : Ω → Z by ϕ(x, y) = y − f(x). Apply theimplicit function theorem to ϕ near the point (a, b).)

X361. (The Inverse Function Theorem, Part 2 — Smoothness). Let X and Y be Banach spaces, let E bean open subset of X , and let f : E → Y be continuous. Let V be an open subset of Y and let g : V → Esuch that for each y ∈ V , f(g(y)) = y . Let U = g[V ] and let ϕ be the restriction of f to U . Notice thatϕ is a one-to-one map from U onto V .

(a) Let b ∈ V such that g is continuous at b . Let a = g(b). Suppose that f is differentiable at a andthat f ′(a) ∈ G L (X,Y ). Prove that g is differentiable at b and that g′(b) = f ′(a)−1 .

For the remainder of this exercise, suppose in addition that g is continuous in V , f is differentiable at eachpoint of U , that the restriction of f ′ to U is continuous, and that for each x ∈ U , f ′(x) ∈ G L (X,Y ).

(b) Prove that g is C1 in V and that U is open in X .(c) Let k ∈ N . Suppose that f is Ck in U . Prove that g is Ck in V .(d) Suppose that f is C∞ in U . Prove that g is C∞ in V

Definition. The rank of a linear map is the dimension of its range.

X362. Let Ω be an open subset of Rn , let f : Ω → Rm be continuously differentiable, let a ∈ Ω, and let rbe the rank of f ′(a). Show that a has a nhd U ⊆ Ω such that for each x ∈ U , the rank of f ′(x) is at leastr . Thus the rank of f ′(x) is a lower semicontinuous function of x . (Hint: Find linear maps S : Rr → Rn

and T : Rm → Rr such that T f ′(a) S ∈ G L (Rr).)

About the Rank Theorem. Let n,m, r ∈ N with r ≤ min n,m . Define L : Rn → Rm by

L(x1, . . . , xr, xr+1, . . . , xn) = (x1, . . . , xr, 0, . . . , 0).

Then L is a linear map and L has rank r . The rank theorem tells us that if Ω is an open subset of Rn andf : Ω → Rm is a Ck function, and if a ∈ Ω such that f ′(x) has rank r for all x in some nhd of a , then itis possible to introduce Ck curvilinear coordinates in Rn near a and in Rm near f(a) in such a way thatnear a , in terms of these coordinates, the map f looks like the linear map L .

X363. Define f : R2 → R2 byf(x, y) = (x2 − y2, 2xy).

(By the way, under the usual identification of R2 with C , f can also be described as the map z 7→ z2 .)(a) Find the rank of f ′(a, b), if (a, b) ∈ R2 \ (0, 0) .(b) Find the rank of f ′(0, 0).(c) Explain how the conclusion of the rank theorem breaks down for f near the point (0, 0).

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Math 651 Extra Problems Autumn, 2005 · 2017-12-22 · Math 651 Extra Problems Autumn, 2005 X7. Let...

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