Real Analysis - Homework solutions

Chris Monico, May 2, 2013

1.1 (a) Rings (resp. σ-rings) are closed under finite (resp. countable) intersections.(b) If R is a ring (resp. σ-ring) then R is an algebra (resp. σ-algebra) iff X ∈ R.(c) If R is a (nonempty) σ-ring then {E ⊂ X : E ∈ R or Ec ∈ R} is a σ-algebra.(d) If R is a σ-ring, then {E ⊂ X : E ∩ F ∈ R for all F ∈ R} is a σ-algebra.

Solution:(a) If R is a ring and E1, E2 ∈ R, then since R is closed under differences, E1 − (E1 − E2) ∈ R. But

E1 − (E1 − E2) = E1 − (E1 ∩ Ec2) = E1 ∩ (Ec1 ∪ E2) = E1 ∩ E2.

It follows inductively that R is closed under finite intersections. Suppose now that R is a σ-ring and{En} ⊂ R. Let A =

⋃Ej and En = A− En. Then En ∈ R for all n ≥ 1 and so F = E1 −

⋃∞j=2 Ej is also

in R. But

F = E1 ∩

∞⋂j=2

Ejc

= E1 ∩

∞⋂j=2

(Ac ∪ Ej)

=

∞⋂j=2

((E1 ∩Ac) ∪ (E1 ∩ Ej)) =

∞⋂j=2

(E1 ∩ Ej) =

∞⋂j=1

Ej .

(b) If R is a ring (resp. σ-ring) and X ∈ R, then for all E ∈ R we have Ec = X −E ∈ R so R is closedunder complements and therefore an algebra (resp. σ-algebra). On the other hand if R is an algebra (resp.σ-algebra), then it’s nonempty so there exists E ∈ R and hence X = E ∪ Ec ∈ R.

(c) Suppose R is a nonempty σ-ring and M = {E ⊂ X : E ∈ R or Ec ∈ R}. Clearly M is closedunder complements. Suppose that {En} ⊂ M. By part (a), R is closed under countable intersections so

A =⋂j≥1Ec

j∈R

Ecj ∈ R. Since R is closed under countable unions, B =⋃j≥1Ej∈R

Ej ∈ R. Now

⋃j≥1

Ej =

⋃j≥1Ec

j∈R

Ej

∪ ⋃

j≥1Ej∈R

Ej

=

⋂

j≥1Ec

j∈R

Ecj

∩ ⋃

j≥1Ej∈R

Ej

cc

= (A−B)c.

Since A,B ∈ R and R is closed under differences, A− B ∈ R, hence⋃Ej = (A− B)c ∈ M. Therefore M

is also closed under countable unions, hence it’s a σ-algebra.(d) Suppose R is a σ-ring and A = {E ⊂ X : E ∩ F ∈ R for all F ∈ R}. Suppose E ∈ A and let

F ∈ R. Then E ∩ F ∈ R and since R is closed under differences, Ec ∩ F = F −E = F − (E ∩ F ) ∈ R, andtherefore A is closed under complements. If {En} ⊂ A and F ∈ R then En ∩ F ∈ R and so (

⋃En) ∩ F =⋃

(En ∩ F ) ∈ R for all n, so A is closed under countable unions as well and therefore is a σ-algebra.

1.3 Let M be an infinite σ-algebra. (a) M contains an infinite sequence of (distinct) disjoint sets.(b) card(M) ≥ c.

Solution:Note: the word ’distinct’ here is not given as part of the problem, but part (a) becomes trivial without it,and it is extremely helpful in solving part (b) anyway.

The elements of M are partially ordered by inclusion. By the Hausdorff Maximal Principle, there isa maximal linearly ordered subset {Eα}α∈A ⊂ M. By way of contradiction, suppose that A is finite, sayA = {1, 2, . . . , n}. Since E1 ⊂ E2 ⊂ · · · ⊂ En, the σ-algebraM({E1, . . . , En}) is clearly finite so there existsa set E ∈ M−M({E1, . . . , En}). We must have E ⊂ En, otherwise E1 ⊂ E2 ⊂ · · · ⊂ En ⊂ En ∪ E wouldcontradict the maximality of {Eα}. Now since

E = E ∩ En = E ∩

E1 ∪n⋃j=2

(Ej − Ej−1)

= (E ∩ E1) ∪n⋃j=2

(E ∩ (Ej − Ej−1)),

it follows that either E ∩E1 6∈ M({E1, . . . , En}) or E ∩ (Ej −Ej−1) 6∈ M({E1, . . . , En}) for some j ≥ 2. Inthe first case,

E ∩ E1 ⊂ E1 ⊂ E2 ⊂ · · · ⊂ En,

would contradict the maximality of {Eα}. In the second case, letting E′ = Ej−1∪ (E∩ (Ej−Ej−1)) we have

E1 ⊂ · · · ⊂ Ej−1 ⊂ E′ ⊂ Ej ⊂ · · · ⊂ En.

But E′ = Ej−1 would imply that E ∩ (Ej − Ej−1) = ∅ ∈ M({E1, . . . , En}), so E′ 6= Ej−1. And E′ = Ejwould imply that E ∩ (Ej −Ej−1) = Ej −Ej−1 ∈M({E1, . . . , En}), so E′ 6= Ej . Therefore, this contradictsthe maximality of {Eα}, and hence A must be infinite.

So there must exist an infinite increasing sequence {En} of distinct sets in M. Letting F1 = E1 andFk+1 = Ek+1 − Ek for all k ≥ 1, it follows that {Fn} is an infinite sequence of distinct disjoint sets in M.

(b) For each x ∈ [0, 1) fix a binary representation x =∑∞k=1 bk2−k and let F1, F2, . . . be an infinite

sequence of distinct disjoint sets in M. Define a function

ψ : [0, 1) −→ M∑bk2−k 7−→

⋃k≥1bk=1

Fk.

Then ψ is an injection and hence card(M) ≥ card([0, 1)) = c.

1.5 IfM =M(E) thenM is the union of all σ-algebras generated by F as F ranges over all countablesubsets of E . Hint: show that the latter object is a σ-algebra.

Solution:Let M′ =

⋃F⊂E

countable

M(F). As per the hint, we’ll first show that M′ is a σ-algebra. Suppose {En} ⊂ M′.

For each n ≥ 1 there is a countable subset Fn ⊂ E such that En ∈ M(Fn). Then F =⋃Fn is a countable

subset of E and for all k ≥ 1, Fk ⊂ F implies M(Fn) ⊂M(F), so Ek ∈M(Fk) ⊂M(F). Since M(F) is aσ-algebra,

⋃Ek ∈M(F). And since F is countable, M(F) ⊂M′.

Now if E ∈ M′ then there exists a countable subset F ⊂ E such that E ∈ M(F), in which caseEc ∈M(F) ⊂M′. Therefore, M′ is a σ-algebra.

For all F ⊂ E , M(F) ⊂ M(E), so M′ ⊂ M(E). On the other hand if E ∈ E then E is contained inM({E}) and {E} is countable, so E ∈M′ and hence E ⊂M′, so that M(E) ⊂M′.

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1.6 Complete the proof of Theorem 1.9.Solution:It’s routine to verify that µ is a complete measure, so we’ll show only the uniqueness. For this, suppose µ′

is another measure defined on M which extends µ. Then for E ∪ F ∈M with E ∈M and F ⊂ N ∈ N , wehave

µ′E ≤ µ′(E ∪ F ) ≤ µ′(E ∪N) = µ(E ∪N) (since E ∪N ∈M )

≤ µE + µN = µE = µ′E (since E ∈M).

Therefore µ′(E ∪ F ) = µE = µ(E ∪ F ), and hence µ′ = µ.

1.9 If (X,M, µ) is a measure space and E,F ∈M then µE + µF = µ(E ∪ F ) + µ(E ∩ F ).Solution:Let E,F ∈M. Then E − F,E ∩ F , and F − E are disjoint elements of M so

µE + µF = (µ(E − F ) + µ(E ∩ F )) + (µ(F − E) + µ(F ∩ E))

= (µ(E − F ) + µ(E ∩ F ) + µ(F − E)) + µ(F ∩ E)

= µ(E ∪ F ) + µ(E ∩ F ).

1.10 Given a measure space (X,M, µ) and E ∈ M, define µE(A) = µ(A ∩ E) for all A ∈ M. ThenµE is a measure.

Solution:Certainly µE∅ = µ(∅ ∩ E) = µ(∅) = 0. Suppose now that {An}n≥1 ⊂M is a disjoint collection. Then since{E ∩An}n≥1 ⊂M is a disjoint collection it follows that

µE

⋃n≥1

An

= µ

E ∩⋃n≥1

An

= µ

⋃n≥1

(E ∩An)

=∑n≥1

µ(E ∩An) =∑n≥1

µEAn.

So µE is countably additive and hence a measure.

1.13 Every σ-finite measure is semifinite.Solution:Suppose (X,M, µ) is a measure space and µ is σ-finite. If µ is actually finite then the statement is vacuouslytrue, so suppose there exists an E ∈M with µE =∞. Since µ is σ-finite, there is a collection {Xn}n≥1 ⊂Mwith X =

⋃Xn and µXn <∞ for all n. Then

∞ = µE = µ(E ∩X) = µ(⋃

(E ∩Xn))≤∑

µ(E ∩Xn),

so∑µ(E ∩ Xn) = ∞. In particular, there is a k ≥ 1 for which µ(E ∩ Xk) > 0. On the other hand,

∞ > µ(Xk) ≥ µ(E ∩Xk) > 0, so E ∩Xk is a measurable subset of E with positive finite measure. Thus µis semifinite.

1.14 If µ is a semifinite measure and µE = ∞ then for every C > 0 there exists F ⊂ E withC < µF <∞.

Solution:Let C = {F ⊂ E : µF < ∞} and α = sup C. By way of contradiction, suppose that α < ∞. For eachn ≥ 1 there is an Fn ∈ C such that α ≥ µFn ≥ α − 1/n. Define Gn =

⋃nk=1 Fk. Then Gn ⊂ E and

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µGn ≥ µFn ≥ α−1/n for all n. By continuity of measure, µ(⋃

n≥1Gn

)= limn→∞ µGn and since α ≥ µGn

for all n ≥ 1, we haveα ≥ limµGn ≥ lim(α− 1/n) = α,

so that µ (⋃Gn) = α, and hence

⋃Gn ∈ C. But

∞ = µE = µ(⋃

Gn

)+ µ

(E −

⋃Gn

)= α+ µ

(E −

⋃Gn

),

so µ (E −⋃Gn) = ∞. Since µ is semifinite there exists F ′ ⊂ (E −

⋃Gn) with 0 < µF ′ < ∞. Then

µ (F ′ ∪⋃Gn) = µF ′ + α > α. Since F ′ ∪

⋃Gn ∈ C, this contradicts the fact that α = sup C, and therefore

α =∞.

1.17 If µ∗ is an outer measure on X and {Aj} is a sequence of disjoint µ∗-measurable sets thenµ∗ (E ∩

⋃Aj) =

∑µ∗(E ∩Aj) for all E ⊂ X.

Solution:Let {Aj} be such a sequence and E ⊂ X. By countable subadditivity,

µ∗(E ∩

(⋃Aj

))= µ∗

(⋃(E ∩Aj)

)≤∑

µ∗(E ∩Aj).

Set Bn =⋃nj=1Aj . For each n ≥ 2 since An is µ∗-measurable we have

µ∗(E ∩Bn) = µ∗(E ∩Bn ∩An) + µ∗(E ∩Bn ∩Acn)

= µ∗(E ∩An) + µ∗(E ∩Bn−1).

By induction, µ∗(E∩Bn) =∑nj=1 µ

∗(E∩An) for all n ≥ 1. By monotonicity of outer measure, µ∗(E ∩

(⋃j≥1Aj

))≥

µ∗(E ∩Bn) =∑nj=1 µ

∗(E ∩An) for all n ≥ 1, and so µ∗(E ∩

(⋃j≥1Aj

))≥∑∞j=1 µ

∗(E ∩An).

1.18 Let A ⊂ P(X) be an algebra, µ0 a premeasure on A and µ∗ the induced outer measure.(a) For all E ⊂ X and ε > 0 there exists A ∈ Aσ with E ⊂ A and µ∗A ≤ µ∗E + ε.(b) If µ∗E <∞ then E is µ∗-measurable iff there exists B ∈ Aσδ with E ⊂ B and µ∗(B−E) = 0.(c) If µ0 is σ-finite, the restriction µ∗E <∞ in (b) is superfluous.

Solution:Let E ⊂ X and ε > 0. By definition, there is a sequence {An} ⊂ A so that E ⊂

⋃An and

∑µ0An ≤ µ∗E+ε.

Let A =⋃An. Then A ∈ Aσ and

µ∗A ≤∑

µ∗An =∑

µ0An ≤ µ∗E + ε.

For (b), suppose first that E is µ∗-measurable with µ∗E < ∞. By Part (a), for each n ≥ 1 thereexists Bn ∈ Aσ such that E ⊂ Bn and µ∗Bn ≤ µ∗E + 1/n. Set B =

⋂Bn. Then E ⊂ B ∈ Aσδ and

µ∗B ≤ µ∗Bn ≤ µ∗E + 1/n for all n ≥ 1, so µ∗B ≤ µ∗E. But since E ⊂ B it follows that µ∗B = µ∗E. SinceE is µ∗-measurable,

µ∗E = µ∗B = µ∗(B ∩ E) + µ∗(B ∩ Ec)= µ∗E + µ∗(B − E).

Since µ∗E <∞, it follows that µ∗(B − E) = 0. Conversely, suppose that there exists B ∈ Aσδ with E ⊂ Band µ∗(B − E) = 0. Then B is µ∗ measurable since B ∈ Aσδ and B − E is µ∗-measurable by the proof ofCaratheodory’s Theorem. So (B − E)c is µ∗-measurable as is

B ∩ (B − E)c = B ∩ (Bc ∪ E) = B ∩ E = E.

Finally, for c suppose that µ0 is σ-finite. Then so is µ∗, so there is a disjoint sequence {Xn} of µ∗-measurable sets with µ∗Xn <∞ and X =

⋃Xn.

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Suppose E ⊂ X is µ∗-measurable and ε > 0. Then each E ∩Xn is measurable with finite measure, andby Part (a) there exists An,k ∈ Aσ such that E ∩Xn ⊂ An,k and µ∗An,k < µ∗(E ∩Xn) + 1

k2n . Since E ∩Xn

is measurable,

µ∗An,k = µ∗(An,k ∩ (E ∩Xn)) + µ∗(An,k ∩ (E ∩Xn)c)

= µ∗(E ∩Xn) + µ∗(An,k − (E ∩Xn)),

and since µ∗(E ∩Xn) < ∞, we thus have that µ∗(An,k − (E ∩Xn)) = µ∗An,k − µ∗(E ∩Xn) < 1k2n . Thus,

with Bk =⋃∞n=1An,k we have Bk ∈ Aσ, E ⊂ Bk, and

µ∗(Bk − E) = µ∗(Bk ∩ Ec) = µ∗

( ∞⋃n=1

(An,k ∩ Ec)

)

≤∞∑n=1

µ∗(An,k − E) ≤∞∑n=1

µ∗(An,k − (E ∩Xn)) ≤∞∑n=1

1

k2n=

1

k.

Now set B =⋂∞k=1Bk. Then E ⊂ B ∈ Aσδ and µ∗(B − E) ≤ µ∗(Bk − E) ≤ 1

k for all k ≥ 1, henceµ∗(B−E) = 0. The converse follows from our proof of the corresponding implication in Part (b) (which didnot make use of the µ∗E <∞ assumption).

1.20 Let µ∗ be an outer measure on X, M∗ the σ-algebra of µ∗-measurable sets, µ = µ∗|M∗ , and µ+

the outer measure induced by µ.(a) If E ⊂ X then µ∗E ≤ µ+E with equality iff ∃A ∈M∗ with E ⊂ A and µ∗A = µ∗E.(b) If µ∗ is induced by a premeasure then µ∗ = µ+.(c) If X = {0, 1} there exists an outer measure µ∗ on X such that µ∗ 6= µ+.

Solution:Let E ⊂ X. Then µ+E = inf{

∑µAn : {An} ⊂ M∗, E ⊂

⋃An}. If {An} ⊂ M∗ is a collection with

E ⊂⋃An then

µ∗E ≤ µ∗(⋃

An

)(by monotonicity)

≤∑

µ∗An (by countable subadditivity)

=∑

µAn (since µ = µ∗|M∗).

It follows that µ∗E ≤ µ+E.Suppose now that µ∗E = µ+E. Then for each k ≥ 1 there exists {An,k} ⊂ M∗ with E ⊂

⋃∞n=1An,k and

µ∗

⋃n≥1

An,k

≤∑n≥1

µAn,k ≤ µ+E +1

k.

Then with A =⋂k≥1

(⋃n≥1An,k

), we have A ∈ M∗, E ⊂ A, and µ∗A ≤ µ+E + 1/k for all k, hence

µ∗A ≤ µ+E = µ∗E. But since E ⊂ A we have also that µ∗E ≤ µ∗A and so µ∗A = µ∗E. Conversely,suppose A ∈ M∗ with A ⊃ E and µ∗A = µ∗E. Then by definition of µ+ and since A ∈ M∗ we haveµ+E ≤ µA = µ∗A, so µ+E = µ∗A.

For (b), suppose µ∗ is induced by a premeasure on an algebra A ⊂ X and let E ⊂ X. As per thehint, we invoke Exercise 18a to conclude that for each n ≥ 1 there exists An ∈ Aσ with An ⊃ E andµ∗An ≤ µ∗E+ 1/n. Set A =

⋂n≥1An ∈ Aσδ ⊂M∗ and it follows that A ⊃ E and µ∗A ≤ µ∗E. The reverse

inequality follows by monotonicity, so µ∗A = µ∗E, and so by Part (a), µ∗E = µ+E.For (c), let X = {0, 1}. Define µ∗∅ = 0, µ∗{0} = 1, µ∗{1} = 2, and µ∗{0, 1} = 2. Then µ∗ is an outer

measure on X. Let M∗ be the σ-algebra of µ∗-measurable sets. Since

µ∗(X ∩ {0}) + µ∗(X ∩ {0}c) = 3 > µ∗X,

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it follows that {0} 6∈ M∗. Similarly, {1} 6∈ M∗, so M∗ = {∅, X} and µ = µ∗|M∗ . Now

µ+{0} = inf{∑

µEn : {En} ⊂ M∗ and {0} ⊂⋃En

}= µX = 2.

But µ∗{0} = 1, so µ∗ 6= µ+.

1.23 Let A be the collection of finite unions of sets of the form (a, b] ∩Q, where −∞ ≤ a < b ≤ ∞ (ifb =∞, then it is understood that (a, b] = (a,∞)).(a) A is an algebra on Q.(b) The σ-algebra generated by A is P(Q).(c) Define µ0 on A by µ0∅ = 0 and µ0A =∞ for A 6= ∅. Then µ0 is a premeasure on A and thereis more than one measure on P(Q) whose restriction to A is µ0.

Solution:As per the hint for (a), let E = {∅}∪{(a, b]∩Q : −∞ ≤ a < b ≤ ∞}. Clearly E is closed under intersectionsand for ∅ 6= A ∈ E we have that A = (a, b] ∩Q for some a < b. The complement of A is taken with respectto Q (since the claim is that A is a σ-algebra on Q), and so Ac = (−∞, a] ∪ (b,∞] is a disjoint union ofmembers of E . Therefore E is an elementary family and so the collection of finite disjoint unions of membersof E is an algebra by Proposition 1.7. But every finite union of sets of the form (a, b] can be written as adisjoint union of such sets, and hence this algebra is precisely A.

For (b), suppose that r ∈ Q. Then {r} =⋂n≥1 ((r − 1/n, r] ∩Q) ∈ M(A). If X ⊂ Q, then X is

countable, say X = {x1, x2, . . . }. But {xj} ∈ M(A) for all j ≥ 1, so X =⋃j≥1{xj} is in M(A). Therefore

M(A) = P(Q).It’s clear that µ0 is a premeasure on A. The induced outer measure is

µ∗E = inf

∑n≥1

µ0En : {En} ⊂ A, E ⊂⋃En

=

{∞, if E 6= ∅0, otherwise.

It follows by Theorem 1.14 that µ = µ∗ is a measure on P(Q) which extends µ0.If ν is the counting measure on P(Q), then ν∅ = 0 = µ0∅. And for a < b, E = (a, b] ∩ Q has infinite

cardinality, so νE = ∞ = µ0E, and therefore ν|E = µ0|E . By finite additivity of ν, it follows that νE =∞ = µ0E for all ∅ 6= E ∈ A, so that ν|A = µ0. But ν{1} = 1 < ∞ = µ{1}, so ν is a different measure onP(Q) which extends µ0.

1.25 Complete the proof of Theorem 1.19.Solution:The only thing left to do is to show that (a) implies (b) and (c) in the general case, having already shownit for for sets of finite measure, so let E ∈ Mµ. For j ∈ Z let Ej = E ∩ (j, j + 1]. Then µEj ≤ µ(j, j +1] = F (j + 1) − F (j) < ∞. Thus there are Gδ sets Vj and null sets Nj so that Ej = Vj − Nj . SinceEj = Ej ∩ (j, j + 1] = (Vj ∩ (j, j + 1])−Nj and (j, j + 1] is Gδ, we may further assume that Vj ⊂ (j, j + 1]and Nj ⊂ (j, j + 1]. With this assumption, it follows that

E =⋃j∈Z

Ej =⋃j∈Z

(Vj ∩N cj ) =

⋃j∈Z

Vj

−⋃j∈Z

Nj

.

A countable union of null sets is null, so we need only show that⋃j∈Z Vj is a Gδ set. Clearly, for each j ∈ Z,

the set V ′j = Vj ∪ (j, j + 1]c is Gδ, and so is⋂j∈Z V

′j . But for each i ∈ Z we have that

(i, i+ 1] ∩⋂j∈Z

V ′j =⋂j∈Z

((Vj ∪ (j, j + 1]c) ∩ (i, i+ 1]) = Vi.

Therefore⋂j∈Z V

′j =

⋃j∈Z Vj is Gδ. The proof that (a) implies (c) is similar.

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1.26 Prove Proposition 1.20.Solution:See Exam 1, Problem #4.

1.27 Prove Proposition 1.22a.Solution:This is straightforward - just argue on the ternary expansions.

1.28 Let F be increasing and right-continuous, and let µF be the associated measure. Then µF {a} =F (a)−F (a−), µF [a, b) = F (b−)−F (a−), µF [a, b] = F (b)−F (a−), and µF (a, b) = F (b−)−F (a).

Solution:Let a, b ∈ R with a < b. Since F is increasing and F (a) <∞, the limit F (a−) exists and limn→∞ F (a−1/n) =F (a−). Since µF (a− 1, a] <∞, we have by continuity of measure that

µF {a} = µF

⋂n≥1

(a− 1/n, a]

= limn→∞

µF (a− 1/n, a] = limn→∞

(F (a)− F (a− 1/n)) = F (a)− F (a−).

The remaining properties are easily shown using this; for example,

µF [a, b) = µF (a, b] + µF {a} − µF (b) = F (b)− F (a) + F (a)− F (a−)− F (b) + F (b−) = F (b−)− F (a−).

1.29 Let E be a Lebesgue measurable set.(a) If E ⊂ N where N is the nonmeasurable set from §1.1, then mE = 0.(b) If mE > 0 then E contains a nonmeasurable set.

Solution:

Suppose E ⊂ N is measurable. In the notation of §1.1, let

Er = {x+ r : x ∈ E ∩ [0, 1− r)} ∪ {x+ r − 1 : x ∈ E ∩ [1− r, 1).

Then for all r ∈ R = Q ∩ [0, 1), since Er ⊂ Nr and the collection {Nr}r∈R is a disjoint collection, it followsthat {Er}r∈R is a disjoint collection. By the translation invariance and finite additivity of Lebesgue measure,we have that each Er is measurable and mEr = mE for all r ∈ R. So

1 ≥ m

(⋃r∈R

Er

)=∑r∈R

mEr =∑r∈R

mE,

and therefore mE = 0.For (b), suppose that E ⊂ [0, 1] is a subset with the property that every subset of E is measurable. Then

for each r ∈ R, the set E ∩Nr is measurable. By the translation invariance and finite additivity of Lebesguemeasure, the set E1−r ∩ N is therefore measurable, and hence must have measure zero by Part (a). Sincethe collection {Nr}r∈R is disjoint we have that

mE = m

(⋃r∈R

(E ∩Nr)

)=∑r∈R

m(E ∩Nr) =∑r∈R

m(E1−r ∩N) = 0.

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1.30 If E ∈ L and mE > 0 then for every α < 1 there is an open interval I such that m(E ∩ I) > αmI.Solution:It’s clear if α ≤ 0, so assume that 0 < α < 1. Suppose first that E ∈ L with mE < ∞. By way of

contradiction suppose that m(E ∩ I) ≤ αmI for every open interval I. Let ε ∈(

0, (1−α)mEα

). Then there is

a disjoint collection {Ik}k≥1 of open intervals such that E ⊂⋃Ik and

∑mI)k ≤ mE + ε. It follows that

mE = m(E ∩

(⋃Ik

))= m

(⋃(E ∩ Ik)

)=

∑m(E ∩ Ik)

≤∑

αmIk

= α∑

mIk ≤ α(mE + ε) < αmE + (1− α)mE = mE,

a contradiction.Now if mE = ∞ then there is a k ∈ Z such that E′ = E ∩ (k, k + 1] has positive measure. So for each

α < 1 there is an open interval I such that m(E′ ∩ I) > αmI, and since m(E ∩ I) ≥ m(E′ ∩ I), we’re done.

1.31 If E ∈ L and mE > 0 then the set E − E = {x− y : x, y ∈ E} contains an interval centered atzero.

Solution:As per the hint, we invoke Problem #30 and let I = (x0 − α, x0 + α) be an interval such that m(E ∩ I) >(3/4)mI. Let 0 ≤ δ < α and suppose BWOC that δ 6∈ E − E. Then x− y 6= δ for all x, y ∈ E. Define

E1 = E ∩ (x0 − α, x0]

E2 = E ∩ (x0, x0 + α).

Then for all x ∈ E1, we have x+ δ ∈ I and x+ δ 6∈ E, so x+ δ ∈ I−E. Therefore E1 + δ ⊂ I−E. Similarly,we have that E2 − δ ⊂ I − E. Then E1 + δ ⊂ I − E implies

mE1 = m(E1 + δ) ≤ m(I − E) = mI −m(I ∩ E).

Similarly,mE2 = m(E2 − δ) ≤ m(I − E) = mI −m(I ∩ E).

It follows thatm(E∩I) = mE1+mE2 ≤ 2mI−2m(I∩E), and som(E∩I) ≤ (2/3)mI < (4/3)(2/3)m(E∩I) =(8/9)m(E ∩ I), a contradiction. Therefore δ ∈ E −E and so [0, α) ⊂ E −E. But z ∈ E −E iff −z ∈ E −E,so in fact, (−α, α) ⊂ E − E.

2.1 Let f : X −→ R and Y = f−1(R). Then f is measurable iff f−1({−∞}) ∈ M, f−1({∞}) ∈ Mand f is measurable on Y .

Solution:Suppose f is measurable. Then {∞}, {−∞} ∈ BR so f−1({−∞}), f−1({∞}) ∈ M. Let B ∈ BR. Thenf−1(B) ∈M. Since R ∈ BR, Y = f−1(R) ∈M hence f−1(B) ∩ Y ∈M so f is measurable on Y .

For the converse, let B ∈ BR. Set B1 = B∩R, B2 = B∩{−∞,∞}. Since f is measurable on Y , f−1(B1) =f−1(B1)∩Y ∈M and by hypothesis, f−1(B2) ∈M, so f−1(B) = f−1(B1∪B2) = f−1(B1)∪f−1(B2) ∈M.

8

2.3 If {fn} is a sequence of [complex-valued] measurable functions on X, then {x ∈ X :lim fn(x) exists } is a measurable set.

Solution:For n,N,m ≥ 1 let

En,N,m = {x ∈ X : |(fn − fN )(x)| < 1/m}= (fn − fN )−1(B1/m(0)),

where B1/m(0) is the open ball of radius 1/m centered at zero. Since fn − fN is measurable, En,N,m ismeasurable for all n,N,m ≥ 1, and hence

E =⋂m≥1

⋃N≥1

⋂n≥N

En,N,m,

is measurable. Furthermore, x ∈ E iff ∀m ≥ 1 ∃N ≥ 1 such that |fn(x) − fN (x)| < 1/m for all n ≥ N , soE = {x ∈ X : lim fn(x) exists }.

2.4 If f : X −→ R and f−1((r,∞]) ∈M for each r ∈ Q then f is measurable.Solution:For all a ∈ R,

(a,∞] =⋃r∈Qr>a

(r,∞],

so the σ-algebra generated by {(r,∞] : r ∈ Q} contains {(a,∞] : a ∈ R}, and hence it contains BR.Therefore f is measurable by Proposition 2.1.

2.5 If X = A ∪ B where A,B ∈ M, a function f on X is measurable iff f is measurable on both Aand B.

Solution:

Suppose f is measurable and B is a Borel set. Then f−1(E) ∈M and hence both f−1(E)∩A ∈M andf−1(E) ∩B ∈M, so f is measurable on A and B.

Conversely, suppose f is measurable on A and B and E is a Borel set. Then f−1(E) ∩ A ∈ M andf−1(E) ∩B ∈M so

f−1(E) = f−1(E) ∩X =(f−1(E) ∩A

)∪(f−1(E) ∩B

)∈M,

so f is measurable.

2.8 If f : R −→ R is monotone then f is Borel measurable.Solution:Without loss of generality, assume f is increasing. Suppose (a, b) is a finite open interval in R. We claimthat f−1(a, b) is an open interval. For this, suppose that x1, x2 ∈ f−1(a, b) and x1 < x2. Let x ∈ (x1, x2).Then since f is increasing,

a < f(x1) ≤ f(z) ≤ f(x2) < b,

so that z ∈ f−1(a, b) proving the claim. But the collection of finite open intervals in R generates BR, so f isBorel measurable by Proposition 2.1.

9

2.9 Let f : [0, 1] −→ [0, 1] be the Cantor-Lebesgue function from §1.5 and let g(x) = f(x) + x.a. g is a bijection from [0, 1] to [0, 2] and h = g−1 is continuous from [0, 2] to [0, 1].b. If C is the Cantor set, m(g(C)) = 1.c. By Exercise 1.29, g(C) contains a Lebesgue nonmeasurable set A. Let B = g−1(A). Then Bis Lebesgue measurable but not Borel.

Solution:

(a) Since f is increasing and a(x) = x is strictly increasing, it follows that g is strictly increasing, henceit is one-to-one. Since f is continuous, so is g, and so it’s clearly onto, hence bijective. Since g is strictlyincreasing and continuous, for 0 ≤ a < b ≤ 1 we have

h−1((a, b)) = g((a, b)) = (g(a), g(b)).

It follows that for every open O ⊂ [0, 1], h−1(O) is open in [0, 2], so h is continuous.(b) There is a collection {Ik} of disjoint closed intervals in [0, 1] so that C = [0, 1] −

⋃k≥1 Ik and∑

mIk = 1. Then [0, 1]−C =⋃Ik, and so g([0, 1]−C) = g (

⋃Ik) =

⋃g(Ik). Since g is one-to-one, {g(Ik)}

is a disjoint collection so

2 = m(g([0, 1])) = m(g(C)) +∑

m(g(Ik)). (0.1)

If Ik = [ak, bk] then f is constant on Ik by construction, so g(x) = f(x) + x = f(ak) + x for all x ∈ Ik, andthus g(Ik) = Ik + f(ak). Therefore m(g(Ik)) = mIk and we have from (0.1) that

2 = m(g(C)) +∑

mIk = m(g(C)) + 1,

hence m(g(C)) = 1.(c) Since B = g−1(A) ⊂ g−1(g(C)) = C and mC = 0, it follows that B is Lebesgue measurable. Since h :

[0, 2] −→ [0, 1] is continuous, it is measurable. If B ⊂ [0, 1] were Borel then h−1(B) = g(B) = g(g−1(A)) = Awould be measurable, so B cannot be Borel.

2.12 Prove Prop. 2.20: If f ∈ L+ and∫f <∞ then {x : f(x) =∞} is a null set and {x : f(x) > 0}

is a σ-finite set.Solution:Let E = {x : f(x) = ∞}. Then E is measurable and for all n ≥ 1, ϕn = nχ

Eis a simple function with

0 ≤ ϕn ≤ f , so ∫f ≥

∫ϕn = nµE.

Thus, µE ≤ 1n

∫f for all n ≥ 1. Since 0 ≤

∫f <∞ it follows that µE = 0.

Now let F = {x : f(x) > 0} and Fn = {x : f(x) > 1/n}. Then F =⋃n≥1 Fn. For each n,

ψn = (1/n)χFn

is a simple function with 0 ≤ ψn ≤ f , so∫f ≥

∫ψn = 1

nµFn, from which it follows that

µFn ≤ n∫f <∞ for all n, so F indeed is a σ-finite set.

2.13 Suppose {fn} ⊂ L+, fn −→ f pointwise, and∫f = lim

∫fn <∞. Then

∫Ef = lim

∫Efn for all

E ∈M. However, this need not be true if∫f = lim

∫fn =∞.

Solution:

Let A ∈M and note that fχA≤ f so

∫Af ≤

∫f <∞. By Theorem 2.15 we have that∫

f =

∫(fχ

A+ fχ

Ac) =

∫A

f +

∫Ac

f.

10

Since fnχA−→ fχ

A, we have from Fatou’s Lemma that∫

A

f =

∫(lim inf fnχ

A)

≤ lim inf

∫A

fn

= lim inf

(∫fn −

∫Ac

fn

)=

∫f − lim sup

∫Ac

fn.

All terms above are finite; from this and another application of Fatou’s Lemma,

lim sup

∫Ac

fn ≤∫f −

∫A

f =

∫Ac

f ≤ lim inf

∫Ac

fn.

Therefore, lim∫Ac fn exists and equals

∫Ac f , and hence lim

∫Efn =

∫Ef for all E ∈M.

To see that this need not be true when lim∫fn = ∞ =

∫f , consider (X,M) = (R,L) and fn =

nχ(0,1/n]

+ 1χ[1,∞)

. Then fn −→ χ[1,∞)

on R and lim∫fn =∞ =

∫f . But lim

∫[0,1]

fn = 1 6= 0 =∫[0,1]

f .

2.14 If f ∈ L+ let λE =∫Ef dµ for E ∈ M. Then λ is a measure on M and for every g ∈ L+,∫

g dλ =∫fg dµ. Hint: first suppose that g is simple.

Solution:First note that λ∅ =

∫∅ f dµ =

∫fχ∅dµ =

∫0 dµ = 0. Suppose that {En} is a disjoint sequence inM. Then

λ(⋃

En

)=

∫⋃En

f dµ =

∫fχ⋃

Endµ

=

∫ (∑fχ

En

)dµ =

∑∫fχ

Endµ (by Thm. 2.15)

=∑∫

En

f dµ =∑

λEn.

Therefore λ is a measure.Suppose that E ∈M. Then∫

χE

dλ =

∫E

dλ = λE =

∫E

f dµ =

∫fχ

Edµ.

It follows from linearity (for nonnegative functions and nonnegative constants) that∫ϕdλ =

∫fϕdµ for

every nonnegative simple function ϕ. Suppose now that g ∈ L+. By Thm. 2.10, there is a sequence {ϕn} ofincreasing, nonnegative simple functions converging pointwise to g. By the Monotone Convergence Theoremwe have ∫

g dλ = lim

∫ϕn dλ = lim

∫fϕn dµ.

But ϕn ≤ ϕn+1 and f ≥ 0 implies that fϕn ≤ fϕn+1, so {fϕn} is an increasing sequence of nonnegativefunctions tending to fg, so again applying the MCT we find that lim

∫fϕn dµ =

∫fg dµ.

2.15 If {fn} ⊂ L+, fn decreases pointwise to f , and∫f1 <∞, then

∫f = lim

∫fn.

Solution:Since fn is a decreasing sequence converging to f , we have f ≤ f1 and consequently

∫f ≤

∫f1 < ∞. Now∫

f +∫

(f1 − f) =∫f1, and since

∫f <∞, it follows that

∫(f1 − f) =

∫f1 −

∫f . It similarly follows that∫

(f1 − fn) =∫f1 −

∫fn. Since {f1 − fn} is an increasing sequence of nonnegative functions tending to

f1 − f , it follows from the MCT that∫f1 −

∫f =

∫(f1 − f) = lim

∫(f1 − fn) =

∫f1 − lim

∫fn.

11

2.16 If f ∈ L+ and∫f < ∞ then for every ε > 0 there exists E ∈ M such that µE < ∞ and∫

Ef >

(∫f)− ε.

Solution:Let ε > 0. By definition there is a simple function ϕ with 0 ≤ ϕ ≤ f and

∫ϕ >

(∫f)− ε. Since

∫ϕ < ∞,

the support of ϕ is finite; that is, if ϕ =∑nj=1 ajχEj

is the canonical representation, then aj = 0 for at

most one value of j, say a1 = 0. Then ∞ >∫ϕ =

∑ajµEj ≥ min{a2, . . . , an}(µE2 + . . . µEn), and so

E = E2 ∪ · · · ∪ En has finite measure. Then∫E

f ≥∫E

ϕ =

∫ϕ >

(∫f

)− ε.

2.17 Assume Fatou’s Lemma and deduce the MCT from it.Solution:Suppose that {fn} is an increasing sequence of nonnegative functions tending to f . By Fatou’s Lemma,∫

f =

∫(lim inf fn) ≤ lim inf

∫fn.

On the other hand, since fn ≤ f for all n ≥ 1,∫fn ≤

∫f for all n ≥ 1 and so lim sup

∫fn ≤

∫f . Thus,

lim sup

∫fn ≤

∫f ≤ lim inf

∫fn,

so lim∫fn exists and equals

∫f .

2.18 Fatou’s Lemma remains valid if the hypothesis that fn ∈ L+ is replaced by the hypothesis thatfn is measurable and fn ≥ −g where g ∈ L+ ∩ L1.

Solution:Suppose {fn} is a measurable sequence, g ∈ L+ ∩ L1 and fn ≥ −g for all n ≥ 1. Set fn = fn + g. Then

fn ≥ 0 so by Fatou’s Lemma∫g +

∫lim inf fn =

∫lim inf fn ≤ lim inf

∫fn =

∫g + lim inf

∫fn.

Since g ∈ L1,∫g is finite so it may be subtracted from both sides to obtain

∫lim inf fn ≤ lim inf

∫fn.

2.19 Suppose {fn} ⊂ L1(µ) and fn −→ f uniformly.(a) If µX <∞ then f ∈ L1(µ) and

∫fn −→

∫f .

(b) If µX =∞ the conclusions of (a) can fail.Solution:Suppose {fn} ⊂ L1, fn −→ f uniformly, and µX < ∞. Then there exists N such that |fn(x) − f(x)| < 1for all n ≥ N and all x ∈ X. Therefore∫

|f | ≤∫|f − fN |+

∫|fN | ≤

∫1 +

∫|fN | = µX +

∫|fN | <∞,

so f ∈ L1(µ). Set g = 1 + |f |. Then∫|g| = µX +

∫|f | < ∞, so g ∈ L1(µ) and |fn| ≤ g for all n ≥ N . So

by LDCT, lim∫fn =

∫f .

For (b), consider the measure space (R,L,m). For an example where the first conclusion fails, considerfn(x) = (1/x)χ

[1,n)(x) for n ≥ 2. Each fn is certainly in L1(m) and fn(x) −→ (1/x)χ

[1,∞)(x) uniformly on

R, but (1/x)χ[1,∞)

(x) 6∈ L1(m).

For an example where the first conclusion holds but the second fails, consider fn = (1/n)χ[0,n)

. Then

fn ∈ L1(m) for all n and fn −→ 0 uniformly and 0 ∈ L1(m). But∫fn = 1 for all n so

∫fn −→ 1 while∫

0 = 0.

12

2.20 If fn, gn, f, g ∈ L1, fn → f , gn → g a.e., |fn| ≤ gn, and∫gn →

∫g then

∫fn →

∫f .

Solution:As in the proof of Theorem 2.24, begin by taking real and imaginary parts, so that it suffices to considerthe case where all functions are real-valued. Then |fn| ≤ gn implies gn + fn ≥ 0 and gn − fn ≥ 0. Sincelim(fn + gn) = f + g, applying Fatou’s Lemma in each of these cases 1 we have∫

g +

∫f =

∫lim(gn + fn) ≤ lim inf

∫(gn + fn) =

∫g + lim inf

∫fn,

and ∫g −

∫f =

∫lim(gn − fn) ≤ lim inf

∫(gn − fn) =

∫g − lim sup

∫fn.

Since both∫g and

∫f are finite, it follows that

lim sup

∫fn ≤

∫f ≤ lim inf

∫fn,

so lim∫fn exists and equals

∫f .

2.22 Let µ be the counting measure on N. Interpret Fatou’s Lemma, MCT, and LDCT as statementsabout infinite series.

Solution:First note that every function f : N −→ C is measurable in this case since each singleton {n} with n ∈ Nis measurable and every subset of N is a countable union of singletons. Furthermore, since µ{n} = 1 for alln ∈ N, ∫

|f |dµ = sup0≤ϕ≤|f |

∫ϕdµ =

∞∑n=1

|f(n)|.

Thus, f ∈ L1 iff∑f(n) is absolutely convergent.

Fatou’s Lemma in this case translates to: If {ak,n} is a doubly-indexed nonnegative series with∑∞n=1 ak,n

absolutely convergent for every k ≥ 1, then

∞∑n=1

(limk

inf ak,n) ≤ limk

inf

∞∑n=1

ak,n.

The Monotone Convergence Theorem says: If {ak,n} is a doubly-indexed nonnegative series for whichak+1,n ≥ ak,n for all k, n ≥ 1 and

∑∞n=1 ak,n is absolutely convergent for every k ≥ 1 then

∞∑n=1

limk→∞

ak,n = limk→∞

∞∑n=1

ak,n.

Finally, the Lebesgue Dominated Convergence Theorem yields the result that: If {ak,n} is a doubly-indexed series of complex numbers with

∑∞n=1 ak,n absolutely convergent for every k ≥ 1 and there exists a

sequence {bn} with∑bn absolutely convergent and |ak,n| ≤ bn for all k, n ≥ 1, then limk→∞ ak,n exists for

each n and∞∑n=1

limk→∞

ak,n = limk→∞

∞∑n=1

ak,n.

1Note: It’s not generally true that lim inf(ak + bk) = lim inf ak + lim inf bk; produce a counterexample to convince yourselfof this fact. However, we are using in this case the result that if lim ak = a then lim inf(ak + bk) = a + lim inf bk. If you’reunfamiliar with this, you should prove it.

13

2.25 Let f(x) = x−1/2 for 0 < x < 1 and 0 otherwise. Let Q = {r1, r2, . . . } be an enumeration of therationals and set g(x) =

∑f(x− rn)/2n.

(a) g ∈ L1(m) and, in particular, g <∞ a.e..(b) g is discontinuous everywhere, unbounded on every interval, and remains so after any modi-fication on a Lebesgue null set.(c) g2 <∞ a.e. but g2 is not integrable on any interval.

Solution:Since f is measurable, so is each f(x−rn)/2n, and hence g is measurable as well. Notice that for each n ≥ 1,

f(x− rn) =

{ 1√x−rn

, if rn < x < 1 + rn,

0, otherwise.

So for each n ≥ 1, ∫f(x− rn)

2ndx =

1

2n

∫ 1+rn

rn

dx√x− rn

=1

2n−1.

By Theorem 2.15,∫|g(x)|dx =

∫ ∑ f(x− rn)

2ndx =

∑∫f(x− rn)

2ndx =

∞∑n=1

1

2n−1= 2,

so indeed g ∈ L1(m). By Prop. 2.20, g is finite a.e.(b) Let x0 ∈ R and suppose first that g(x0) is finite. Let ε > 0 and 0 < δ < 1. By the density of Q in R

there exists rn ∈ Q such that x0 < rn < x0 + δ. Let x′ ∈ (rn, x0 + δ) such that

1

2n√x′ − rn

> ε+ g(x0).

Then

g(x′) ≥ f(x′ − rn)

2n=

1

2n√x′ − rn

> ε+ g(x0), and |x′ − x0| < δ.

Since 0 < δ < 1 was otherwise arbitrary, g is discontinuous at x0. On the other hand, if g(x0) =∞ and δ > 0then there is an x′ ∈ (x0, x0 + δ) for which g(x′) is finite since g is finite a.e.. Therefore g is discontinuouson R.

Let I ⊂ R be a (nondegenerate) interval and rN ∈ Q a rational point in the interior of I. Let ε ∈ (0, 1)

such that (rN , rN + ε) ⊂ I, and let α > 12N ε

. Then for all x ∈(rN , rN +

(1

2Nα

)2)we have

g(x) =

∞∑n=1

f(x− rn)

2n≥ f(x− rN )

2N=

1

2N√x− rN

>1

2N(

12Nα

) = α.

Thus, for all sufficiently large α, m(g−1((α,∞))

)≥(

12Nα

)2> 0. This shows that g is unbounded on I and

remains so after any modification on a Lebesgue null set.(c) If g is finite on E then so is g2. Therefore g2 < ∞ a.e. Again, let I ⊂ R be an interval and rN ∈ Q

be a point in the interior of I. Let ε ∈ (0, 1) so that (rN , rN + ε) ⊂ I. Then∫I

g2 =

∫I

(∑ f(x− rn)

2n

)2

dx ≥∫I

(f(x− rN )

2N

)2

dx ≥∫ rN+ε

rN

dx

4N (x− rN )=

1

4N

∫ rN+ε

rN

dx

x− rN=∞,

so g2 is not integrable over I.

14

2.26 If f ∈ L1(m) and F (x) =∫ x−∞ f(t) dt then F is continuous on R.

Solution:Let x′ ∈ R and let {xn} be a sequence converging to x′. Set fn = fχ

(−∞,xn]. Then the sequence {fn}

converges pointwise to either fχ(−∞,x′)

or fχ(−∞,x′]

. In any case |fn| ≤ |f | and |f | ∈ L1(m), so by the LDCT

F (x′) =

∫ x′

−∞f(t) dt = lim

n→∞

∫fn = lim

n→∞

∫ xn

−∞f = lim

n→∞F (xn).

Since {xn} was an arbitrary sequence converging to x′, limx→x′ F (x) = F (x′), so F is continuous at x′, andhence continuous on R.

2.27 Let fn(x) = ae−nax − be−nbx where 0 < a < b.(a)

∑∞n=1

∫∞0|fn(x)|dx =∞.

(b)∑∞n=1

∫∞0fn(x) dx = 0.

(c)∑∞n=1 fn ∈ L1([0,∞),m) and

∫∞0

∑∞n=1 fn(x) dx = log(b/a).

Solution:

For (a), we have

∞∑n=1

∫ ∞0

|fn| ≥∞∑n=1

∫ ∞1an

∣∣ae−nax − be−nbx∣∣ dx ≥∞∑n=1

∣∣∣∣∣∫ ∞

1an

(ae−nax − be−nbx) dx

∣∣∣∣∣=

∞∑n=1

∣∣∣∣ 1

ne− 1

neb/a

∣∣∣∣=

∣∣∣∣1e − 1

eb/a

∣∣∣∣ ∞∑n=1

1

n=∞.

For (b), notice that for every n ≥ 1,∫ ∞0

fn =

∫ ∞0

(ae−nax − be−nbx) dx =

[−e−nax

n+e−nbx

n

∣∣∣∣∞0

= 0,

so∑∞n=1

∫fn = 0.

Finally for (c), let f =∑∞n=1 fn. Then for each x > 0 we have

f =∞∑n=1

(ae−nax − be−nbx

)= a

∞∑n=1

e−nax − b∞∑n=1

e−nbx (since both series conv. absolutely)

= a

∞∑n=1

(1

eax

)n− b

∞∑n=1

(1

ebx

)n=

a

eax − 1− b

ebx − 1

Note that for x > 0 and 0 < a < b we have

eax − 1

a= x+

ax2

2!+a2x3

3!+ · · · ≤ x+

bx2

2!+b2x3

3!+ · · · = ebx − 1

b,

and so f(x) = aeax−1 −

bebx−1 ≥ 0. Therefore,

gn(x) =

(a

eax − 1− b

ebx − 1

)χ

[1/n,∞)(x) ≥ 0,

15

and lim gn = f , so by the Monotone Convergence Theorem,∫|f | =

∫f = lim

n→∞

∫gn = lim

n→∞log

(1− e−b/n

1− e−a/n

)= log lim

n→∞

(1− e−b/n

1− e−a/n

)= log(b/a) <∞.

2.28 Compute the following limits and justify the calculations.a. limn→∞

∫∞0

(1 + (x/n))−n sin(x/n) dx.

b. limn→∞∫ 1

0(1 + nx2)(1 + x2)−n dx.

c. limn→∞∫∞0n sin(x/n)[x(1 + x2)]−1 dx.

d. limn→∞∫∞an(1 + n2x2)−1 dx.

Solution:

a. Notice that for x ≥ 0, (1 + xn )n ≥ 1 +

(n1

)xn +

(n2

)x2

n2 = 1 + x+ n(n−1)x2

2n2 . Since n(n−1)2n2 = 1

2 −12n ≥

14 for

n ≥ 2, it follows that (1 + xn )n ≥ 1 + x+ x2/4 for n ≥ 2 and all x ≥ 0. Therefore∣∣∣∣∣ sin(x/n)(

1 + xn

)n∣∣∣∣∣ ≤ 1

1 + x+ x2

4

and

∫ ∞0

dx

1 + x+ x2

4

<∞,

so by the LDCT,

limn→∞

∫ ∞0

(1 +

x

n

)−nsin(x/n) dx =

∫ ∞0

limn→∞

(1 +

x

n

)−nsin(x/n) dx =

∫ ∞0

0 dx = 0.

b. With fn = (1 + nx2)(1 + x2)−n we have for n ≥ 2 that

|fn| ≤1 + nx2

1 + nx2 +(n2

)x4≤ 1,

and 1 ∈ L1([0, 1],m). Since lim fn = χ{0}

, it follows from the LDCT that

limn→∞

∫ 1

0

fn dx =

∫ 1

0

χ{0}

dx = 0.

c. Let fn = n sin(x/n)x(1+x2) . Since | sinα| ≤ |α| for all α ∈ R, we have that | sin(x/n)| ≤ |x/n| and so

|fn| ≤∣∣∣∣ n · xnx(1 + x2)

∣∣∣∣ =1

1 + x2= g(x),

and g ∈ L1([0,∞),m). Furthermore, fn → xx(1+x2) = 1

1+x2 , so by LDCT (with dominating function g),

lim

∫ ∞0

fn dm =

∫ ∞0

dx

1 + x2= tan−1 x

∣∣∞0

=π

2.

d. Here we simply compute the limit directly:

limn→∞

∫ ∞a

n

1 + (nx)2dx = lim

n→∞

[tan−1(nx)

∣∣∞a

= limn→∞

(π2− tan−1(an)

)=

0, if a > 0,

π/2, if a = 0,π, if a < 0.

How does this accord with the various convergence theorems? Let

f(x) = limn→∞

n

1 + (nx)2=

{0, if x 6= 0,∞, if x = 0.

Since∫∞af = 0 for all a, it is possible that the convergence theorems may apply for a > 0, but the hypotheses

of MCT and LDCT cannot be satisfied for a ≤ 0. Indeed, {fn} is not monotone on any interval containingzero since {fn(0)} is an increasing sequence while for all x 6= 0, {fn(x)} is eventually decreasing.

16

If a > 0 then |fn| ≤ 1/x2 on [a,∞), so LDCT may be applied in this case, but there is no functiong ∈ L1([a,∞)) that dominates {fn} on [a,∞) if a ≤ 0.

Fatou’s Lemma, however, does apply and it supplies us with the conclusion that

0 =

∫ ∞a

f ≤ lim inf

∫fn,

which is rather weak since the sequence {fn} is nonnegative.

2.29 Show that∫∞0xne−x dx = n! by differentiating the equation

∫∞0e−tx dx = 1/t. Similarly show

that∫∞−∞ x2ne−x

2

dx = (2n)!√π

4nn! by differentiating the equation∫∞−∞ e−tx

2

dx =√π/t.

Solution:

We first show by induction that for all n ≥ 1 and t ∈ [1, 2], n!/tn+1 =∫∞0xne−tx dx. Evaluation at t = 1

then produces the desired result.Let f(x, t) = e−tx. Then ∂f

∂t = −xe−tx exists for t ∈ [1, 2]. With g(x) = xe−x we have g ∈ L1([0,∞))

and∣∣∣∂f∂t ∣∣∣ ≤ g for all (x, t) ∈ [0,∞)× [1, 2]. So, by Theorem 2.27,

−1

t2=

∂

∂t

∫ ∞0

f(x, t) dx =

∫ ∞0

∂f

∂tdx = −

∫ ∞0

xe−tx dx.

Suppose now that n ≥ 1 and n!/tn+1 =∫∞0xne−tx dx. Then

(n+ 1)!

tn+2= − ∂

∂t

[n!

tn+1

]= − ∂

∂t

∫ ∞0

xne−tx dx.

For all t ∈ [1, 2], ∣∣∣∣ ∂∂t [xne−tx]∣∣∣∣ =

∣∣xn+1e−tx∣∣ ≤ xn+1e−x ∈ L1([0,∞)),

so we may again apply Theorem 2.27 to conclude that

(n+ 1)!

tn+2=

∫ ∞0

− ∂

∂t

[xne−tx

]dx =

∫ ∞0

xn+1e−tx dx.

The second part of the problem is similar.

2.31 Derive the following:a. For a > 0,

∫∞−∞ e−x

2

cos ax dx =√πe−a

2/4.

b. For a > −1,∫ 1

0xa(1− x)−1 log xdx = −

∑∞k=1(a+ k)−2.

c. For a > 1,∫∞0xa−1(ex − 1)−1 dx = Γ(a)ζ(a), where ζ(a) =

∑∞n=1 n

−a.

d. For a > 1,∫∞0e−axx−1 sinx dx = arctan(a−1).

e. For a > 1,∫∞0e−axJ0(x) dx = (a2 + 1)−1/2, where J0(x) =

∑∞n=0(−1)nx2n4−n(n!)−2 is the

Bessel function of order 0.Solution:

a. Using the power series expansion of cos ax we have

e−x2

cos ax =

∞∑n=0

e−x2 (ax)2n

(2n)!(−1)n.

Furthermore,

∞∑n=0

∣∣∣∣e−x2 (ax)2n

(2n)!(−1)n

∣∣∣∣ = e−x2∞∑n=0

(ax)2n

(2n)!≤ e−x

2∞∑k=0

(ax)k

k!= e−x

2

eax = eax−x2

.

17

Since eax−x2 ∈ L1(R) it follows from Theorem 2.25 and the result of Exercise 2.29 that∫ ∞

−∞

∞∑n=0

e−x2 (ax)2n

(2n)!(−1)n dx =

∞∑n=0

(−1)na2n

(2n)!

∫ ∞−∞

x2ne−x2

dx

=

∞∑n=0

(−1)na2n

(2n)!· (2n)!

√π

4nn!

=√π

∞∑n=0

(−a2

4

)n1

n!=√πe−a

2/4.

b. Using the formula for the geometric series, we have for 0 < x < 1 that xa log x1−x =

∑∞n=0 x

n+a log x =

−∑∞n=0 x

n+a log(1/x). Since each function xn+a log(1/x) is nonnegative on (0, 1), we have by Theorem 2.15that ∫ 1

0

xa log x

1− xdx = −

∫ 1

0

∞∑n=0

xn+a log(1/x) dx

= −∞∑n=0

∫ 1

0

xn+a log(1/x) dx. (0.2)

For j ≥ 1, let gj = xn+a log(1/x)χ[1/j,1]

. Then {gj} is an increasing sequence of functions, so by MCT and

integration by parts,∫ 1

0

xn+a log(1/x) dx =

∫(0,1]

limj→∞

gj dm(x) = limj→∞

∫(0,1]

gj dm(x)

= limj→∞

∫ 1

1/j

xn+a log(1/x) dx

= limj→∞

[−xn+a+1 log x

n+ a+ 1+

xn+a+1

(n+ a+ 1)2

∣∣∣∣11/j

.

=1

(n+ a+ 1)2( a > −1 was used here)

Note: this type of argument shows that the usual technique for dealing with such ‘improper’ integralsworks, provided the integrand is nonnegative for almost all values sufficiently close to the limit of integration;henceforth, I will assume this rather than detail out the steps as I’ve done here.

Combining with (0.2), we have∫ 1

0

xa log x

1− xdx = −

∞∑n=0

1

(n+ a+ 1)2= −

∞∑k=1

1

(a+ k)2.

c. Using the formula α1−α = α+ α2 + α3 + . . . which is valid for |α| < 1, we have

xa−1

ex − 1= xa−1

(e−x

1− e−x

)=

∞∑k=1

xa−1e−kx, for x > 0.

In this last expression, the summands are nonnegative, so∫ ∞0

xa−1

ex − 1dx =

∞∑k=0

∫ ∞0

xa−1e−kx dx =

∞∑k=1

∫ ∞0

(uk

)a−1e−u

du

k

=

∞∑k=1

1

ka

∫ ∞0

ua−1e−u du

=

∞∑k=1

Γ(a)

ka= Γ(a)ζ(a).

18

d. Using the power series expansion for sinx we have

x−1e−ax sinx =

∞∑k=0

(−1)kx2k

(2k + 1)!e−ax.

Letting fk = (−1)kx2k

(2k+1)! e−ax, it follows that

∞∑k=0

∫ ∞0

|fk| =∞∑k=0

1

(2k + 1)!

∫ ∞0

x2ke−ax dx =

∞∑k=0

1

(2k + 1)!a2k+1

∫ ∞0

t2ke−t dt

=

∞∑k=0

Γ(2k + 1)

(2k + 1)!a2k+1

=

∞∑k=0

1

(2k + 1)a2k+1<∞, for a > 1.

So by Theorem 2.25,∫ ∞0

( ∞∑k=0

fk

)=

∞∑k=0

∫ ∞0

fk =

∞∑k=0

(−1)k

(2k + 1)!

∫ ∞0

x2ke−ax dx

=

∞∑k=0

(−1)kΓ(2k + 1)

(2k + 1)!a2k+1

=

∞∑k=0

(−1)k

(2k + 1)a2k+1= arctan(a−1).

e. Let

fn =(−1)nx2n

4n(n!)2e−ax.

Then∞∑n=0

∫ ∞0

|fn| =∞∑n=0

1

4n(n!)2

∫ ∞0

x2ne−ax dx =

∞∑n=0

1

4n(n!)2a2n+1

∫ ∞0

t2ne−t dt

=

∞∑n=0

Γ(2n+ 1)

4n(n!)2a2n+1

=

∞∑n=0

(2n)!

4n(n!)2a2n+1.

If you are familiar with the well-known bound(2nn

)≤ 22n you could use it here. Otherwise, deduce it

inductively, or sloppily by:

(2n)!

4n(n!)2=

(2n)(2n− 1)(2n− 2)(2n− 3) . . . (2)(1)

(2n)(2n)(2n− 2)(2n− 2) . . . (2)(2)=

(2n− 1

2n

)(2n− 3

2n− 2

). . .

(3

4

)(1

2

)≤ 1.

(The right-hand side here cannot be taken as 1/2 if n = 0). It thus follows that∑∞n=0

∫∞0|fn| <∞. So by

Theorem 2.25 we have that∫ ∞0

(∑fn

)=

∞∑n=0

∫ ∞0

fn =

∞∑n=0

(−1)n

4n(n!)2

∫ ∞0

x2ne−ax dx

=

∞∑n=0

(−1)nΓ(2n+ 1)

4n(n!)2a2n+1

=1√

a2 + 1.

(For the last equality, you may either look it up or derive it).

19

2.45 If (Xj ,Mj) is a measurable space for j = 1, 2, 3 then ⊗31Mj =M1 ⊗M2 ⊗M3. Moreover, if µj

is a σ-finite measure on (Xj ,Mj) then µ1 × µ2 × µ3 = (µ1 × µ2)× µ3.Solution:By Prop. 1.3, M1 ⊗M2 is generated by E1 = {E1 × E2 : Ej ∈ Mj}. Since E3 = M3 generates M3, wehave by Prop. 1.4 that (M1 ⊗M2)⊗M3 is generated by

E = {A×B : A ∈ E1, B ∈ E3}= {(E1 × E2)× E3 : Ej ∈Mj}.

Under the natural identification where we take (X1 ×X2)×X3 = X1 ×X2 ×X3, we thus have

E = {E1 × E2 × E3 : Ej ∈Mj},

and this set generates ⊗31Mj by Proposition 1.3.

Suppose now that µ1, µ2, µ3 are σ-finite. Then for all E1 ∈M1, E2 ∈M2, E3 ∈M3 we have

(µ1 × µ2)× µ3((E1 × E2)× E3) = (µ1 × µ2)(E1 × E2)µ3(E3)

= µ1(E1)µ2(E2)µ3(E3)

= (µ1 × µ2 × µ3)(E1 × E2 × E3).

It follows by countable additivity that ((µ1 × µ2) × µ3)(A) = (µ1 × µ2 × µ3)(A) for every set A which is afinite disjoint union of measurable cuboids (rectangular parallelpipeds). The collection A of all such sets Ais an algebra and the σ-algebra generated by A is ⊗3

1Mj ; since the measures (µ1×µ2)×µ3 and µ1×µ2×µ3

are σ-finite and agree on A, they are equal by the uniqueness assertion in Theorem 1.14.

2.46 Let X = Y = [0, 1], M = N = B[0,1], let µ be the Lebesgue measure and ν the countingmeasure. If D = {(x, x) : x ∈ [0, 1]} is the diagonal in X × Y , then

∫∫χ

Ddµdν,

∫∫χ

Ddν dµ

and∫χ

Dd(µ× ν) are all unequal.

Solution:For all y ∈ [0, 1] we have

∫χ

D(x, y) dµ(x) =

∫{y} 1 dµ(x) = 0, so∫∫

χD

dµdν =

∫∫χ

D(x, y) dµ(x) dν(y) =

∫0 dν(y) = 0.

For all x ∈ [0, 1],∫χ

D(x, y) dν(y) =

∫{x} 1 dν(y) = ν{x} = 1, so∫∫

χD

dν dµ =

∫∫χ

D(x, y) dν(y) dµ(x) =

∫[0,1]

1 dµ(x) = 1.

Finally,∫χ

Dd(µ× ν) = (µ × ν)(D) which we will now show is ∞, by arguing on the outer measure of

D. Suppose that {An × Bn} is a countable collection of measurable rectangles with D ⊂⋃

(An × Bn).Then D =

⋃(D ∩ An × Bn), and so

⋃(An ∩ Bn) ⊃ [0, 1]. It follows that there is an N ≥ 1 for which

m(AN ∩ BN ) > 0. In particular, mAN > 0 and BN is infinite so that νBN = ∞. It follows that the outermeasure of D is infinite and since D is measurable, (µ× ν)(D) =∞.

2.49 Prove Theorem 2.39 by using Thm. 2.37 and Prop. 2.12 together with the following lemmas.a. If E ∈M×N and (µ× ν)(E) = 0 then ν(Ex) = µ(Ey) = 0 for almost all x, y.b. If f is L-measurable and f = 0 λ-a.e., then fx and fy are integrable for almost all x, y and∫fx dν =

∫fy dµ = 0 for almost all x, y.

Solution:a. One may prove the stronger result for E ∈ M⊗ N with (µ × ν)(E) = 0 trivially by invoking Theorem2.36; I think this is what was intended, since it’s the first thing I’m going to do in Part (b) anyway.

20

b. Let E = {(x, y) : f(x, y) 6= 0}. Then λE = 0 and since λ is the completion of µ × ν, there is a set

E ∈M⊗N such that E ⊃ E and (µ× ν)(E) = 0. By Theorem 2.36,

0 = (µ× ν)(E) =

∫νEx dµ(x) =

∫µEy dν(y).

Therefore νEx = 0 a.e. [µ] and µEy = 0 a.e. [ν]. Since Ex ⊂ Ex it follows by the completeness of νthat νEx = 0 a.e. [µ]. Similarly, from the completeness of µ it follows that µEy = 0 a.e. [ν]. Since{y : fx(y) 6= 0} = Ex and νEx = 0 a.e. [µ], it follows that for almost all x, fx = 0 a.e.. By Prop. 2.11,we therefore have that for almost all x, fx is measurable, |fx| = 0 a.e. so fx is integrable, and

∫fx dν = 0.

Similarly, we conclude that for almost all y, fy is integrable, and∫|fy|dµ = 0.

Thm. 2.39: Assume the hypotheses of the Theorem and suppose that f is L-measurable and f ≥ 0. ByProp. 2.12, there is an M⊗N -measurable function f so that f = f a.e. [λ]. We may further assume that

f ≥ 0 (by possibly modifying f on a µ × ν-null set). Set g = f − f . Then g = 0 a.e. [λ], so by Part (b),

gx, gy are integrable, and in particular measurable, for almost all x, y. fx, f

y are measurable by Prop. 2.34,and so fx, f

y are measurable for almost all x, y respectively. Furthermore, by Theorem 2.37, the functionsx 7→

∫fx dν and y 7→

∫fy dµ are measurable. By Part (b), we have

∫gx dν =

∫gy dµ = 0 for almost all x, y

resp., so the functions

x 7→∫fx dν =

∫(fx + gx) dν =

∫fx dν, for almost all x

y 7→∫fy dµ =

∫(fy + gy) dµ =

∫fy dµ, for almost all y

are measurable for almost all x, y respectively. By Theorem 2.37,∫∫f dν dµ =

∫∫(fx + gx) dν dµ =

∫∫fx dν dµ =

∫f d(µ× ν),

and similarly,∫∫

f dµdν =∫f d(µ× ν), so it remains only to show that

∫f d(µ× ν) =

∫f dλ.

If E isM⊗N measurable then (µ×ν)(E) = λ(E), so∫χ

Ed(µ× ν) =

∫χ

Edλ. By linearity, we have the

same result for M⊗ N-measurable simple functions. Let {ϕn} be a sequence of simple functions increasing

to f . Using the Monotone Convergence Theorem twice, we deduce that∫f d(µ× ν) = lim

∫ϕn d(µ× ν) = lim

∫ϕn dλ =

∫f dλ,

and since f = f a.e. [λ], it follows that∫f dλ =

∫f dλ.

Tonelli’s Theorem now follows from Fubini’s as in the proof of Theorem 2.37.

3.8 ν � µ iff |ν| � µ iff ν+ � µ and ν− � µ.Solution:Suppose ν � µ and µE = 0. Let X = P ∪N be a Hahn decomposition w.r.t. ν, so that ν+E = ν(E ∩ P )and ν−E = −ν(E ∩N). Then

µE = 0⇒ µ(E ∩ P ) = 0⇒ ν(E ∩ P ) = 0⇒ ν+E = 0,

andµE = 0⇒ µ(E ∩N) = 0⇒ ν(E ∩N) = 0⇒ ν−E = 0,

so |ν|E = ν+E + ν−E = 0, and thus |ν| � µ.Now if |ν| � µ and µE = 0 then ν+E + ν−E = 0, and so ν+E = ν−E = 0 since ν+, ν− are positive

measures. Therefore ν+ � µ and ν− � µ.Finally, if ν+ � µ and ν− � µ and µE = 0 then ν+E = 0 and ν−E = 0 so νE = ν+E − ν−E = 0, and

thus ν � µ.

21

3.9 Suppose {νj} is a sequence of positive measures. If νj ⊥ µ for all j then (∑νj) ⊥ µ. If νj � µ

for all j then (∑νj)� µ.

Solution:

Suppose first that νj ⊥ µ for all j. Then for each j there is a measurable Aj so that νj ≡ 0 on Aj andµ ≡ 0 on Acj . Set A =

⋂Aj . Then (

∑νj) ≡ 0 on A and Ac =

⋃Acj , so for every measurable E ⊂ Ac,

µE = µ(E ∩Ac) ≤∑

µ(E ∩Acj) = 0.

For the second part, suppose νj � µ for all j, and E has µE = 0. Then νjE = 0 for all j, so(∑νj)E =

∑νjE = 0, and hence (

∑νj)� µ.

3.10 Theorem 3.5 may fail when ν is not finite.Solution:As per the hint, consider dν(x) = dx/x and dµ(x) = dx on (0, 1). We have ν � µ since µE = 0 ⇒ νE =∫E

(1/x) dx = 0.On the other hand, suppose ε = 1. Then for all δ > 0, µ((0, δ/2)) < δ and

ν((0, δ/2)) =

∫(0,δ/2)

dx

x=

∫ δ/2

0

dx

x=∞ > 1 = ε,

so the conclusion of Theorem 3.5 does not hold in this case.

3.11 (a) Every finite subset of L1(µ) is uniformly integrable.(b) If {fn} ⊂ L1(µ) converges in the L1 metric to f ∈ L1(µ), then {fn} is uniformly integrable.

Solution:For (a), suppose {f1, . . . , fn} ⊂ L1(µ) and ε > 0. By Corollary 3.6, for each 1 ≤ j ≤ n there is a δj > 0 s.t.

µE < δj =⇒∣∣∣∣∫E

fj dµ

∣∣∣∣ < ε.

Then δ = min{δ1, . . . , δn} answers the ε-challenge for uniform integrability.(b) Suppose {fn} ⊂ L1(µ) and fn −→ f ∈ L1(µ) in the L1 metric. Let ε > 0. Then there is a δ1 > 0 so

that

µE < δ1 =⇒∣∣∣∣∫E

f dµ

∣∣∣∣ < ε/2.

Since fn → f in L1, there is an N so that∫|fn − f |dµ < ε/2 for all n ≥ N . Thus, for E with µE < δ1 and

n ≥ N , ∣∣∣∣∫E

fn dµ

∣∣∣∣ =

∣∣∣∣∫E

f dµ+

∫E

(fn − f) dµ

∣∣∣∣ ≤ ∣∣∣∣∫E

f dµ

∣∣∣∣+

∫E

|fn − f |dµ < ε/2 + ε/2.

By the first part, there is a δ2 > 0 responding to the ε-challenge for uniform integrability of {f1, . . . , fN−1},so δ = min{δ1, δ2} does the job for every fn.

3.16 Suppose µ, ν are σ-finite measures on (X,M) with ν � µ and let λ = µ+ ν. If f = dν/dλ, then0 ≤ f < 1 µ-a.e. and dν/dµ = f/(1− f).

Solution:Since λ� µ, we have by Prop 3.9 that dν

dµ = f dλdµ .

22

Since dλdµ = 1 + dν

dµ = 1 + f dλdµ and f, dλ

dµ can be taken to be finite everywhere, it follows that

(1− f)dλ

dµ= 1. (0.3)

Let E = {x : f(x) ≥ 1}. Then

µE =

∫E

1 dµ =

∫E

(1− f)dλ

dµdµ =

∫E

(1− f) dλ ≤∫E

0 dλ = 0.

But since µ is a positive measure, it follows that µE = 0. Therefore, 1 − f > 0 µ-a.e. and from (0.3) wededuce that dλ

dµ = 11−f , and so dν

dµ = f dλdµ = f

1−f .

Extra: Suppose µ is a σ-finite measure on (X,M) and {En} a sequence of measurable sets. Define ν onM by νE =

∑µ(E ∩ En). Find the Radon-Nikodym derivative dν

dµ .

Solution:Clearly ν is σ-finite and ν � µ, so the Radon-Nikodym derivative in question exists. Furthermore, for everyE ∈M,

νE =∑

µ(E ∩ En)

=∑∫

χE∩En

dµ =∑∫

E

χEn

dµ =

∫ (∑χ

En

)dµ,

so that dνdµ =

∑χ

En.

3.25 If E is a Borel set in Rn define DE(x) = limr→0m(E ∩B(r, x))/mB(r, x), if the limit exists.a. Show that DE(x) = 1 a.e. on E and DE(x) = 0 a.e. on Ec.b. Find examples of E and x s.t. DE(x) is a given number α ∈ (0, 1) and an example for whichDE(x) does not exist.

Solution:a. This is a direct application of Thm. 3.22: let E ∈ BRn and define νA = m(A ∩ E) for all Borel sets A(i.e., ν = m|E). Then ν � m and dν

dm = χE. Since the family {B(r, x)}r>0 shrinks nicely to x, it follows by

Theorem 3.22 that DE(x) = limr→0 νB(r, x)/mB(r, x) = χE(x) for m-almost all x ∈ Rn.

b. Let α ∈ (0, 1) and E = {(t sin θ, t cos θ) : t > 0, 0 ≤ θ ≤ 2πα} ⊂ R2, and x = (0, 0). For each r > 0,m(E ∩B(r, x)) = 2πr2α, so DE(x) = α.

For an example in which the limit does not exist, consider the subset E of R given by

E =⋃n≥1

[1

22n+1,

1

22n

]=

[1

8,

1

4

]∪[

1

32,

1

16

]∪ . . . ,

and x = 0. Consider the sequence rk = 2−k. Then mB(rk, 0) = 12k−1 . Now if k is even, say k = 2k′, then

m(E ∩B(rk, 0)) =∑2n≥k

m

[1

22n+1,

1

22n

]=∑n≥k′

1

22n+1=

1

2 · 4k′· 4

3=

1

3 · 2k−1,

and so m(E ∩B(rk, 0))/mB(rk, 0) = 1/3 when k is even. Now for odd k, say k = 2k′ + 1 we have

m(E ∩B(rk, 0)) =∑2n≥k

m

[1

22n+1,

1

22n

]=

∑n≥k′+1

1

22n+1=

1

2 · 4k′+1· 4

3=

1

3 · 2k,

and so m(E∩B(rk, 0))/mB(rk, 0) = 1/6 when k is odd. Therefore, limk→∞m(E∩B(rk, 0))/mB(rk, 0) doesnot exist and hence DE(0) is undefined.

23

3.26 If λ and µ are positive, mutually singular Borel measures on Rn and λ+ µ is regular then so areλ and µ.

Solution:For compact K ⊂ Rn, (λ+ µ)K <∞, and so λK <∞, µK <∞ since λ and µ are positive.

Let A ∈ BRn so that µA = λAc = 0. Then for all E ∈ BRn ,

λE = λ(E ∩A) = (λ+ µ)(E ∩A)

= inf{(λ+ µ)U : U is open and U ⊃ E ∩A}≥ inf{λU : U is open and U ⊃ E ∩A}≥ λ(E ∩A) = λE.

So we have λE = inf{λU : U is open and U ⊃ E ∩A}, Below, we will show that for each ε > 0 there is anopen set O ⊃ Ac such that λO < ε. With this, it follows that for each open U ⊃ E ∩ A, the set O ∪ U isopen, O ∪ U ⊃ E and λ(O ∪ U) ≤ λU + ε so that inf{λG : G is open and G ⊃ E} ≤ λE by above, whilethe reverse inequality follows from monotonicity. The corresponding property for µ follows by interchangingµ and λ in this argument, so showing the existence of such O will complete the proof.

Let ε > 0. Since λ+ µ is regular and Rn is a countable union of compact sets, λ+ µ is σ-finite. So thereis a countable disjoint collection {Xj}j≥1 with Ac =

⋃Xj and (λ + µ)Xj < ∞. For each j ≥ 1, let Oj be

an open set with Oj ⊃ Xj such that

(λ+ µ)Oj ≤ (λ+ µ)Xj +ε

2j,

and set O =⋃Oj . Then O is open, O ⊃ Ac and

λOj = λ(Oj ∩A) = (λ+ µ)(Oj ∩A) = (λ+ µ)Oj − (λ+ µ)(Oj ∩Ac) ≤ (λ+ µ)Xj +ε

2j− µXj =

ε

2j.

Therefore λO ≤∑λOj < ε.

3.30 Construct an increasing function on R whose set of discontinuities is Q.Solution:One method is to enumerate the rationals Q = {x1, x2 . . . } and construct a Borel measure µ for whichµE = µ(E ∩Q) for all Borel sets E and µ{xj} = 2−j . Then apply Theorem 1.16 to get an increasing F withµF = µ.

Another method is to construct an increasing function f : [0,∞) −→ [0,∞) which is discontinuous on Q;the result then follows by extending f to an odd function via f(−x) = −f(x) for x > 0. To that end, let

f(x) =∑

0≤ pq≤x

1

q3,

where the sum is taken over all rational pq with gcd(p, q) = 1. Notice first that for x > 0,

0 ≤ f(x) ≤∞∑q=1

∑0≤p≤qx

1

q3=

∞∑q=1

bqxcq3≤ x

∞∑q=1

1

q2=xπ2

6,

so the sum converges. Clearly f is increasing. If x = ab is a positive rational, then for all x > ε > 0,

|f(x)− f(x− ε)| =∑

0≤ pq≤x

1

q3−

∑0≤ p

q≤x−ε

1

q3≥ 1

b3,

so f is discontinuous at x.

24

Suppose x is irrational and ε > 0. Choose N ∈ N so that∑∞q=N

1q2 < ε. For each b = 1, 2, . . . , N let nb ∈ N

for which∣∣x− nb

b

∣∣ = δb is minimal. Since x is irrational, each δb is positive. Let δ = min{1, δ1, δ2, . . . , δN}.Then

∣∣∣x− pq

∣∣∣ < δ ⇒ q > N , so for 0 < y < δ,

0 ≤ f(x)− f(x− δ) =∑

x−δ< pq≤x

1

q3<

∞∑q=N

∑qx−qδ<p≤qx

1

q3≤∞∑q=N

1 + qδ

q3=

∞∑q=N

1

q3+ δ

∞∑q=N

1

q2< ε+ δε ≤ 2ε.

Similarly, 0 ≤ f(x+ δ)− f(x) ≤ 2ε, so f is continuous at x.

3.33 If F is increasing on R then F (b)− F (a) ≥∫ baF ′(t) dt.

Solution:Let −∞ < a < b <∞ and assume WOLG that F is NBV by properly redefining it outside the interval [a, b].Let µF be the Borel measure associated with F and

µF = λ+ ρ, λ ⊥ m, ρ� m,

its Lebesgue-Radon-Nikodym decomposition. Both λ and ρ are regular measures, so by Theorem 1.16 thereexist right-continuous increasing functions G,H such that λ = µG, ρ = µH . Both G,H ∈ NBV , so byProposition 3.30, G′ = 0 a.e. and for x ∈ [a, b], H(x) =

∫ x−∞H ′(t) dt =

∫ xaH ′(t) dt. But µF = µG + µH =

µG+H , so F = G+H a.e. and since G′ and H ′ exist a.e., it follows that F ′ = G′ +H ′ = H ′ a.e., so

F (b)− F (a) = µF (a, b] = µG(a, b] + µH(a, b] ≥ µH(a, b] =

∫ b

a

H ′(t) dt =

∫ b

a

F ′(t) dt.

3.37 Suppose F : R −→ C. There is a constant M s.t. |F (x)− F (y)| ≤M |x− y| for all x, y ∈ R iff Fis AC and |F ′| ≤M a.e..

Solution:Suppose first that such a constant M > 0 exists. Let ε > 0 and set δ = ε/M . Suppose {(ai, bi)}Ni=1 is a

disjoint collection of open intervals for which∑Ni=1(bi − ai) < δ. Then

N∑i=1

|F (bi)− F (ai)| ≤MN∑i=1

(bi − ai) < Mδ = ε,

so F is absolutely continuous. Furthermore, for all h > 0,

|F (x+ h)− F (x)|h

=|F (x+ h)− F (x)|

(x+ h)− x≤M,

so |F ′(x)| ≤M for all x ∈ R.Conversely, suppose F is AC and |F ′| ≤M a.e.. Let x, y ∈ R with X < y. Then

|F (y)− F (x)| =∣∣∣∣∫ y

x

F ′(t) dt

∣∣∣∣ ≤ ∫ y

x

|F ′(t)|dt ≤∫ y

x

M dt = M |y − x|.

3.39 If {Fj} is a sequence of nonnegative increasing functions on [a, b] so that F (x) =∑Fj(x) < ∞

for all x ∈ [a, b], then F ′(x) =∑F ′j(x) for almost all x ∈ [a, b].

Solution:As per the hint, it suffices to assume (by properly extending the Fj ’s and F to R) that the Fj ’s and F areNBV. Let µF and µFj

be the associated Borel measures. These are positive σ-finite Borel measures, so theyadmit Lebesgue-Radon-Nikodym decompositions:

dµF = dλ+ f dm, and dµFj= dλj + fj dm,

25

where all of these measures are positive and λ ⊥ m and λj ⊥ m. Since µF and∑µFj agree on h-intervals,

they are equal, and so

dµF =∑

dµFj=(∑

dλj

)+(∑

fj dm).

Furthermore, since (∑

dλj) ⊥ m and (∑fj dm) � m by Exercise 3.8, it follows from the uniqueness

assertion in the Lebesgue-Radon-Nikodym Theorem that dλ =∑

dλj and f dm =∑fj dm; in particular,

f =∑fj m-almost everywhere. Thus, with Er = (x, x+ r] we have by Theorem 3.22 for m-almost all x,

F ′(x) = limr→0

F (x+ r)− F (x)

r= limr→0

µFErmEr

= f(x).

But for almost all x, f(x) =∑fj(x) and another application of Theorem 3.22 yields for almost all x that

f(x) =∑

fj(x) =∑(

limr→0

µFjEr

mEr

)=∑

F ′j(x).

4.3 Every metric space is normal.Solution:Suppose (X, ρ) is a metric space. If x, y ∈ X with x 6= y then r = ρ(x, y) > 0 and B(r/2, x) is an open setcontaining x but not y, so X is T1.

Now suppose A,B are disjoint closed subsets of X. Let U = {x ∈ X : ρ(x,A) < ρ(x,B)} andV = {x ∈ X : ρ(x,B) < ρ(x,A)}. We claim that U is open. If x ∈ U then r = ρ(x,B) − ρ(x,A) > 0 andB(r/2, x) ⊂ U since ρ(z, x) < r/2 implies

ρ(z,A) ≤ ρ(z, x) + ρ(x,A) < r/2 + ρ(x,A) =ρ(x,B)− ρ(x,A)

2+ ρ(x,A) =

ρ(x,B) + ρ(x,A)

2< ρ(x,B).

The same argument applied to V shows that V is open. By construction, we have that U ∩ V = ∅ andA ⊂ U , B ⊂ V , so indeed the space is normal.

4.5 Every separable metric space is second countable.Solution:Suppose (X, ρ) is separable and {xn} is a countable dense subset. LetB = {B(1/m, xn) : m,n ∈ N}.CertainlyB is a countable collection of open sets, and we claim it is a base for the metric space topologyon X. To show this, we will appeal to Proposition 4.2.

Let U ⊂ X be an open set. For each y ∈ U there is an ry > 0 so that B(ry, y) ⊂ U and there exists ny

so that ρ(xny, y) < ry/2. Let my ∈ N such that 1

my<

ry2 − ρ(xny

, y). Then xny∈ B

(1my, xny

)and for all

z ∈ B(

1my, xny

)we have that

ρ(y, z) ≤ ρ(y, xny) + ρ(xny

, z)

<ry2

+1

my< ry − ρ(xny

, y) < ry,

so z ∈ B(ry, y) ⊂ U and hence B(

1my, xny

)⊂ U . Therefore, U =

⋃y∈U B

(1my, xny

), soB is a base for the

metric space topology on X by Prop. 4.2.

26

Spr.2012 prelim # 8: Suppose f is continuous and real-valued on [a, b] and∫ baxnf(x) dx = 0 for all n = 0, 1, . . . . Show

that f ≡ 0 on [a, b].Solution:

Let ε > 0. Since polynomial functions are dense in C([a, b],R), there is a polynomial p(x) such that

‖f − p‖u < ε. From the hypothesis, it follows by linearity that∫ bap(x)f(x) dx = 0, so

0 =

∫ b

a

p(x)f(x) dx =

∫ b

a

(p(x)− f(x))f(x) dx+

∫ b

a

f(x)2 dx,

and so ∣∣∣∣∣∫ b

a

f(x)2 dx

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

(f(x)− p(x))f(x) dx

∣∣∣∣∣ ≤∫ b

a

|f(x)− p(x)||f(x)|dx ≤ ε∫ b

a

|f(x)|dx.

Since ε > 0 is arbitrary and f is fixed and∫ ba|f(x)|dx < ∞, it follows that

∫ baf(x)2 dx = 0, so f = 0 a.e.

on [a, b]. But since f is continuous, this implies f ≡ 0 on [a, b].

5.1 If X is a normed vector space over K (= R or C) then addition and scalar multiplication arecontinuous from X ×X and K ×X to X. Moreover, the norm is continuous from X to [0,∞); infact,

∣∣‖x‖ − ‖y‖∣∣ ≤ ‖x− y‖.Solution:

Let A : X ×X −→ X be given by A(x1, x2) = x1 + x2. Then A is a linear map from the NVS X ×X tothe NVS X. Furthermore, for all (x1, x2) ∈ X ×X,

‖A(x1, x2)‖ = ‖x1 + x2‖ ≤ ‖x1‖+ ‖x2‖ ≤ 2 max{‖x1‖, ‖x2‖} = 2‖(x1, x2)‖,

so A is bounded, hence continuous, by Prop. 5.2.Since K is a NVS over itself, K ×X is a NVS and M : K ×X −→ X given by M(λ, x) = λx is a linear

map. We will show that M is continuous at (0, 0), and result then follows from Prop. 5.2. Let ε > 0, andδ = min{ε, 1}. Then for all (λ, x) ∈ K ×X with ‖(λ, x)‖ < δ we have max{|λ|, ‖x‖} < δ ≤ ε and δ2 ≤ δ, so

‖M(λ, x)‖ = ‖λx‖ = |λ|‖x‖ < δ2 ≤ δ ≤ ε,

so M is continuous at (0, 0).Finally we show that ‖ · ‖ is continuous. Let ε > 0, set δ = ε and suppose x, y ∈ X such that ‖x− y‖ < δ.

Then ‖x‖ = ‖x− y+ y‖ ≤ ‖x− y‖+ ‖y‖, so ‖x‖− ‖y‖ ≤ ‖x− y‖. Also, ‖y‖ = ‖y− x+ x‖ ≤ ‖y− x‖+ ‖x‖,so ‖y‖ − ‖x‖ ≤ ‖y − x‖ = ‖x− y‖, and therefore

∣∣‖x‖ − ‖y‖∣∣ ≤ ‖x− y‖ < ε, so ‖ · ‖ is uniformly continuouson X.

5.2 L(X,Y ) is a vector space and the function ‖ · ‖ defined by (5.3) is a norm on it.Solution:The space Y X of all functions from X to Y is a vector space, so it suffices to show that L(X,Y ) is nonempty,closed under addition and closed under scalar multiplication (that is, that L(X,Y ) is a subspace of Y X).Certainly the function Z(x) = 0 is in L(X,Y ) so it’s nonempty. If T1, T2 ∈ L(X,Y ) then there are positiveconstants C1, C2 such that ‖T1x‖ ≤ C1‖x‖ and ‖T2x‖ ≤ C2‖x‖ for all x ∈ X. Then T1 +T2 is linear and forall x ∈ X,

‖(T1 + T2)x‖ = ‖T1x+ T2x‖ ≤ ‖T1x‖+ ‖T2x‖ ≤ (C1 + C2)‖x‖,

so T1 + T2 is bounded, hence T1 + T2 ∈ L(X,Y ). If α ∈ K then for all x ∈ X we have

‖(αT1)x‖ = ‖α(T1x)‖ = |α|‖T1x‖ ≤ (|α|C1)‖x‖,

27

so αT ∈ L(X,Y ). Therefore L(X,Y ) is a vector space.By definition, ‖T‖ ≥ 0 for all T ∈ L(X,Y ) and ‖T‖ = 0 iff ‖Tx‖ ≤ 0‖x‖ = 0 for all x ∈ X, so ‖T‖ = 0

iff ‖T‖ = 0. Suppose T1, T2 ∈ L(X,Y ). Then for all x ∈ X,

‖(T1 + T2)x‖ ≤ ‖T1x‖+ ‖T2x‖ ≤ ‖T1‖‖x‖+ ‖T2‖‖x‖ = (‖T1‖+ ‖T2‖)‖x‖,

so ‖T1 + T2‖ ≤ ‖T1‖+ ‖T2‖. Finally, if α ∈ K then

‖αT1‖ = sup{‖(αT1)x‖ : ‖x‖ = 1}= sup{|α|‖T1x‖ : ‖x‖ = 1}= |α| sup{‖T1x‖ : ‖x‖ = 1}= |α|‖T1‖,

so ‖ · ‖ is a norm on L(X,Y ).

5.3 Complete the proof of Prop. 5.4.Solution:We have that {Tn} is a Cauchy sequence in L(X,Y ) and Tx = limTnx for all x ∈ X. We need to show thatT ∈ L(X,Y ) and ‖Tn − T‖ → 0. If x1, x2 ∈ X and α1, α2 ∈ K then

T (α1x1 + α2x2) = limn→∞

Tn(α1x1 + α2x2) = limn→∞

(α1Tnx1 + α2Tnx2)

= α1 limn→∞

Tnx1 + α2 limn→∞

Tnx2,

since addition and scalar multiplication on X are continuous by Problem 1. But the last expression equalsα1Tx1 + α2Tx2, so T is linear.

Since {Tn} is Cauchy in L(X,Y ) and ‖Tm − Tn‖ ≥∣∣‖Tm‖ − ‖Tn‖∣∣, it follows that {‖Tn‖} is Cauchy in

R, hence convergent. From continuity of the norm on X we have that

‖Tx‖ = ‖ limn→∞

Tnx‖ = limn→∞

‖Tnx‖ ≤ ‖x‖ limn→∞

‖Tn‖,

so T is bounded, and hence T ∈ L(X,Y ).Let ε > 0 and N ∈ N such that ‖Tn − Tm‖ < ε for all m,n ≥ N . Then for all x ∈ X with ‖x‖ = 1 we

have that for all n ≥ N ,

‖(Tn − T )x‖ = ‖Tnx− Tx‖ = limm→∞

‖Tnx− Tmx‖

= limm→∞

‖(Tn − Tm)x‖ ≤ ‖x‖ limm→∞

‖Tn − Tm‖ ≤ ε.

Therefore, ‖Tn − T‖ ≤ ε for all n ≥ N , and so ‖Tn − T‖ → 0. In particular, since∣∣‖Tn‖ − ‖T‖∣∣ ≤ ‖Tn − T‖,

we also have that ‖Tn‖ → ‖T‖.

5.6 Suppose that X is a finite dimensional vector space. Let e1, . . . , en be a basis for X and define‖∑n

1 ajej‖1 =∑n

1 |aj |.a. ‖ · ‖1 is a norm on X.b. The map (a1, . . . , an) 7→

∑n1 ajej is continuous from Kn with the usual Euclidean topology to

X with the topology defined by ‖ · ‖1.c. {x ∈ X : ‖x‖1} is compact in the topology defined by ‖ · ‖1.d. All norms on X are equivalent.

Solution:

a. It’s clear that for all x ∈ X, ‖x‖1 ≥ 0 and ‖x‖1 = 0 iff x = 0. If x =∑ajej and α ∈ K then

‖αx‖1 =∥∥∥∑αajej

∥∥∥1

=∑|αaj | = |α|

∑|aj | = α‖x‖1.

28

Finally, if x =∑ajej and y =

∑bjej then

‖x+ y‖1 =∑|aj + bj | ≤

∑(|aj |+ |bj |) = ‖x‖1 + ‖y‖1,

so ‖ · ‖1 is a norm on X.

b. If we endow the vector space Kn with the usual norm, ‖(a1, . . . , an)‖ =(|a1|2 + · · ·+ |an|2

)1/2, then

Kn is a NVS and this norm induces the Euclidean toplogy on Kn. The function L : Kn −→ X given byL(a1, . . . , an) =

∑ajej is linear, so by Prop. 5.2 is suffices to show that L is continuous at 0. Let ε > 0 and

set δ = ε/n. Let a ∈ Kn with ‖a‖ < δ. Then

δ > ‖a‖ =(|a1|2 + · · ·+ |an|2

)1/2 ⇒ δ2 > |a1|2 + · · ·+ |an|2 ⇒ δ2 > |aj |2 for all 1 ≤ j ≤ n,

so ε/n > |aj | for all j and hence ‖L(a)‖1 =∑n

1 |aj | < ε, so L is continuous at zero, and hence continuous.c. Let F = {(a1, . . . , an) ∈ Kn :

∑|aj | = 1}. Then F is a bounded subset of Kn and we will show

that it is closed by showing that F c is open. If b ∈ F c then either∑|bj | < 1 or

∑|bj | > 1. In the

first case, let ε = 1 −∑|bj | > 0. Then for all c ∈ Kn with ‖c − b‖ < ε/n we have |cj | < |bj | + ε/n for

all j and so ‖c‖1 =∑|cj | < ε +

∑|bj | = 1. Therefore, B(ε/n, b) ⊂ F c. Similarly, if

∑|bj | > 1 then

B(ε′/n, b) ⊂ F c where ε′ =∑|bj | − 1. So F c is open and hence F is closed. Since Kn = Rn or Kn = Cn

and F ⊂ Kn is closed and bounded, F is compact. Since the function L in part (b) is continuous, it followsthat L(F ) = {x ∈ X : ‖x‖1 = 1} is compact in X.

d. Let ‖ · ‖ be an arbitrary norm on X. Let C2 = max{‖e1‖, . . . , ‖en‖} and x ∈ X. Then

‖x‖ = ‖x1e1 + · · ·+ xnen‖ ≤ |x1|‖e1‖+ · · ·+ |xn|‖en‖ ≤ |x1|C2 + · · ·+ |xn|C2 = C2‖x‖1.

We claim now that ‖ · ‖ is continuous with respect to the topology induced by ‖ · ‖1. Let x ∈ X andε > 0. With δ = ε/C2 we have that for all y ∈ X, if ‖y − x‖1 < δ then ‖y − x‖ ≤ C2‖y − x‖1 < ε, so ‖ · ‖ iscontinuous at x, and hence continuous on X (in fact, uniformly continuous).

By part (c), the set B = {x ∈ X : ‖x‖1 = 1} is compact in the ‖ · ‖1 topology, and since ‖ · ‖ iscontinuous with respect to the ‖ · ‖1 topology, it has a minimum on B, say C1. Since 0 6∈ B, we have C1 > 0and so for all 0 6= x ∈ X, ∥∥∥∥ x

‖x‖1

∥∥∥∥ ≥ C1, so ‖x‖ ≥ C1‖x‖1.

Therefore, ‖ · ‖ is equivalent to ‖ · ‖1.

5.7 Let X be a Banach space.a. If T ∈ L(X,X) and ‖I − T‖ < 1 where I is the identity operator, then T is invertible; in fact,the series

∑∞0 (I − T )n converges in L(X,X) to T−1.

b. If T ∈ L(X,X) is invertible and ‖S − T‖ < ‖T−1‖−1, then S is invertible. Thus the set ofinvertible operators is open in L(X,X).

Solution:

Notice that for all A ∈ L(X,X) and all x ∈ X, ‖A2x‖ = ‖A(Ax)‖ ≤ ‖A‖‖Ax‖ ≤ ‖A‖2‖x‖, so ‖A2‖ ≤‖A‖2, and inductively, ‖An‖ ≤ ‖A‖n for all n ≥ 1. Since ‖I − T‖ < 1, it follows that

∑∞0 ‖(I − T )n‖ ≤∑∞

0 ‖I − T‖n < ∞, so the series∑∞

0 (I − T )n is absolutely convergent, and hence convergent by Theorem5.1 since L(X,X) is a Banach space by Prop 5.4.

For N ≥ 0 let SN =∑Nn=0(I − T )n. Then

TSN = (I − (I − T ))SN = SN −N∑n=0

(I − T )n+1 = I − (I − T )N+1.

Since T is continuous,T ( lim

N→∞SN ) = lim

N→∞TSN = I − lim

N→∞(I − T )N+1.

29

But ∥∥∥ limN→∞

(I − T )N+1∥∥∥ = lim

N→∞‖(I − T )N+1‖ ≤ lim

N→∞‖I − T‖N+1 = 0,

so limN→∞(I − T )N+1 = 0. Therefore

I = T ( limN→∞

SN ) = T

∞∑n=0

(I − T )n, (0.4)

and so, if T is bijective, then∑∞n=0(I − T )n = T−1. We have already that

∑∞n=0(I − T )n ∈ L(X,X), so it

remains only to show that T is bijective.Suppose BWOC that x ∈ kerT and x 6= 0. Then

‖x‖ = ‖(I − T )x‖ ≤ ‖I − T‖‖x‖ < ‖x‖,

a contradiction, so kerT = {0} and T is therefore one-to-one. It follows immediately from (0.4) that T issurjective.

b. ‖ST−1 − I‖ = ‖(S − T )T−1‖ ≤ ‖S − T‖‖T−1‖ < 1, so by part (a), ST−1 is invertible, say A =(ST−1)−1. Then S(T−1A) = (ST−1)A = I, and since T−1A ∈ L(X,X), S is invertible.

5.13 Suppose ‖ · ‖ is a seminorm on X and let M = {x ∈ X : ‖x‖ = 0}. Then M is a subspace of Xand the map x+M 7→ ‖x‖ is a norm on X/M.

Solution:

M is clearly nonempty since 0 ∈ M. Suppose x1, x2 ∈ M. Then ‖x1 + x2‖ ≤ ‖x1‖ + ‖x2‖ = 0, so‖x1 + x2‖ = 0 and hence x1 + x2 ∈ M. Additionally, if α ∈ K then ‖αx1‖ = |α|‖x1‖ = 0, so αx1 ∈ M andtherefore M is a subspace of X.

Denote the map in question by ‖x +M‖ = ‖x‖. We first need to see that this map is well-defined onX/M. Suppose x1 +M = x2 +M. Then x1 − x2 ∈M, so ‖x1 − x2‖ = 0. It follows that

‖x1‖ = ‖x1 − x2 + x2‖ ≤ ‖x1 − x2‖+ ‖x2‖ = ‖x2‖,‖x2‖ = ‖x2 − x1 + x1‖ ≤ ‖x2 − x1‖+ ‖x1‖ = ‖x1‖,

so ‖x1‖ = ‖x2‖ and therefore the map is well-defined.Finally, to see that this is a norm on X/M, suppose x1 +M, x2 +M∈ X/M. Then

‖(x1 +M) + (x2 +M)‖ = ‖(x1 + x2) +M‖ = ‖x1 + x2‖ ≤ ‖x1‖+ ‖x2‖ = ‖x1 +M‖+ ‖x2 +M‖.

If α ∈ K then ‖α(x1 +M)‖ = ‖αx1 +M‖ = ‖αx1‖ = |α|‖x1‖ = |α|‖x1 +M‖. Finally, if ‖x+M‖ = 0 then‖x‖ = 0, so x ∈M and hence x+M = 0 +M, so ‖ · ‖ is a norm on x/M.

5.17 A linear functional f on a NVS X is bounded iff f−1({0}) is closed.Solution:Suppose f is bounded. Since {0} is closed in K and f is continuous, f−1({0}) is closed in X.

(A hint is given to use Exercise 12(b) for the converse, but I don’t immediately see how to use it; so hereis a direct proof instead.)

Suppose that f−1({0}) is closed. If f−1({0}) = X, then f = 0 and hence bounded. So assume there isan x0 ∈ X such that f(x0) 6= 0. By scaling, we may assume that f(x0) = 1, and by linearity we have thatf−1({1}) = x0 + f−1({0}), so that f−1({1}) is closed. BWOC assume that f is not bounded. Then foreach n ∈ N there exists xn ∈ X such that ‖xn‖ = 1 and |f(xn)| > n. Set yn = xn

f(xn)so that f(yn) = 1 and

‖yn‖ < 1/n. Since {yn} is contained in the closed subset f−1({1}) and yn → 0, it follows that 0 ∈ f−1({1}),a contradiction. Therefore f is bounded.

30

5.19(a) Let X be an infinite dimensional vector space. There is a sequence {xj} in X such that ‖xj‖ = 1for all j and ‖xj − xk‖ ≥ 1/2 for j 6= k.

Solution:Let x1 be an arbitrary element of X with ‖x1‖ = 1. We inductively construct xn as follows: suppose thatx1, . . . , xn−1 have been constructed which satisfy the conclusions. Let M = Span {x1, . . . , xn−1}. By Ex.18b, M is closed. Since X is infinite dimensional, M is a proper subspace. So, by Ex. 12b with ε = 1/2,there is an element xn ∈ X such that ‖xn‖ = 1 and ‖x +M‖ ≥ 1/2. In particular, since −xi ∈ M for1 ≤ i < n, it follows that

‖xn − xi‖ ≥ inf{‖x+ y‖ : y ∈M} = ‖xn +M‖ ≥ 1/2, for all 1 ≤ i < n.

5.22 Suppose X,Y are NVS and T ∈ L(X,Y ).a. Define T † : Y ∗ −→ X∗ by T †f = f ◦ T . Then T † ∈ L(Y ∗, X∗) and ‖T †‖ = ‖T‖.b. Applying the construction in (a) twice, one obtains T †† ∈ L(X∗∗, Y ∗∗). If X and Y are

identified with their natural images X, Y in X∗∗, Y ∗∗, then T ††|X = T .c. T † is injective iff the range of T is dense in Y .d. If the range of T † is dense in X∗ then T is injective; the converse is true if X is reflexive.

Solution:a. It’s straightforward to verify that T † is linear. To see that it’s bounded, let g ∈ Y ∗. Then for all x ∈ Xwe have

‖(T †g)(x)‖ = ‖(g ◦ T )(x)‖ = ‖g(T (x)‖ ≤ ‖g‖‖T (x)‖ ≤ ‖g‖‖T‖‖x‖,

so ‖T †g‖ ≤ ‖g‖‖T‖, and hence ‖T †‖ ≤ ‖T‖, so T † is bounded.Claim: ‖T †‖ ≥ ‖Tx‖ for all x ∈ X with ‖x‖ = 1. If Tx = 0 the claim holds trivially, so assume x0 ∈ X

with ‖x0‖ = 1 and Tx0 6= 0. Set y0 = Tx0. By Thm 5.8b, there exists g0 ∈ Y ∗ such that ‖g0‖ = 1 andg0(y0) = ‖y0‖. Then

‖T †‖ = sup{‖T †g‖ : ‖g‖ = 1}≥ ‖T †g0‖ = ‖g0 ◦ T‖= sup{‖(g0 ◦ T )(x)‖ : ‖x‖ = 1}≥ ‖g0(T (x0))‖ = ‖g0(y0)‖ = ‖y0‖ = ‖Tx0‖.

This proves the claim and it follows that ‖T †‖ ≥ sup{‖Tx‖ : ‖x‖ = 1} = ‖T‖, hence ‖T †‖ = ‖T‖.b. Let x ∈ X. Then T ††x = x ◦ T †. For all g ∈ Y ∗ we have that

(x ◦ T †)(g) = x(T †g) = (T †g)(x) = g(Tx).

Therefore T †† = x ◦ T † = T x, so T ††|X = T , after identification of X with X and Y with Y .c. Suppose first that T (X) is dense in Y . Suppose g ∈ Y ∗ and T †g = 0. Then for all x ∈ X,

0 = (T †g)(x) = g(T (x)). Since g is continuous and vanishes on the dense subset T (X) of Y , it is identicallyzero, hence T † is injective.

Conversely, suppose T † is injective. BWOC, suppose that T (X) is not dense in Y . Then M = T (X)is a closed proper subspace of Y by Ex. 5.5. Let y0 ∈ Y −M, By Thm 5.8a, there exists g ∈ Y ∗ suchthat g|M = 0 and g(y0) 6= 0. Then for all x ∈ X, (T †g)(x) = g(T (x)) = 0, so T †g = 0 ∈ X∗. But sinceg 6= 0 ∈ Y ∗, it follows that T † is not injective, a contradiction.

d. If the range of T † is dense in X∗, then by part (c), T †† is injective. Under the (bijective) identificationin part (b), T ††|X = T , so T is injective.

Suppose X is reflexive and T is injective. Then X = X∗∗ so T †† = T after identification of X withX = X∗∗ and Y with Y . Since T †† is injective, by part (c), the range of T † is dense in X∗.

5.27 There exist meager subsets of R whose complements have Lebesgue measure zero.Solution:For each n ≥ 1, there is an open subset On ⊂ R such that Q ⊂ On and mOn < 1/n. Set Fn = Ocn. Then

31

each Fn is closed and Fn ∩ Q = ∅, so F on = ∅, hence Fn is nowhere dense. Let X =⋃n≥1 Fn. Then X is a

countable union of nowhere dense sets, hence meager, and

mXc = m

⋂n≥1

F cn

= m

⋂n≥1

On

= 0.

Extra prob. If X,Y are NVS’s and T ∈ L(X,Y ) then Γ(T ) is closed in X × Y .Solution:We will show that the complement is open. Let (x0, y0) ∈ Γ(T )c. Then r = ‖Tx0 − y0‖ > 0. Let δ′ > 0such that ‖x‖ < δ′ implies ‖Tx‖ < r/3 and set δ = min{δ′, r/3}. Then for all (x, y) ∈ X × Y such that‖(x, y)− (x0, y0)‖ < δ we have ‖x− x0‖ < δ and ‖y − y0‖ < δ, so

‖Tx− y‖ = ‖T (x− x0) + Tx0 − (y − y0)− y0‖≥ ‖Tx0 − y0‖ − ‖T (x− x0)‖ − ‖y − y0‖= r − ‖T (x− x0)‖ − ‖y − y0‖≥ r − r/3− r/3 = r/3.

Therefore B(δ, (x0, y0)) ⊂ Γ(T )c and so Γ(T )c is open, hence Γ(T ) is closed.

5.32 Let ‖ · ‖1, ‖ · ‖2 be norms on a vector space X such that ‖ · ‖1 ≤ ‖ · ‖2. If X is complete withrespect to both norms then the norms are equivalent.

Solution:Let X1 = (X, ‖ · ‖1), X2 = (X, ‖ · ‖2) and define T : X2 −→ X1 by T (x) = x. Clearly T is linear and for allx ∈ X, ‖Tx‖1 = ‖x‖1 ≤ ‖x‖2, so ‖T‖ ≤ 1. Thus, T ∈ L(X2, X1) and T is bijective, so by Corollary 5.11,T is an isomorphism, and hence T−1 ∈ L(X1, X2). In particular, T−1 is bounded, so there is a C > 0 suchthat for all x ∈ X, ‖x‖2 = ‖T−1x‖2 ≤ C‖x‖1, and therefore the norms are equivalent.

5.54 For every nonempty set A, `2(A) is complete.Solution:Notice first that for f ∈ `2(A),

‖f‖2 = 〈f, f〉 =∑a∈A

f(a)f(a) =∑a∈A|f(a)|2.

Suppose {fn}n≥1 is a Cauchy sequence in `2(A). Then for each ε > 0 there is an N ∈ N such that for allm,n ≥ N

ε2 > ‖fm − fn‖2 =∑a∈A|fm(a)− fn(a)|2,

and in particular, for each a0 ∈ A we have ε > |fm(a0) − fn(a0)| for all m,n ≥ N . Therefore {fn(a0)} is aCauchy sequence in C for each a0 ∈ A, so we define f(a) = limn→∞ fn(a) for all a ∈ A.

Claim: f ∈ `2(A). Since {fn} is Cauchy and∣∣‖fm‖ − ‖fn‖∣∣ ≤ ‖fm − fn‖ for all m,n ≥ 1, it follows that

‖fn‖ is Cauchy in R, hence convergent. By continuity of the norm, ‖f‖ = limn→∞ ‖fn‖ <∞, so f ∈ `2(A).Furthermore,

limn→∞

‖f − fn‖ = ‖ limn→∞

(f − fn)‖ = 0,

so fn → f in `2(A).

32

5.55 Let H be a Hilbert space.a. (The polarization identity) For all x, y ∈ H,

〈x, y〉 =1

4

(‖x+ y‖2 − ‖x− y‖2 + i‖x+ iy‖2 − i‖x− iy‖2

).

b. If H ′ is another Hilbert space, a linear map from H to H ′ is unitary iff it is isometric andsurjective.

Solution:For (a), fix x, y ∈ H and let α denote the RHS of the given equation. Then

α =1

4(〈x+ y, x+ y〉 − 〈x− y, x− y〉+ i〈x+ iy〉 − i〈x− iy, x− iy〉)

=1

4(2〈x, y〉+ 2〈y, x〉) +

i

4(2〈x, iy〉+ 2〈iy, x〉)

=1

2

(〈x, y〉+ 〈x, y〉

)+i

2

(〈x, iy〉+ 〈x, iy〉

)=

1

2(2 Re 〈x, y〉) +

i

2(2 Re 〈x, iy〉)

= Re 〈x, y〉+ iRe (−i〈x, y〉)= Re 〈x, y〉+ i Im 〈x, y〉 = 〈x, y〉.

For (b), suppose that U : H −→ H ′ is unitary. Then U is surjective by definition and for all x ∈ H,‖Ux‖2 = 〈Ux,Ux〉 = 〈x, x〉 = ‖x‖2, so U is an isometry.

Conversely, suppose U : H −→ H ′ is a surjective isometry. Then U is one-to-one since Ux = 0⇔ ‖Ux‖ =0⇔ ‖x‖ = 0⇔ x = 0. So U is bijective and therefore has a linear inverse. For each y ∈ H ′ there is a uniquex ∈ H such that Ux = y, so that ‖y‖ = ‖Ux‖ = ‖x‖, and so ‖U−1y‖ = ‖x‖ = ‖y‖, so U−1 is a boundedlinear map. Finally, to see that U is unitary it remains only to show that it preserves the inner product. Forall x, y ∈ H we have by (a) that

〈Ux,Uy〉 =1

4

(‖Ux+ Uy‖2 − ‖Ux− Uy‖2 + i‖Ux+ iUy‖2 − i‖Ux− iUy‖2

)=

1

4

(‖U(x+ y)‖2 − ‖U(x− y)‖2 + i‖U(x+ iy)‖2 − i‖U(x− iy)‖2

)=

1

4

(‖x+ y‖2 − ‖x− y‖2 + i‖x+ iy‖2 − i‖x− iy‖2

)= 〈x, y〉.

6.2 a. If f and g are measurable functions on X then ‖fg‖1 ≤ ‖f‖1‖g‖∞. If f ∈ L1 and g ∈ L∞,‖fg‖1 = ‖f‖1‖g‖∞ iff |g(x)| = ‖g‖∞ a.e. on the set where f(x) 6= 0.b. ‖ · ‖∞ is a norm on L∞.c. ‖fn − f‖∞ → 0 iff there exists E ∈M such that µEc = 0 and fn → f uniformly on E.d. L∞ is a Banach space.e. The simple functions are dense in L∞.

Solution:

a. Since |g| ≤ ‖g‖∞ a.e., ‖fg‖1 =∫|f ||g| ≤

∫|f |‖g‖∞ = ‖f‖1‖g‖∞. Let E = {x ∈ X : f(x) 6= 0}.

It’s clear that if |g(x)| = ‖g‖∞ a.e. on E then ‖fg‖1 = ‖f‖1‖g‖∞, so we will show the converse. Supposef ∈ L1, g ∈ L∞ and

∫|fg| =

∫|f |‖g‖∞. Then

0 =

∫(|f |‖g‖∞ − |fg|) =

∫|f |(‖g‖∞ − |g|) =

∫E

|f |(‖g‖∞ − |g|),

since ‖g‖∞ − |g| is finite on X. Furthermore, since ‖g‖∞ − |g| ≥ 0 a.e., the integrand is nonegative a.e., so|f |(‖g‖∞ − |g|) = 0 a.e. on E, and hence ‖g‖∞ = |g| a.e. on E since f 6= 0 on E.

33

b. The only norm property which is not immediately obvious is the triangle inequality, so supposef, g ∈ L∞. Then |f | ≤ ‖f‖∞ a.e. and |g| ≤ ‖g‖∞ a.e. imply |f |+|g| ≤ ‖f‖∞+‖g‖∞ a.e., but |f+g| ≤ |f |+|g|,so |f + g| ≤ ‖f‖∞ + ‖g‖∞ a.e., hence ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞.

c. Suppose ‖fn − f‖∞ → 0. Then for all k ∈ N there is an Nk so that ‖fn − f‖∞ ≤ 1/k for all n ≥ Nk.So, for each k ∈ N there is an Ek ∈ M such that µEck = 0 and |fn − f | ≤ 1/k on Ek for all n ≥ Nk. SetE =

⋂∞k=1Ek. Then µEc = µ (

⋃∞k=1E

ck) = 0.

We will now show that fn → f uniformly on E. Let ε > 0. Then there is a k ∈ N such that 1/k ≤ ε andfor all n ≥ Nk we have |fn − f | ≤ 1/k on Ek for all n ≥ Nk. Since 1/k ≤ ε and E ⊂ Ek, it follows that|fn − f | ≤ ε on E for all n ≥ Nk. The converse is clear.

d. The only thing that remains to show is that L∞ is complete. Suppose that {fn} is a Cauchy sequencein L∞. For each k ∈ N there exists Nk such that ‖fm − fn‖∞ < 1/k for all m,n ≥ Nk. Let Ek ∈ M sothat |fm − fn| < 1/k on Ek for all m,n ≥ Nk and µEck = 0. Set E =

⋂∞k=1Ek. Then µEc = 0 and for all

x ∈ E, {fn(x)} is Cauchy, so let f(x) = limn→∞ fn(x)χE(x). Then fn → f uniformly on E, so by part (c),

‖fn − f‖∞ → 0. In particular, there is an N so that ‖fN − f‖∞ < 1 and hence ‖f‖∞ ≤ ‖fN‖+ 1 <∞, sof ∈ L∞ and L∞ is therefore complete.

e. Let f ∈ L∞ and ε > 0. It suffices to show that there is a simple function ϕ ∈ L∞ with ‖f −ϕ‖∞ < ε.Suppose first that f is real-valued. Let n ∈ N such that 1/n < ε and let M = ‖f‖∞. For each integerj ∈ [−Mn,Mn] let

Ej =

{x ∈ X :

j

n≤ f(x) ≤ j + 1

n

},

and set ϕ =∑mNj=−Mn(j/n)χ

Ej. Then ϕ ∈ L∞ is simple and for all x with |f(x)| ≤ M there is a unique j

such that x ∈ Ej and so |f(x)− ϕ(x)| ≤ 1/n < ε. Thus, |f − ϕ| < ε a.e., so ‖f − ϕ‖∞ < ε.For the general case, let ϕ1, ϕ2 be simple functions for which ‖Re (f)−ϕ1‖∞ < ε/2 and ‖ Im (f)−ϕ2‖∞ <

ε/2. Then ϕ = ϕ1 + iϕ2 is a simple function in L∞ and ‖f − ϕ‖∞ = ‖Re (f) − ϕ1 + i( Im (f) − ϕ2)‖∞ ≤‖Re (f)− ϕ1‖∞ + ‖i( Im (f)− ϕ2)‖∞ < ε.

6.5 Suppose 0 < p < q <∞. Then Lp 6⊂ Lq iff X contains sets of arbitrarily small positive measure.Lq 6⊂ Lp iff X contains sets of arbitrarily large finite measure. What about the case q =∞?

Solution:

Suppose X contains sets of arbitrarily small positive measure. Let F1 ⊂ X be a measurable set with0 < µF1 < 1/2. Inductively, let Fn+1 be a measurable set with 0 < µFn+1 < (1/4)µFn. For n ≥ 1 setEn = Fn −

⋃∞k=n+1 Fk. Then for i < j,

Ei ∩ Ej = Fi ∩

( ∞⋂k=i+1

F ck

)∩ Fj ∩

∞⋂k=j+1

F ck

⊂ ( ∞⋂k=i+1

F ck

)∩ Fj = ∅ (since )i+ 1 ≤ j.

Furthermore µEn ≥ µFn −∑∞k=n+1 µFk ≥ µFn − (1/2)µFn > 0 and µEn ≤ µFn < 2−n. Therefore {En} is

a disjoint sequence with 0 < µEn < 2−n.Set ϕ =

∑∞n=1

1(nµEn)1/q

χEn

. Since the En’s are disjoint, this sum converges everywhere on X. Finally,

we have

‖ϕ‖qq =

∫|ϕ|q =

∞∑n=1

µEnnµEn

=

∞∑n=1

1

n=∞,

so ϕ 6∈ Lq. But

‖ϕ‖pp =

∫|ϕ|p =

∞∑n=1

µEn(nµEn)p/q

=

∞∑n=1

(µEn)1−p/q

np/q<

∞∑n=1

(µEn)1−p/q <

∞∑n=1

2−n(1−p/q) <∞,

since 21−p/q > 1. Therefore ϕ ∈ Lp and so Lp 6⊂ Lq.

34

Conversely, suppose Lp 6⊂ Lq and let f ∈ Lp with f 6∈ Lq. For each positive integer n let En = {x ∈ X :|f(x)| > n}. By Prop. 6.10, if f ∈ L∞ then f ∈ Lq, so we must have ‖f‖∞ =∞ and therefore µEn > 0 forall n ≥ 1. On the other hand, for each n ≥ 1,

‖f‖pp =

∫|f |p ≥

∫En

|f |p ≥∫En

np = npµEn,

so limn→∞ µEn ≤ limn→∞‖f‖ppnp = 0, so µEn → 0 as n→∞.

For the second question, suppose first that X contains sets of arbitrarily large finite measure. LetF1 ⊂ X with 1 ≤ µF1 <∞. Inductively, let Fn+1 ⊂ X with ∞ > µFn+1 >

∑nk=1 µFk + 1. Set E1 = F1 and

En+1 = Fn+1 −⋃nk=1 Fk for n ≥ 1. The sequence {En} is disjoint and µEn ≥ µFn+1 −

∑nk=1 µFk > 1. Let

f =∑∞n=1

1(nµEn)1/p

χEn

. Then

‖f‖pp =

∫|f |p =

∞∑n=1

µEnnµEn

=∞,

so f 6∈ Lp. But

‖f‖qq =

∫|f |q =

∞∑n=1

µEn(nµEn)q/p

=

∞∑n=1

(µEn)1−q/p

nq/p<

∞∑n=1

1

nq/p<∞,

so f ∈ Lq and hence Lq 6⊂ Lp.Conversely, suppose Lq 6⊂ Lp and let f ∈ Lq − Lp. Let En = {x ∈ X : |f(x)| > 1/n}. Since

‖f‖qq =∫|f |q > µEn

nq for all n ≥ 1, it follows that µEn ≤ nq‖f‖qq <∞ for all n. Furthermore, the sets En areincreasing, so with E =

⋃n≥1En we have µE = limn→∞ µEn. But E = {x ∈ X : f(x) 6= 0}, so if µE <∞

then by Prop. 6.12 applied to the measure space (E,M|E , µ|E), we’d have ‖f‖p <∞, a contradiction. Thuslimn→∞ µEn =∞. (Alternatively, apply Holder as in the proof of Theorem 6.12 to the function fχ

E.)

What about the case q = ∞? The first assertion continues to hold with q = ∞. Suppose X containssets of arbitrarily small positive measure. As above, let {En} be a disjoint sequence with 0 < µEn < 2−n.Set f =

∑nχ

En. Then f 6∈ L∞ but ‖f‖pp =

∑npµEn <

∑∞n=1

np

2n < ∞, so f ∈ Lp and hence Lp 6⊂ L∞.

Conversely, if Lp 6⊂ L∞ then let f ∈ Lp − L∞. Set En = {x ∈ X : |f(x)| ≥ n}. Since f 6∈ L∞ we have

µEn > 0 for all n and ‖f‖pp ≥∫En|f |p ≥ npµEn so µEn ≤

‖f‖ppnp → 0 as n → ∞, so X contains sets of

arbitrarily small positive measure.For the second statement, the “only if” fails with q = ∞. Let X = {x0}, M = {∅, X} and µ∅ = 0,

µX = ∞. The constant function f = 1 is in L∞ but not Lp, so L∞ 6⊂ Lp bu X does not contain setsof arbitrarily large finite measure. The converse, however, does still hold with q = ∞. If X contains setsof arbitrarily large positive measure, let {En} be a disjoint sequence with 1 ≤ µEn < ∞ for all n. Letf =

∑∞n=1 χEn

. Then |f | ≤ 1 on X so f ∈ L∞, but ‖f‖pp =∫|f |p = µ (

⋃En) = ∞, so f 6∈ Lp, hence

L∞ 6⊂ Lp.

6.6 Suppose 0 < p0 < p1 ≤ ∞. Find examples of functions f on (0,∞) (with Lebesgue measure) suchthat f ∈ Lp iff (a) p0 < p < p1, (b) p0 ≤ p ≤ p1, (c) p = p0.

Solution:Assume first that p1 <∞. For (a), set f = x−1/p1χ

(0,1)+ x−1/p0χ

[1,∞). Then∫

|f |p =

∫ 1

0

dx

xp/p1+

∫ ∞1

dx

xp/p0.

Both integrals are nonnegative, so ‖f‖p <∞ iff both integrals are finite. The first is finite iff p < p1 and thesecond is finite iff p > p0, hence f ∈ Lp iff p0 < p < p1.

For (b), set

f =1

x1/p1 | log x|2/p1χ

(0,1/2)+

1

x1/p0 | log x|2/p0χ

[2,∞).

It is a calculus exercise to verify that∫|f |p < ∞ iff p0 ≤ p ≤ p1. For (c), take p1 = p0 in the function

defined f defined in (b).

If p1 =∞, for (a) take f = | log(x−1)|x1/p0

χ[1,∞)

. For (b) take f = 1x1/p0 | log x|2/p0 χ[2,∞)

.

35

6.7 If f ∈ Lp ∩ L∞ for some p <∞ so that f ∈ Lq for all q > p, then ‖f‖∞ = limq→∞ ‖f‖q.Solution:It’s clear if ‖f‖∞ = 0, so assume ‖f‖∞ > 0. Let 0 < ε < ‖f‖∞, and set A = {x ∈ X : |f(x)| > ‖f‖∞ − ε}.Then f ∈ Lp implies that µA <∞, and by construction, µA > 0. For all q ≥ p,

‖f‖q =

(∫|f |q

)1/q

≥(∫

A

|f |q)1/q

≥ (‖f‖∞ − ε)(µA)1/q,

so lim infq→∞ ‖f‖q ≥ (‖f‖∞ − ε) lim infq→∞(µA)1/q = ‖f‖∞ − ε. Since this holds for all ε > 0, it followsthat lim infq→∞ ‖f‖q ≥ ‖f‖∞.

For the reverse inequality, we appeal to Prop. 6.10 with r = ∞ to conclude that for all q > p, ‖f‖q ≤‖f‖p/qp ‖f‖1−p/q∞ . Therefore, lim supq→∞ ‖f‖q ≤ lim supq→∞ ‖f‖

p/qp ‖f‖1−p/q∞ = ‖f‖∞.

6.10 Suppose 1 ≤ p <∞. If fn, f ∈ Lp and fn → f a.e., then ‖fn − f‖p → 0 iff ‖fn‖p → ‖f‖p.Solution:

Suppose first that ‖fn‖p → ‖f‖p. Then |fn − f |p ≤ (|fn|+ |f |)p ≤ (2 max{|f |, |fn|})p ≤ 2p(|f |p + |fn|p).By assumption, limn→∞

∫2p(|f |p + |fn|p) =

∫(lim 2p(|f |p + |fn|p) < ∞, so by the Generalized LDCT it

follows that

lim ‖f − fn‖pp = lim

∫|f − fn|p =

∫lim |f − fn|p = 0.

The converse is easy: ‖f − fn‖p → 0 implies that fn → f in Lp. Since the norm ‖ · ‖p is continuous on Lp,it follows that limn→∞ ‖fn‖p = ‖ limn→∞ fn‖p = ‖f‖p.

6.12 If p 6= 2 the Lp norm does not arise from an inner product on Lp except in the trivial cases whendimLp ≤ 1.

Solution:Let 1 ≤ p ≤ ∞, and suppose that (X,M, µ) is a measure space and dimLp > 1. If every E ∈ M hadµE ∈ {0,∞} then the dimension of Lp would be zero, so there must exist a set E1 ∈M with 0 < µE1 <∞.Furthermore, if every E ∈ M with 0 < µE < ∞ satisfied µ(E4E1) = 0, then every function in Lp wouldequal a constant multiple of χ

E1a.e., which would give dimLp = 1. Thus, there is a set E2 ∈ M with

0 < µE2 <∞ and E1 ∩ E2 = ∅.Suppose now that p < ∞. Set f1 = χ

E1/(µE1)1/p and f2 = χ

E2/(µE2)1/p. Then ‖f1‖p = ‖f2‖p = 1.

Suppose that the norm on Lp does arise from an inner product. Then the parallelogram law holds, so that

‖f1 + f2‖2p + ‖f1 − f2‖2p = 2(‖f1‖2p + ‖f2‖2p) = 4.

Computing the LHS we have

‖f1 + f2‖2p + ‖f1 − f2‖2p =

(∫ ∣∣∣∣ χE1

(µE1)1/p+

χE2

(µE2)1/p

∣∣∣∣p)2/p

+

(∫ ∣∣∣∣ χE1

(µE1)1/p−

χE2

(µE2)1/p

∣∣∣∣p)2/p

=

(∫ ( χE1

µE1+χ

E2

µE2

))2/p

+

(∫ ( χE1

µE1−χ

E2

µE2

))2/p

(since E1 ∩ E2 = ∅)

= 22/p + 22/p = 21+2p .

So we must have 21+2p = 4, so that 1 + 2

p = 2 and hence p = 2.On the other hand, if p =∞, set f1 = χ

E1and f2 = χ

E2. Then the parallelogram law is violated since

‖f1 + f2‖2∞ + ‖f1 − f2‖2∞ = 2 6= 4 = 2(‖f1‖2∞ + ‖f2‖2∞).

So the inner product on L∞ also does not arise from an inner product.

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6.13 Lp(R,m) is separable for 1 ≤ p <∞. However, L∞(R,m) is not separable.Solution:

Let 1 ≤ p < ∞, and S = {∑ni=1 riχ(ci,di)

: n ∈ N, ri, ci, di ∈ Q}. Clearly S is countable and we will

show that S is dense in Lp(R,m).Let f ∈ Lp(R,m) and ε > 0. By Prop. 6.7 there is a simple function ϕ =

∑ni=1 aiχEi

with mEi < ∞for 1 ≤ i ≤ n such that ‖f − ϕ‖p < ε/3. Let M = max{|ai − aj | : 1 ≤ i < j ≤ n}. Then ϕ ∈ L1, so byTheorem 2.26 there is a step function ψ with finite measure support such that

∫|ϕ − ψ| <

(ε3

)p 1Mp−1 . It

follows that‖ϕ− ψ‖1/p1 ‖ϕ− ψ‖1−1/p∞ ≤ ‖ϕ− ψ‖1/p1 M1−1/p < ε/3.

But ϕ− ψ ∈ L1 ∩ L∞ so by Prop. 6.10 it follows that

‖ϕ− ψ‖p ≤ ‖ϕ− ψ‖1/p1 ‖ϕ− ψ‖1−1/p∞ < ε/3.

Suppose ψ =∑ni=1 r

′iχ(c′i,d

′i)

with the intervals (c′i, d′i) disjoint. For each 1 ≤ i ≤ n let ri, ci, di ∈ Q such that

‖riχ(ci,di)

− r′iχ(c′i,d′i)‖p < ε/(3n) (it’s clear that they exist, but the argument is straightforward if you want

to do it). Then s =∑ni=1 riχ(ci,di)

∈ S and ‖s− ψ‖p ≤∑ni=1 ‖riχ(ci,di)

− r′iχ(c′i,d′i)‖p < ε/3 and so

‖f − s‖p ≤ ‖f − ϕ‖p + ‖ϕ− ψ‖p + ‖ψ − s‖p < ε.

Therefore S is dense in Lp(R,m).Let {fn} be a countable sequence in L∞(R,m). For each n ≥ 1, let

an =

{−1, if fn ≥ 0 a.e. on [n, n+ 1),

1, otherwise.

Define g(x) =∑∞n=1 anχ[n,n+1)

. By construction, ‖g‖∞ = 1 so g ∈ L∞(R,m) and ‖g − fn‖∞ ≥ 1 for all n,

so {fn} is not dense in L∞(R,m). Therefore no countable sequence in L∞(R,m) is dense and so L∞(R,m)is not separable.

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