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Real Analysis Homework: #1

Yingwei Wang ∗

Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 Banach space

Question: Let (xn) ⊂ X be a Banach space, and∑

∞

n=1‖xn‖ is convergent. Proof that

∑∞

n=1xn is

convergent in X.

Proof: Suppose∑

∞

n=1‖xn‖ = M < ∞, then ∀ ε > 0,∃N, s.t.

∑∞

n=N ‖xn‖ < ε.

Let sn =∑n

i=1xi, then ∀n > m > N ,

‖sn − sm‖ = ‖n∑

i=m

xi‖ 6

∞∑

i=m

‖xi‖ <∞∑

i=N

‖xi‖ < ε.

Hence, (sn) is a Cauchy sequence and must converge to an element in X.

2 Continuous function

Question: f : (X, τX ) → (Y, τY ) is continuous ⇔ ∀x0 ∈ X and any neighborhood V of f(x0), thereis a neighborhood U of x0 such that f(U) ⊂ V .

Proof: “⇒”: Let x0 ∈ X f(x0) ∈ Y . For each open set V containing f(x0), since f iscontinuous, f−1(V ) which containing x0 is open. Then, there is a neighborhood U of x0 such thatx0 ∈ U ⊂ f−1(V ), that is to say f(U) ⊂ V .

“⇐”: Let V ∈ Y . ∀y ∈ V , choose Vy satisfy y ∈ Vy ⊂ V, Vy ∈ τY . Then V =⋃

Vy.

Let x = f−1(y), then ∀Vy,∃Ux ∈ τX , s.t. x ∈ Ux and f(Ux) ⊂ Vy.

Let U =⋃

Ux, then U ∈ τX and f(U) = V .

Therefore, f : (X, τX) → (Y, τY ) is continuous.

∗

E-mail address: [email protected]; Tel : 765 337 3504

I

Yingwei Wang Real Analysis

3 Closure

3.1

Question: Topological space (X, τ), x ∈ X,E ⊂ X, x ∈ E ⇔ ∀ neighborhoodV of x, E⋂

V 6= ∅.

Proof: Actually, this question is equal to the following one:

(*) Topological space (X, τ), x /∈ E ⇔ There exist an open set V containing x that E⋂

V = ∅.

We just need to prove (*).

“⇒ ” If x /∈ E, then V = X\E is an open set containing x and E⋂

V = ∅.

“⇐” If there exist an open set V containing x that E⋂

V = ∅, then X\V is a closed setcontaining E.

By the definition of closer E, that is the intersection of all closed sets containing E, the setX\V must contain E.

Thus, x /∈ E.

3.2

Question: Metric space (X, d), x ∈ X,E ⊂ X, x ∈ E ⇔ there is a sequence (en)∞n=1in E such

that limn→∞ d(en, x) = 0.

Proof: Similarly as Section 3.1, the above question is equal to the following one:

(**) Metric space (X, d), x /∈ E ⇔ ∀e ∈ E,∃ ε0, s.t. d(e, x) > ε0.

We just need to prove (**).

“⇒ ” Suppose x /∈ E. Since X\E is an open set, then ∃B(x, δ) ⊂ X\E. That is to say∀e ∈ E, d(e, x) > δ.

“⇐” If ∀e ∈ E,∃ ε0, s.t. d(e, x) > ε0, then B(x, ε0

2) is an open set that containing x and

B(x, ε0

2)⋂

E = ∅. So X\B(x, ε0

2) is a closed set containing E.

By the definition of E, E ⊂ X\B(x, ε0

2).

Since x ∈ B(x, ε0

2), x /∈ E.

4 Distance between point and set

If E is a nonempty subset of a metric space X, define the distance from x ∈ X to E by

d(x,E) = infz∈E

d(x, z). (4.1)

4.1

(a) Prove that d(x,E) = 0 ⇔ x ∈ E.

Proof: If infz∈E d(x, z) = 0, then ∀ε, ∃ zε, s.t. d(x, zε) < ε.

II


Choosing ε = 1

n, then we can get a sequence (zn)∞n=1

satisfying limn→∞ d(x, zn) = 0.

According to Section 3.2, x ∈ E.

“⇐” If x ∈ E, by the previously Section 3.2, there is a sequence (zn)∞n=1in E such that

limn→∞ d(zn, x) = 0.

Then d(x,E) = infz∈E d(x, z) = limn→∞ d(zn, x) = 0.

4.2

(b) Prove that x 7→ d(x,E) is a uniformly continuous function on X, by showing that

|d(x,E) − d(y,E)| ≤ d(x, y).

for all x ∈ X, y ∈ X.

Proof: By the definition of (4.1), we can get:

∀x, y ∈ X,∀ε > 0,∃zx ∈ E, s.t. d(x,E) = d(x, zx) + ε, ∃zy ∈ E, s.t. d(y,E) = d(y, zy) + ε.

Without loss of generality, we suppose that d(x,E) ≤ d(y,E). Then when ε is sufficient small,d(y, zy) ≤ d(y, zx).

According to triangle inequality,

d(x, y) ≥ d(y, zx) − d(x, zx) ≥ d(y, zy) − d(x, zx) = d(x,E) − d(y,E).

Besides, in the case that d(x,E) ≥ d(y,E), we can get d(x, y) ≥ d(y,E) − d(x,E). So we canconclude that |d(x,E) − d(y,E)| ≤ d(x, y).

∀x0 ∈ X,∀ε > 0, just choose δ = ε2, then ∀x ∈ B(x0, δ), we have |d(x,E) − d(x0, E)| ≤

d(x, x0) < δ < ε. That is to say x 7→ d(x,E) is a uniformly continuous function on X.

III


Yingwei Wang ∗


1 Banach space

Question: Let C([a, b]) denote the linear space of continuous function f : [a, b] → R. Show thatC[a, b] is a Banach space with respect to the norm

‖f‖ = max{|f(t)| : t ∈ [a, b]}.

Proof: Let (fn)∞n=1 be a Cauchy sequence in C([a, b]). For ∀t ∈ [a, b],∀m,n ∈ N, |fm(t)−fn(t)| ≤‖fn − fm‖, which means (fn(t))∞n=1 is a Cauchy sequence in R and must converge to an elementin R. So we can define a function f : [a, b] → R that

f(t) = limn→∞

fn(t), ∀t ∈ [a, b].

First, we will show that fn → f when n → ∞. ∀ε > 0, there is an N s.t. ∀n,m ≥ N we have‖fn − fm‖ < ε. For each x ∈ [a, b] we have

|fn(x) − f(x)| = limm→∞

|fn(x) − fm(x)|

≤ limm→∞

‖fn − fm‖

≤ ε,

for all n ≥ N . Then for all n ≥ N ,

‖fn − f‖ = maxx∈[a,b]

{|fn(x) − f(x)|} ≤ ε.

Thus fn → f .

∗


I


Second, we will show that f ∈ C([a, b]). Choose a point t0 ∈ [a, b]. Since fn ∈ C([a, b]), for theprevious ε, ∃ δn, s.t. ∀t ∈ (t0 − δ, t0 + δ), |fn(t) − fn(t0)| < ε. For t ∈ (t0 − δ, t0 + δ), n > N wehave

|f(t) − f(t0)| ≤ |f(t) − fn(t)| + |fn(t) − fn(t0)| + |fn(t0) − f(t0)|

< ε + ε + ε

< 3ε.

Thus f ∈ C([a, b]) and C([a, b]) is a Banach space.

2 Banach space

Question: Let l1(N) be the linear space of all the functions f : N → R with the property that|f |1 :=

∑∞n=1 |f(n)| < ∞. Prove that (l1(N), ‖ · ‖1) is a Banach space.

Proof: Let (fi)∞i=1 be a Cauchy sequence in l1(N). For ∀n ∈ N,∀i, j ∈ N, |fi(t) − fj(t)| ≤

‖fi − fj‖, which means (fi)∞i=1(n) is a Cauchy sequence in R and must converge to an element in

R. So we can define a function f : N → R that

f(n) = limi→∞

fi(n), ∀n ∈ N.

First, we will show that fi → f when i → ∞. ∀ε > 0, there is an I s.t. ∀i, j ≥ I we have‖fi − fj‖ < ε. For each n ∈ N we have

|fi(n) − f(n)| = limj→∞

|fi(x) − fj(x)|

≤ limj→∞

‖fi − fj‖

≤ ε,

for all i ≥ I. Then for all i ≥ N ,

‖fi − f‖ = maxn∈N

{|fi(n) − f(n)|} ≤ ε.

Thus fi → f .

Second, we will show that f ∈ l1(N).

‖f‖1 =

∞∑

n=1

|f(n)|

=∞∑

n=1

| limi→∞

fi(n)|

≤ limi→∞

∞∑

n=1

|fi(n)| < ∞.

II


Then f ∈ l1(N) and (l1(N), ‖ · ‖1) is a Banach space.

3 Application of Baire Theorem

Question: Let f : R → R be a smooth function (i.e. C∞). Suppose that for each t ∈ R there isnt ∈ N such that f (nt)(t) = 0. Prove that there is an interval I of positive length such that therestriction of f to I is a polynomial.

Proof: Define Tn := {t ∈ R : f (n)(t) = 0}. It is easy to verify that for each n the restrictionof f to Tn is a polynomial with at most (n + 1) degree. Then we just need to show that ∃n0, s.t.(Tn0

)◦ 6= ∅.

First, we claim that ∀n ≥ 1, Tn is a closed set. Consider the set T cn = R\Tn: ∀t ∈ T c

n, f (n) 6= 0.Since f ∈ C∞, f (n) is continuous, we know that ∃ δ, s.t. ∀x ∈ (t − δ, t + δ), f (n)(x) 6= 0. That isto say (t − δ, t + δ) ⊂ T c

n. Then R\Tn is an open set while Tn is a closed set.

Second, we claim that∞⋃

n=1Tn = R. ∀t ∈ R, ∃ nt ∈ N, s.t. f (nt)(t) = 0, which means t ∈ Tnt

.

Thus,∞⋃

n=1Tn = R.

According to the Baire Category Theorem, ∃n0, s.t. (Tn0)◦ 6= ∅. That is to say ∃ interval I ⊂

(Tn)◦, s.t. the restriction of f to I is a polynomial.

4 Outer measure

Question: Let m∗(A) denote the outer measure of A ⊂ R. Show that if B ⊂ R and m∗(B) = 0,then m∗(A ∪ B) = m∗(A).

Proof: On one hand,A ⊂ (A ∪ B) ⇒ m∗(A) ≤ m∗(A ∪ B).

On the other hand,

m∗(A ∪ B) ≤ m∗(A) + m∗(B) = m∗(A) + 0 = m∗(A).

Thus,m∗(A ∪ B) = m∗(A).

5 Continuity of function

Question: Let (xn)∞n be an enumeration of Q. Define f : R → R by f(x) =∑ 1

2n . where thesummation is extended over all n such that xn < x. Prove that f is discontinuous at each rationalnumber and that f is continuous at each irrational number.

III


Proof: Define Ax = {i ∈ N : xi < x & xi ∈ {xn}∞n = Q}. Then f(x) =

∑

n∈Ax

12n , ∀x ∈ R.

5.1 Rational points ⇒ Discontinuity

Fix xr ∈ Q, r ∈ N. Let xn ∈ Q and xn > xr. It is obviously that r ∈ (Axn\ Axr

). Then

|f(xn) − f(xr)| =

∣

∣

∣

∣

∣

∣

∑

p∈Axn

1

2p−

∑

q∈Axr

1

2q

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∑

i∈(Axn\Axr )

1

2i

∣

∣

∣

∣

∣

∣

>1

2r.

which means limx∈Q, x→x+

r

f(x) 6= f(xr). That is to say f is discontinuous at each rational number.

5.2 Irrational points ⇒ Continuity

Since∞∑

n=1

12n = 1, we can know that ∀ε > 0, ∃N ∈ N, s.t.

∞∑

n=N+1

12n < ε. Let the set Sε =

{x1, x2, · · · , xN}. We can rearrange the elements of the set Sε such that x1 < x2 < · · · < xN .

∀α ∈ (R \ Q), we can choose δ > 0 as the following way:

δ =

12(x1 − α) if α < x1,12(min{α − xi, xi+1 − α}) if xi < α < xi+1,12(α − xN ) if α > xN .

Then (Ax+δ \ Ax−δ) ∩ ASε= ∅. That is to say ∀n ∈ (Ax+δ \ Ax−δ), n ≥ N + 1.

So ∀x ∈ (x− δ, x+ δ), |f(x)− f(α)| ≤ |f(x+ δ)− f(x− δ)| =∑

n∈(Ax+δ\Ax−δ)

12n <

∞∑

n=N+1

12n < ε,

which means f is continuous at the point α ∈ (R \ Q).

IV


Yingwei Wang ∗


1 Measure inequality

Question: Let (X,A, µ) be a measure space. Let (Ak)∞

k=1be a sequence of sets in A. Prove that

µ

(

∞⋃

n=1

∞⋂

k=n

Ak

)

≤ limk→∞

inf µ(Ak).

Proof: Let Bn =∞⋂

k=n

Ak, then Bn = Bn+1 ∩ An, Bn ⊂ Bn+1. So

µ(

m⋃

n=1

Bn) = µ(Bm) =

∞⋂

k=m

Ak ⊂ Ak, ∀k ≥ m

⇒ µ

(

m⋃

n=1

Bn

)

≤ µ(Ak), ∀k ≥ m

⇒ µ

(

∞⋃

n=1

Bn

)

≤ limk→∞

inf µ(Ak).

2 Example of measure

Question: Let µ be a measure on (R,B), where B are the Borel sets, such that µ([0, 1)) = 1 andµ(x + B) = µ(B) for all x ∈ R and B ∈ B. Prove that

(1) µ([0, 1/n)) = 1/n for all integers n ≥ 1 and

(2) µ([a, b)) = b − a for all real numbers a < b.

∗


I


Proof: (1) By the assumption that µ(x + B) = µ(B), we can know that µ([0, 1/n)) =µ([1/n, 2/n)) = · · · = µ([(n − 1)/n, 1)). Then

1 = µ([0, 1]) =

n∑

k=1

µ([(k − 1)/n, k/n)) = nµ([0, 1/n))

⇒ µ([0, 1/n)) = 1/n.

(2) Since for any a < b, µ([a, b)) = µ([0, b − a)), we only need to consider the cases thata = 0, b > 0.

If b ∈ Q, then b = pq

where p, q ∈ N, q 6= 0.

µ([0, b)) = µ([0,p

q)) =

p−1∑

k=0

µ([k

q,k + 1

q)) = p µ([0,

1

q)) =

p

q= b.

If b ∈ R\Q, then ∃ (bn)∞n=1 ⊂ Q s.t. limn→∞

bn = b. Then

µ([0, b)) = limn→∞

µ([0, bn)) = limn→∞

bn = b.

3 Lebesgue outer measure

Question: Let m∗ be the Lebegue outer measure on R. For any two sets A,B ⊂ R, prove theinequality:

m∗(A) + m∗(B) ≤ 2m∗(A△B) + 2m∗(A ∩ B). (3.1)

Proof: For the set A, we have

m∗(A) ≤ m∗(A\B) + m∗(A ∩ B) (since A = (A\B) ∪ (A ∩ B))

≤ m∗(A△B) + m∗(A ∩ B). (since (A\B) ⊂ (A△B)) (3.2)

Similarly, for the set B, we have

m∗(B) ≤ m∗(A△B) + m∗(A ∩ B). (3.3)

By (3.2)-(3.3), we can get (3.1).

4 Zero measure set

Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebesgue measurable set.Suppose that m(A ∩ [a, b]) < (b − a)/2 for all a < b real numbers. Show that m(A) = 0.

II


Proof: On one hand, by the definition of Lebesgue measure,

m(A) = inf{∞∑

n=1

l(In) : In ⊂ R are open intervals, A ⊂∞⋃

n=1

In},

we know that ∀ε > 0, ∃ {In}∞

n=1, s.t.

m(A) ≥∞∑

n=1

l(In) − ε,

where A ⊂∞⋃

n=1

In and Ii ∩ Ij = ∅, i 6= j.

We can choose ε = 1

4m(A) in the above inequality, then we can get

m(A) ≥4

5

∞∑

n=1

l(In). (4.1)

On the other hand, by the assumption, m(A ∩ In) < 1

2l(In),∀n ∈ N+. So we have

m(A) =

∞∑

n=1

m(A ∩ In) ≤1

2

∞∑

n=1

l(In). (4.2)

From (4.1)-(4.2) we can conclude that m(A) = 0.

5 Measure function

Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebegue measurableset with m(A) < ∞. Show that the function f : R → [0,∞), f(x) = m(A ∩ (−∞, x)) iscontinuous. Deduce that for every β ∈ [0,m(A)] there is a Lebesgue measurable set B ⊂ A suchthat m(B) = β.

Proof:

5.1 The continuity of the auxiliary function g

Since m(A) < ∞, by the Proposition 15 on Page 63 in Royden’s book, given ∀ε > 0, there is afinite union U of open intervals such that

m∗(U△A) < ε. (5.1)

III


Suppose U =n⋃

i=1

(ai, bi), and a = a1, b = bn, then U ⊂ [a, b].

On the interval [a, b], we can define the function g such that

g(x) = m(A ∩ [a, x)), ∀x ∈ [a, b].

Then for ∀∆x > 0,

A ∩ [a, x + ∆x) = (A ∩ [a, x))⋃

(A ∩ [x, x + ∆x))

⇒ g(x + ∆x) ≤ g(x) + ∆x

⇒ g(x + ∆x) − g(x) ≤ ∆x

For ∆x < 0, we can get the similar result:

g(x + ∆x) − g(x) ≥ −∆x

Then we have|g(x + ∆x) − g(x)| ≤ |∆x|

which means g ∈ C([a, b]).

5.2 The continuity of f

For ∀x ∈ [a, b], choose ∆x < ε, then

|g(x) − f(x)| ≤ m(U△A) < ε

⇒ |f(x + ∆x) − f(x)| ≤ |f(x + ∆x) − g(x + ∆x)| + |g(x + ∆x) − g(x)| + |g(x) − f(x)| < 3ε.

which means f ∈ C([a, b]).

For x ≤ a, |f(x)| ≤ m(U△A) < ε, so f ∈ C((−∞, a]). Similarly, we have f ∈ C([b,∞)).Thus, f ∈ C(R).

5.3 The Intermediate Value Theorem

Since g(a) = 0, g(b) = m(A), by the Intermediate Value Theorem of continuous functions, for∀β ∈ [0,m(A)] there is an x0 ∈ [a, b] such that g(x0) = β. Then we can choose B = A ∩ [a, x0].

IV


Yingwei Wang ∗


1 Measurable function

Question: Let (X,A, µ) be a measure space. Let (fn)∞n=1 be a sequence of measurablefunctions: fn : X → R. Show that the set of points x where the limit lim

n→∞fn exists

(finite or infinite) is a measurable set.

Proof: Let Ak = {x ∈ X : |fn(x) − fm(x)| < 1k , ∀m,n > k}, A = lim

k→∞Ak, where

m,n, k ∈ N. It is easy to know that each Ak is measurable set since fn is a measurablefunction. We have this observation: Ak+1 ⊂ Ak ⇒ A ⊂ Ak,∀k. Thus, A is a measurableset.

Let B = {x ∈ R : limn→∞

fn(x) exists}. I will show that A = B.

On one hand, let x ∈ A, then for ∀k ∈ N, |fn(x) − fm(x)| < 1k , ∀m,n > k, which

mean fn(x) is a Cauchy sequence in R. So limn→∞

fn(x) exists and then x ∈ B.

On the other hand, let x ∈ B, then ∀ε > 0, ∃N ∈ N s.t. |fn(x)−fm(x)| < ε, ∀m,n >

N , which means x ∈ AN . Then if ε → 0, N → ∞, x ∈ A.

Now we can conclude that B is a measurable set.

2 Differential function

Question: Let f : R → R be a differential function. Prove that f ′ is Lebesgue measurable.

∗


I


Proof: Define a sequence of functions: gn(x) = f(x+1/n)−f(x)1/n , x ∈ R, n ∈ N. Then

f is measurable

⇒ gn(x) is measurable ,∀n,

⇒ f ′(x) = limn→∞

gn is measurable.

3 The sets with measure zero

Question: Let f : R → R be defined by f(x) = x5 + sin x. Suppose that a set A ⊂ R hasLebesgue measure zero. Show that f(A) has Lebesgue measure zero.

Proof: I want to prove that for any function f ∈ C1(R), the image of the set withLebesgue measure zero has also Lebesgue measure zero.

Let AN = [−N,N ] ∩ A, then A =∞⋃

N=1AN , m(A) = 0 ⇒ m(AN ) = 0,∀N ∈ N. We

will focus the problem on each AN .

By the definition of Lebesgue measure, for ∀ε > 0, there exists a sequence of intervals(In)∞n=1, Ii ∩ Ij = ∅, i 6= j and AN ⊂ ∪In, s.t.

∑

n

l(In) < ε, since m(AN ) = 0. (3.1)

Let In = (an, bn), In = [an, bn], then m(In) = m(In). Then we have

m(f(AN )) <∑

n

m(f(In)) =∑

n

m(f(In)). (3.2)

Since f is continuous, we can know that

m(f(In)) = maxx∈In

f(x) − minx∈In

f(x).

Since In is a closed interval and f ∈ C1(In), we can find x1, x2 ∈ In s.t. f(x1) =maxx∈In

f(x), f(x2) = minx∈In

f(x). Without loss of generality, we can assume that x1 ≤ x2.

Then

m(f(In)) = f(x1) − f(x2)

≤ |f ′(ξ)|(x2 − x1), ξ ∈ [x1, x2],

≤ maxx∈In

f ′(x) l(In)

≤ maxx∈In

f ′(x) ε.

II


Since maxx∈Inf ′(x) < ∞ and let ε → 0, we can get m(f(In)) = 0.

By (3.2) we have m(f(AN )) = 0. So

m(f(A)) = m(f(∪AN )) ≤∑

N

m(f(AN )) = 0.


Question: Let f : R → R be a Lebesgue measurable function. Define g : R → R byg(x) =

∑∞n=1

f(x)n

n!(n+3) . Prove that g is Lebesgue measurable.

Proof: First, we claim that ∀a ∈ R, the series sn =∑n

k=1ak

k!(k+3) is convergent.

∀a ∈ R,∃N ∈ N, s.t. |a| < N . Then for ∀k > 2N , we have∣

∣

∣

∣

ak

k!(k + 3)

∣

∣

∣

∣

≤|a|N

1 · · ·N·

|a|N

(N + 1) · · · (2N)·

|a|k−2N

(2N + 1) · · · k·

1

k + 3.

Let M = |a|N

1···N , we can choose k > M , s.t. |a|N

1···N · 1k+3 < 1. Since |a|N

(N+1)···(2N) < 1,

|a|k−2N

(2N+1)···k <(

12

)k−2N, we have

∣

∣

∣

∣

ak

k!(k + 3)

∣

∣

∣

∣

<

(

1

2

)k−2N

.

Since for fixed N ,∑∞

k=1

(

12

)k−2Nis convergent, we have for fixed a,

∑∞k=1

ak

k!(k+3) isalso convergent.

Second, let fn =∑n

k=1f(x)n

n!(n+3) which is measurable and

g(x) = limn→∞

fn(x), for ∀x ∈ R.

Hence, we know that g(x) is also measurable.


Question: Let f : R → R be a Lebesgue measurable function. Suppose that A ⊂ R is aBorel set. Show that the set {x ∈ A : f(x) > x} is Lebesgue measurable.

Proof: Define a function g : R → R by g(x) = f(x) − x, then g(x) is measurable (bythe Theorem 6 in the page 259 of Royden’s book). So the set B = {x ∈ R : g(x) > 0} ismeasurable. Then, {x ∈ A : f(x) > x} = A ∩ B is measurable.

III


Yingwei Wang ∗


Note: In this paper, {f(x) satisfies some property} = {x : f(x) satisfies some property}

1 Integrable function

1.1 a

Question: Show that if f is integrable then the set {f(x) 6= 0} is of σ-finite measure.

Proof: On one hand,

{f(x) 6= 0} = {|f(x)| > 0} =

∞⋃

n=1

{|f(x)| ≥1

n}. (1.1)

On the other hand,

∞ >

∫

|f |dµ ≥

∫

{|f(x)|≥ 1

n}|f |dµ ≥

1

nµ{|f(x)| ≥

1

n}, ∀n ∈ N, (1.2)

which means

µ{|f(x)| ≥1

n} < ∞,∀n ∈ N. (1.3)

From (1.1) and (1.3), we can know that the set {f(x) 6= 0} is of σ-finite measure.

1.2 b

Question: Show that if f is integrable, f ≥ 0, then f = lim ϕn for some increasing sequenceof simple functions each of which vanishes outside a set of finite measure.

Proof: By proposition 7 on Page 260 in Royden’s book, since the set {f(x) 6= 0} is ofσ-finite measure, we can find a sequence (ϕn) of simple functions defined on {f(x) 6= 0} withϕn+1 ≥ ϕn such that f = lim ϕn and each ϕn vanishes outside a set of finite measure.

Then we can define ϕn(x) = 0 on the set {f(x) = 0}, and get the conclusion.

∗


I


1.3 c

Question: Show that if f is integrable with respect to µ, then given ǫ > 0 there is a simplefunction ϕ such that

∫

|f − ϕ|dµ < ǫ.

Proof: By the assumption, f+ and f− are nonnegative integrable functions. By (b), thereare increasing sequence (φn) and (ϕm) such that f+ = lim φn and f− = lim ϕm. By theMonotone Convergence Theorem, we have

∫

f+dµ = lim

∫

φn,∫

f−dµ = lim

∫

ϕm.

So given ǫ > 0, there are φN and ϕM such that∫

f+dµ −

∫

φNdµ <ǫ

2,

∫

f−dµ −

∫

ϕMdµ <ǫ

2.

Let ϕ = φN − ϕM . Then ϕ is also a simple function and satisfies∫

|f − ϕ|dµ

≤

∫

|f+ − φN |dµ +

∫

|f− − ϕM |dµ

=

(∫

f+dµ −

∫

φNdµ

)

+

(∫

f−dµ −

∫

ϕMdµ

)

< ǫ.

2 Measurable set

Question: Let f : R → R be defined by f(x) = x3 − 2x2 + x − 1. Show that if A ⊂ R isLebesgue measurable, then f(A) ⊂ R is Lebesgue measurable.

Proof: Since f(x) is a polynomial, then it is measurable and there exists a sequence ofstep functions (ϕn) s.t. ϕn(x) → f(x),∀x ∈ R.

By assumption, A ⊂ R is Lebesgue measurable, which means ∀E ⊂ R,

m(E) = m(E ∩ A) + m(E ∩ Ac).

II


For each step function ϕn(x), if A is measurable, then ϕn(A) is just a set of points so itis measurable. Thus f(A) = lim

n→∞ϕn(A) is measurable.

3 Limits

Question: Let (X,A, µ) be a measure space with µ(X) < ∞. Let f : X → (−1, 1) be ameasurable function. Consider the sequence

an =

∫

X(1 + f + · · · + fn)dµ.

Show that either (an) converges to a finite number or otherwise limn→∞

an = ∞. (In other words

(an) cannot have two distinct limit points.)

Proof: Since ∀x ∈ X, |f(x)| < 1, the series∑∞

n=0 |f(x)|n is convergent for ∀x ∈ X. Itfollows that all of the functions g(x) =

∑∞n=0(f(x))n and h0(x) =

∑∞n=0(f(x))2n, h1(x) =

∑∞n=0(−f(x))2n+1 are absolutely convergent.

Let A = {f ≥ 0}, B = {f < 0}, then X = A ∪ B.

a+n =

∫

A

n∑

j=0

(f(x))jdµ

a−n =

∫

B

n∑

j=0

(f(x))jdµ =

{

∫

B

∑kj=0 f2jdµ −

∫

B

∑kj=0(−f)2j+1dµ if n = 2k + 1,

∫

B

∑kj=0 f2jdµ −

∫

B

∑kj=0(−f)2j−1dµ if n = 2k.

Hence,

limn→∞

an = limn→∞

(a+n + a−n ) =

∫

Ag(x)dµ +

∫

Ah0(x)dµ −

∫

Bh1(x)dµ.

which means limn→∞

an can not have two distinct limit points.

4 Convergence

Question: Show that the following sequence is convergent in R.

an = n

∫ 1

1/n

cos(x + 1/n) − cos x

x3/2dx.

Proof: Consider the function

fn(x) =cos(x + 1/n) − cos x

1/n

X[1/n,1](x)

x3/2, x ∈ [0, 1],

III


where XA(x) is the characteristic function.

It is obvious that

limn→∞

fn = −sin x

x3/2, x ∈ [0, 1].

Let f(x) = − sinxx3/2

, then |f(x)| ≤ xx3/2

= 1x1/2

since sin x ≤ x, x ∈ [0, 1].

Let g(x) = 1x1/2

, then ∃N ∈ N, s.t. ∀n > N, |fn(x)| < g(x). Then by Lebesgue dominatedconvergence theorem, we have

limn→∞

an = limn→∞

∫ 1

0fn(x)dx =

∫ 1

0lim

n→∞fn(x)dx =

∫ 1

0−

sin x

x3/2dx <

∫ 1

0

1

x1/2dx = 2.

So an is convergent.

5 Infinitive sum

Question: Let (X,A, µ) be a complete measure space. Let (fn)∞n=1 be a sequence of A-measurable functions, fn : X → R. Suppose that

∞∑

n=1

∫

X|fn|dµ < ∞. (5.1)

Show that the series∑∞

n=1 fn(x) is absolutely convergent µ-almost everywhere on X.

Proof: First, claim that

∫

X

(

∞∑

n=1

|fn|

)

dµ =∞∑

n=1

∫

X|fn|dµ. (5.2)

Let Fn =n∑

i=1|fi|, F∞ =

∞∑

i=1|fi|, ∀x ∈ X. Then the statement (5.2) becomes

∫

XF∞dµ = lim

n→∞

n∑

i=1

∫

X|fi|dµ = lim

n→∞

∫

XFndµ. (5.3)

Since Fn(x) ≤ Fn+1(x) and Fn → F∞, by the Monotone Convergence Theorem, we canget (5.3).

Second, by the assumption (5.1) and the claim (5.2), we can get

∫

X

(

∞∑

n=1

|fn|

)

dµ =∞∑

n=1

∫

X|fn|dµ < ∞, (5.4)

which means the measure µ ({∑∞

n=1 |fn| = ∞}) = 0. That is to say the series∑∞

n=1 fn(x) isabsolutely convergent µ-almost everywhere on X.

IV


Yingwei Wang ∗


1 Integral function

Question: Let f : R → R be a Lebesgue integrable function. Prove that g(x) is alsoLebesgue integrable.

Proof: First, if f(x) is a non-negative simple function, then

f(x) =

n∑

k=1

ckχEi(x), ∀x ∈ R,

g(x) =

n∑

k=1

ckχEi+{1}(x), ∀x ∈ R.

Hence, g(x) is also a non-negative simple function and Lebesgue integrable.

Second, if f(x) is a non-negative measurable function, then there exists a sequenceof non-negative simple monotonic increasing functions {ϕk(x)}∞k=1 s.t.

limk→∞

ϕk(x) = f(x), ∀x ∈ R.

It is obvious that {ϕk(x − 1)}∞k=1 is also a monotonic increasing sequence and

limk→∞

ϕk(x − 1) = f(x − 1), ∀x ∈ R.

∗


I


Hence,

∫

R

g(x) dm

=

∫

R

f(x − 1) dm

= limk→∞

∫

R

ϕk(x − 1) dm

= limk→∞

∫

R

ϕk(x) dm

=

∫

R

f(x)dm < ∞.

So we have g(x) is Lebesgue integrable on R.

Finally, if f(x) is an arbitrary function on R, we can rewrite f as f = f+ − f−,where both f+ and f− are non-negative integrable function, and then get the sameconclusion about g(x).

2 Zero function

Question: Let f : [0, 1] → R be a Lebesgue measurable function. Suppose that forany function g : [0, 1] → R, fg is Lebesgue integrable and

∫

[0,1] fg dm = 0. Prove thatf = 0 almost everywhere with respect to the Lebesgue measure.

Proof: Suppose that there exists a set A ⊂ [0, 1], m(A) > 0, s.t. ∀x ∈ A, f(x) 6= 0.Without loss of generality, we can assume that f(x) > 0,∀x ∈ A.

∀ε > 0, we can find a sequence of intervals In = (a, b) ⊂ [0, 1] such that A ⊂ ∪In

and m(A\ ∪ In) < ε. Choose g(x) as follows:

g(x) =

{

−(x − an)(x − bn), x ∈ [an, bn],0, x ∈ [0, 1]\ ∪ In.

It is easy to verify that g(x) is continuous and satisfies: g(x) > 0, x ∈ ∪In, andg(x) = 0, x ∈ [0, 1]\ ∪ In. Hence, fg ≥ 0, x ∈ [0, 1] and fg > 0, x ∈ A. So we have

∫

[0,1]fg dm =

∫

∪In

fgdm =

∫

A

fgdm +

∫

∪In\Afgdm,

II


On one hand,∫

Afgdm > 0. On the other hand, since m(A\ ∪ In) < ε, we can

choose ε such that∣

∣

∣

∫

∪In\Afgdm

∣

∣

∣<

∫

Afgdm. It follows that

∫

[0,1] fg dm > 0, which

contradicts that∫

[0,1] fg dm = 0,∀g ∈ C1([0, 1]).

Thus, f = 0, a.e. x ∈ [0, 1].

Note: Another method is by the statement that “Any measurable function can beapproximated by a sequence of continuous functions”.

3 σ-finite

Question: Let (X,A), µ be a measure space and let f : X → R be A-measurable.Suppose that µ is σ-finite. Suppose also that there is c ≥ 0 such that for all E ∈ Awith µ(E) < ∞ one has |

∫

Ef dµ| ≤ c. Show that f is Lebesgue integrable on X.

Proof: Since µ is σ-finite, we can choose a sequence of sets (Xn) such that X = ∪Xn,X1 ⊂ X2 · · · ⊂ Xn ⊂ · · · ⊂ X, lim

n→∞Xn = X and

µ(Xn) < ∞,∀n ∈ N.

Define fn(x) = f(x)χXn,∀x ∈ X, then

limn→∞

fn(x) = f(x) and |fn(x)| ≤ |f(x)|, ∀x ∈ X.

Since µ(Xn), by assumption,

∣

∣

∣

∣

∫

X

fn dµ

∣

∣

∣

∣

=

∣

∣

∣

∣

∫

Xn

fdµ

∣

∣

∣

∣

≤ c, ∀n ∈ N.

By Lebesgue dominated control theorem, we have

∫

X

f dµ = limn→∞

∫

X

fn dµ ≤ limn→∞

∣

∣

∣

∣

∫

X

fn dµ

∣

∣

∣

∣

≤ c,

which implies f is Lebesgue integrable on X.

III


4 Limits

Question: Let (X,A), µ be a measure space. Let (fn)∞n=1 be a sequence of A-measurablefunctions, fn : X → [0,+∞]. Show that

∫

X

lim infn→∞

fndµ ≤ lim infn→∞

∫

X

fndµ.

Proof: Let gn(x) = infk≥n

fn(x), f(x) = lim infn→∞

fn(x), ∀x ∈ X. Then gn are nonneg-

ative and gn increases to f , limn→∞

gn(x) = f(x). Therefore,

∫

X

gndµ ≤ infk≥n

∫

X

fkdµ.

If we take the limit as n → ∞, on the left side we obtain∫

Xfdµ by the monotone

convergence theorem, while on the right side we obtain lim infn→∞

∫

Xfndµ.

IV


Yingwei Wang ∗


1 Limits

Question: Compute the following limits and justify your calculation.

(i) limn→∞

∫∞0

(

1 + xn

)−nsin x

ndx.

Solution: Let fn(x) =(

1 + xn

)−nsin x

n , then |fn(x)| ≤ e−x,∀x ∈ [0,∞). Besides, limn→∞

fn(x) =

0. Then we have

limn→∞

∫ ∞

0fndx = 0.

(ii) limn→∞

∫ n0

(

1 − xn

)nex/2dx.


1 − xn

)nex/2χ[0,n], f(x) = e−x/2 then lim

n→∞fn(x) = f(x).

Let gn = 2f − fn, then gn monotonic increasingly converges to f as n → ∞. Besides, whenn is sufficient large, gn ≥ 0. By Monotonic Convergence Theorem, we can see that

limn→∞

gn(x) = f(x).

So we have

limn→∞

∫ n

0

(

1 −x

n

)nex/2dx = lim

n→∞

∫ ∞

0fndx =

∫ ∞

0fdx = 2.

(ii) limn→∞

∫ n0

(

1 + xn

)ne−2xdx.


1 + xn

)ne−2xχ[0,n], f(x) = e−x then lim

n→∞fn(x) = f(x) and fn(x) ≤

f(x),∀x ∈ [0,∞).

So we have

limn→∞

∫ n

0

(

1 +x

n

)ne−2xdx = lim

n→∞

∫ ∞

0fndx =

∫ ∞

0fdx = 1.

∗


I


2 Limits

Question: Let (X,A, µ) be measure space and let f : X → [0,+∞) be A-measurable. Supposethat 0 < c =

∫

X fdµ < ∞. Show that

limn→∞

∫

Xn ln

(

1 +

(

f

n

)α)

dµ =

+∞, if 0 < α < 1,c, if α = 1,0, if 1 < α < ∞.

Proof: Let

fn(x) = n ln

(

1 +

(

f

n

)α)

= ln

(

1 +

(

f

n

)α)n

= ln

(

(

1 +fα

nα

)nα)n1−α

= n1−α · ln

(

(

1 +fα

nα

)nα)

If 0 < α < 1, thenlim

n→∞fn(x) = +∞ · fα = +∞.

By Fatou’s Lemma, we have

limn→∞

∫

Xfn(x)dµ ≥ lim

n→∞

∫

Xfn(x)dµ ≥

∫

Xlim

n→∞fn(x)dµ = +∞.

If α = 1, thenlim

n→∞fn(x) = f(x).

Besides, ∀x ∈ X, fn(x) ≤ f(x). By Lebesgue Dominated Control Theorem, we have

limn→∞

∫

Xfn(x)dµ =

∫

Xf(x)dµ = c.

If 1 < α < ∞, thenlim

n→∞fn(x) = 0 · fα = 0.

By Monotonic Convergence Theorem, we have

limn→∞

∫

Xfn(x)dµ =

∫

X0 dµ = 0.

II


3 Limits

Question: Let (X,A, µ) be measure space. Let (fn)∞n=1 be sequence of A-measurable func-tions, fn : X → [0,+∞). Suppose that limn→∞ fn(x) = f(x) for all x ∈ X and thatlimn→∞

∫

X fn(x)dµ =∫

X f(x)dµ < ∞. Show that limn→∞

∫

A fn(x)dµ =∫

A f(x)dµ for ev-ery set A ∈ A.

Proof: For any set A ∈ A, define gn(x) = fn(x)χA, hn(x) = fn(x)χX\A, the we have

limn→∞

fn(x) = f(x),

limn→∞

gn(x) = f(x)χA,

limn→∞

hn(x) = f(x)χX\A,

gn(x) + hn(x) = fn(x), ∀n and ∀x.

On one hand, by Fatou’s Lemma, we have∫

XfχAdµ ≤ lim

n→∞

∫

Xgndµ, (3.1)

∫

XfχX\Adµ ≤ lim

n→∞

∫

Xhndµ, (3.2)

⇒

∫

Xfdµ ≤ lim

n→∞

∫

Xgndµ + lim

n→∞

∫

Xhndµ. (3.3)

On the other hand,

limn→∞

∫

Xgndµ + lim

n→∞

∫

Xhndµ = lim

n→∞

(∫

Xgndµ +

∫

Xhndµ

)

≤ limn→∞

∫

Xfndµ =

∫

Xfdµ. (3.4)

By (3.3) and (3.4), we can see that

limn→∞

∫

Xgndµ + lim

n→∞

∫

Xhndµ =

∫

Xfdµ. (3.5)

Then by (3.1), (3.2) and (3.5), we can see that∫

XfχAdµ = lim

n→∞

∫

Xgndµ, (3.6)

∫

XfχX\Adµ = lim

n→∞

∫

Xhndµ, (3.7)

So we can get

limn→∞

∫

Xfn(x)χAdµ =

∫

Xf(x)χAdµ

⇒ limn→∞

∫

Afn(x)dµ =

∫

Af(x)dµ.

III


4 The absolute continuity of integral

Question: Let f : R → [0,∞] be a Lebesgue integrable function. Show that for each α > 0there is β > 0 such that if B ⊂ R is Lebesgue measurable and m(B) < β, then

∫

B fdm < α.

Proof: Since f is nonnegative and Lebesgue integrable, ∀α > 0,∃ a nonnegative simplefunction ϕ(x), such that

∫

R

f − ϕdm =

∫

R

fdm −

∫

R

ϕdm <α

2.

Let ϕ(x) < M , then we can choose B ⊂ R s.t. m(B) < α2M , then

∫

Bfdm ≤

∫

Bf − ϕdm +

∫

Bϕdm

≤

∫

R

f − ϕdm + M · m(B)

<α

2+ M ·

α

2M= α.

IV


Yingwei Wang ∗


Note: In this paper, {f(x) satisfies some property} = {x : f(x) satisfies some property}

1 Integral function

Question: Let f : (0, 1) → R be a Lebesgue integrable function. Suppose that∫(0,x)

fdm for all x ∈ (0, 1). (1.1)

Show that f(x) = 0 a.e.

Proof: First I want to claim that∫ 10 fdm = 0. Let fn = fχ(0,1−1/n), then |fχ(0,1/n)(x)| ≤

|f(x)|, ∀n ∈ N and x ∈ (0, 1) By Lebesgue dominated convergence theorem,∫ 1

0fdm = lim

n→∞

∫(0,1)

fndm = limn→∞

∫(0,1−1/n)

fdm = 0.

Second, let A = {f(x) > 0}. Suppose m(A) > 0, then ∃ closed set B ⊂ A, m(B) > 0.Let C = [0, 1]\B, then C is an open set, i.e. C = ∪(an, bn).∫

Bfdm +

∫C

fdm =

∫(0,1)

fdm = 0,

⇒

∫C

fdm 6= 0, (since

∫B

fdm > 0)

⇒

∫∪(an,bn)

fdm 6= 0,

⇒ ∃ (an, bn), s.t.

∫(an,bn)

fdm 6= 0,

⇒

∫(0,bn)

fdm −

∫(0,an)

fdm 6= 0,

∗


I


which contradicts with (1.1).

So m(A) = 0. Similarly, we can prove that m({f(x) < 0}) = 0. That is to say f(x) = 0a.e.

2 Convergence of L1 norm and pointwise

Question: Let (fn) be a convergent sequence in L1(X,A, µ) with limit f . Show that (fn)has a subsequence (fnk

) such that fnk→ f for almost x ∈ X.

Proof: Since (fn) be a convergent sequence in L1(X,A, µ), then (fn) is a Cauchy se-quence in L1(X,A, µ), which means

∫X|fn − fm|dµ → 0, as n,m → ∞.

⇒ ∀ε > 0, ∃N , s.t. ∀n,m > N ,

m({|fn − fm| 6= 0}) < ε.

⇒ ∀i,∃ki ∈ N s.t. ∀n,m > ki,

m({|fn − fm| ≥1

2i}) <

1

2i.

We can assume that k1 < k2 < · · · < ki < ki+1 < · · · , and define

Ei = {|fki− fki+1

| ≥1

2i},

then m(Ei) < 12i .

Let

E =

∞⋂j=1

∞⋃i=j

Ei,

then m(E) = 0.

∀x ∈ X\E, ∃j s.t.

x ∈ X

∞⋃i=j

Ei.

Then ∀i ≥ j,

|fki+1(x) − fki(x)| <

1

2i,

II


which means∑

|fki+1(x) − fki(x)| is convergent, so fk1

+∑

(fki+1(x) − fki(x)) is also

convergent.

Denote limki→∞

fki(x) = g(x),∀x ∈ X\E, then |fnk

− g| → 0, as n → ∞, ∀x ∈ X\E. So

∫X|fnk

− g|dµ =

∫X\E

|fnk− g|dµ → 0,

⇒ ‖fnk− g‖L1 → 0.

Since ‖fnk− f‖L1 → 0, by the completeness of L1(X,A, µ), f = g for almost all x ∈ X.

So we have a subsequence (fki) which is convergent to f for almost all x ∈ X.

3 Integral functions and continuous functions

Question: Let f : [0, 1] → R be a Lebesgue integrable function. Show that there is asequence of continuous functions fn : [0, 1] → R which converges pointwise to f almosteverywhere.

Proof: First, I want to introduce the Lusin’s Theorem (Problem 3.31 on Page 74 inRoyden’s book):

Theorem 3.1 (Lusin). Let f be a measurable real-valued function on an interval [a, b].Then given δ, there is a continuous function ϕ on [a, b] such that m({f 6= ϕ}) < δ.

By this theorem, for f ∈ L1([0, 1]), ∀n ∈ N, ∃ a sequence of continuous functionsϕn ∈ C([0, 1]) s.t.

m({ϕn 6= f}) <1

2n.

Then we have∫X|ϕn − f |dm =

∫{ϕn 6=f}

|ϕn − f |dm → 0, as n → ∞,

since m({δn 6= f}) → 0 as n → ∞.

It implies that ϕn converges to f in the L1 norm. The conclusion of the last problemtells us that (ϕn) has a subsequence (ϕnk

) such that ϕnk→ f for almost x ∈ X.

III


Yingwei Wang ∗


1 Absolutely continuous

Question: Let f, g : [a, b] → R be two absolutely continuous functions. Prove or disprovethat h(x) = ef(x)g(x) is absolutely continuous on [a, b].

Proof: First, I want to show that if both f and g are absolutely continuous, then fg isalso absolutely continuous.

There exists M > 0 such that |f(x)| ≤ M and |g(x)| ≤ M for ∀x ∈ [a, b]. Given ε > 0,there exits δ > 0 such that

n∑i=1

|f(xi) − f(yi)| <ε

2M, and

n∑i=1

|g(xi) − g(yi)| <ε

2M,

for any finite collection {(xi, yi)} of disjoint intervals in [a, b] with∑n

i=1 |xi − yi| < δ.

Then

n∑i=1

|f(xi)g(xi) − f(yi)g(yi)|

≤

n∑i=1

|f(xi)||g(xi) − g(yi)| +

n∑i=1

|g(yi)||f(xi) − f(yi)|

<ε

2+

ε

2= ε.

Second, I want to show that if q : [a, b] → [a′, b′] is absolutely continuous on [a, b] andp : [a′, b′] → R is Lipschitz continuous on [a′, b′], then p(q(x)) is absolutely continuous on[a, b].

∗


I


There exists N > 0 such that for ∀x, y ∈ [a′, b′],

|p(x) − p(y)| ≤ N |x − y|.

Given ε, there exits δ > 0 such that

n∑i=1

|q(xi) − q(yi)| <ε

N

for any finite collection {(xi, yi)} of disjoint intervals in [a, b] with∑n

i=1 |xi − yi| < δ.

Thenn∑

i=1

|p(q(xi)) − p(q(yi))|

≤ N

n∑i=1

|q(xi) − q(yi)|

< ε.

Finally, since ex is Lipschitz continuous in any finite interval [a′, b′], so h = efg isabsolutely continuous if both f and g are absolutely continuous.

2 Bounded variation

Question: Let f ∈ BV [a, b]. Show that T baf ≥

∫ ba |f ′(x)|dx.

Proof: By Jordan Theorem, f(x) can be written as the difference of two monotoneincreasing functions on [a, b],

f(x) = g(x) − h(x),

where

g(x) =1

2(T x

a f + f(x)),

h(x) =1

2(T x

a f − f(x)).

It is easy to verify that both g and h are increasing, so g′ and h′ are nonnegative.

By Lebesgue Theorem, ∫ b

ag′(x)dx ≤ g(b) − g(a),

∫ b

ah′(x)dx ≤ h(b) − h(a).

II


Since f ′ = g′ − h′, |f ′| ≤ g′ + h′. Then we have

∫ b

a|f ′(x)|dx ≤

∫ b

ag′(x) + h(x)dx ≤ g(b) − g(a) + h(b) − h(a) = T b

af.

3 Bounded variation

Question: Let f ∈ BV [a, b]. Show that f has at most countably many points of disconti-nuity.

Proof: First, I want to show that ∀x0 ∈ [a, b], both limx→x+

0

f(x) and limx→x−

0

f(x)exist.

By Jordan Theorem, f = g − h where g and h are monotone increasing functions on[a, b]. Let G = supx∈[a,x0) g(x) and H = supx∈[a,x0) h(x). It is obvious that A,B < ∞.

Given ε > 0, there exists δ > 0 such that

G −ε

2< g(x0 − δ) ≤ G,

H −ε

2< h(x0 − δ) ≤ H.

Then for ∀x ∈ (x0 − δ, x0), we have

G −ε

2< g(x) ≤ G ⇒ 0 ≤ G − g(x) <

ε

2,

H −ε

2< h(x) ≤ H ⇒ 0 ≤ H − h(x) <

ε

2.

It follows that|G − H − f(x)| ≤ (G − g(x)) + (H − h(x)) < ε,

for ∀x ∈ (x0 − δ, x0). So limx→x+

0

f(x) = G−H. Similarly we can know that limx→x−

0

f(x)

also exists.

Second, I want to show that the set of discontinuities of f is at most countably.

Let E = {x ∈ [a, b] : f(x+) 6= f(x−)}, E1 = {x ∈ [a, b] : g(x+) > g(x−)}, E2 = {x ∈[a, b] : h(x+) > h(x−)}. Since g and h are monotone increasing and f = g − h, we haveE = E1 ∪ E2.

For ∀x ∈ E1, ∃rx ∈ Q such that g(x−) < rx < g(x+). Besides, if x1 < x2, theng(x1+) ≤ g(x2−) so rx1

6= rx2. Thus x → rx is a bijection between E1 and a subset of Q,

which means E1 is at most countably.

Similarly, E2 is at most countably and further E = E1 ∪ E2 is also at most countably.

III


4 L1 space

Question: Let (fn) be a sequence of nonnegative functions in L[0, 1] such that∫ 10 fn(t)dt

and∫ 11/n dt ≤ 1/n for all n ≥ 1. If g(x) = supn fn(x) show that g /∈ L1[0, 1].

Proof: Suppose that g ∈ L1[0, 1], I want to get a contradiction.

By the absolutely continuity of the integral, for ε = 13 , there exists δ such that for any

measurable subset E ⊂ [0, 1], if m(E) < δ, then

∫E

gdm <1

3. (4.1)

Choosing an integer n satisfying n ≥ 3 and 1/n < δ, then since∫ 10 fn(t)dt and

∫ 11/n dt ≤

1/n, we have ∫ 1/n

0fndt = 1 −

∫ 1

1/nfndt ≥ 1 − 1/n,

for all n > 3.

By the definition of g, we have

∫ 1/n

0gdt ≥

∫ 1/n

0fndt ≥ 1 − 1/n ≥ 1 − 1/3 ≥

2

3. (4.2)

But by (4.1), we have ∫ 1/n

0gdt <

1

3. (4.3)

It is obvious that (4.2) contradicts with (4.3).

IV


Yingwei Wang ∗


1 Convergence series

Question: Let (an)∞n=1 be sequence of non-negative numbers such that

∞∑

n=1

an < ∞. (1.1)

Let r1, r2, . . . be an enumeration of the rational numbers. Show that the series

∞∑

n=1

a2n

|x − rn|(1.2)

converges for almost all x ∈ R.

Proof: I found two methods to prove this. Hopefully, both of them are correct.

1.1 Method one: integral

Consider the new series:∞∑

n=1

an√

|x − rn|(1.3)

If (1.3) converges, then (1.2) is sure to converge.

If |x − rn| ≥ 1 for some n, then

an√

|x − rn|≤ an. (1.4)

∗


I


So the convergence of the series is obvious.

If |x − rn| < 1 for some n, then rn − 1 < x < rn + 1. Consider the integral

∫ rn+1

rn−1

an√

|x − rn|dx

= an

(∫ rn

rn−1

1√rn − x

dx +

∫ rn+1

rn

1√x − rn

dx

)

= an

(∫

1

0

1√tdt +

∫

1

0

1√tdt

)

= 4an, (1.5)

which means∑∞

n=1an√|x−rn|

is convergent almost everywhere for x ∈ (rn − 1, rn + 1).

In sum, (1.3) converges for x ∈ R, a.e. and so does (1.2).

1.2 Method two: measure

Consider the setEε

n = {x : |x − rn| < εan}, (1.6)

then the measure of this set ism(Eε

n) = 2εan. (1.7)

For ∀x ∈ R\ ∪ Eεn, we have

an

|x − rn|≤ 1

ε

⇒ a2n

|x − rn|≤ an

ε

⇒∞

∑

n=1

a2n

|x − rn|≤ 1

ε

∞∑

n=1

an. (1.8)

As ε → 0, m(∪Eεn) → 0. Hence,

∑∞n=1

a2n

|x−rn|converges for almost all x ∈ R.

II


2 Bounded variation functions

Question: Find all functions f ∈ BV [0, 1] with the property that

f(x) + (T x0 f)1/2 = 1,∀x ∈ [0, 1]. (2.1)

Solution: Let x = 0 in Eqn.(2.1), we can get

f(0) = 1.

Since F (x) = (T x0 f)1/2 is an increasing function, f(x) should be a decreasing function.

Hence, F (x) = (T x0 f)1/2 = (f(0) − f(x))1/2. So

f(x) + (f(0) − f(x))1/2 = 1,

⇒ (f(0) − f(x)) = (1 − f(x))2,

⇒ (1 − f(x)) = (1 − f(x))2,

⇒ f(x) = 1, ∀x ∈ [0, 1].

3 Absolutely continuous function

Question: Let f : [0, 1] → [0, 1], f(x) =√

x. Show that f ∈ AC[0, 1].

Proof: Since√

x = 1

2

∫ t0

t−1/2dt, by the Absolutely continuity of the integral (Propo-sition 4.14 in Page 88 of Royden’s book), it is easy to know that f(x) =

√x ∈ AC[0, 1].

4 Non-absolutely continuous function

Question: Construct f : [0, 1] → [0, 1], f ∈ AC[0, 1] such that√

f /∈ AC[0, 1].

Solution: Let f(x) be as following

f(x) =

{

x2 sin2( 1

x), 0 < x ≤ 1,0, x = 0.

(4.1)

It is easy to know that f(x) ∈ AC[0, 1] since f ′(x) = 2x sin2(1/x)−2 sin(1/x) cos(1/x)is bounded in [0, 1].

III


However, for√

f ,√

f(x) =

{

x sin( 1

x), 0 < x ≤ 1,0, x = 0,

(4.2)

we choose the partition

0 <1

nπ + π/2<

1

nπ<

1

(n − 1)π + π/2< · · · <

1

π + π/2<

1

π<

1

π/2< 1.

Then

n∑

k=0

|√

f(xk + 1) −√

f(xk)| = | sin 1 − 2

π| +

n∑

k=0

1

kπ + π/2+

1

nπ + π/2.

So T 10

√f = ∞ and

√f /∈ BV [0, 1] and further

√f /∈ AC[0, 1].

IV


Yingwei Wang ∗


1 Open sets

Question: Show that every open subset U of R can be written uniquely as the union of acountable family of mutually disjoint open intervals.

Proof: Let U ∈ R be open. For x1 ∈ U , consider the interval I1 = (a1, b1) where

a1 = inf{t ∈ R : (t, x) ⊂ U},

b1 = sup{t ∈ R : (x, t) ⊂ U}.

Then for x2 = U\I1, consider the interval I2 = (a2, b2) where

a2 = inf{t ∈ R : (t, x) ⊂ U\I1},

b2 = sup{t ∈ R : (x, t) ⊂ U\I1}.

In this way, we get a sequence of disjoint open intervals {In}∞

n=1 such that ∪∞

n=1In = U .

2 Measurable sets

Question: Let f : R → R be a continuous functions with the property that m∗(f(A)) = 0whenever m∗(A) = 0, A ⊂ R. Show that if A ⊂ R is Lebesgue measurable, then f(A) isLebesgue measurable.

Proof: That A is measurable means ∀ε > 0, ∃ disjoint open intervals {In}∞

n=1 s.t.A ⊂ ∪In and m(∪In − A) < ε.

Since m∗(A) = inf{∑

l(In) : A ⊂ ∪In}, we have m∗(∪In) → m∗(A) and m∗(∪In−A) →0 as ε → 0.

By the assumption, m∗(f(∪In − A)) → 0, so m(∪f(In) − f(A)) < ε, for ∀ε > 0.

Since ∪f(In) is an open set and f(A) ⊂ ∪f(In), we know that f(A) is measurable.

∗


I


3 Absolutely continuity

Question: Let f : R → R ∈ AC[a, b] for all a < b, a, b ∈ R. Show that if A ⊂ R is Lebesguemeasurable, then f(A) is Lebesgue measurable.

Proof: It is suffices to show that f ∈ AC[a, b] for all a < b, a, b ∈ R implies that f hasthe property that m(f(E)) = 0 whenever m(E) = 0, A ⊂ R.

That f is absolutely continuous means ∀ε > 0, ∃ δ such that for disjoint intervals{Ii = (xi, yi)}

ni=1 satisfying

n∑

i=1

(yi − xi) < δ,

thenn

∑

i=1

|f(yi) − f(xi)| < ε.

Suppose E ⊂ R and m(E) = 0. Let G = ∪(xi, yi) ⊃ E. Choose ci, di ∈ [xi, yi] such that

f([xi, yi]) = [f(ci), f(di)],

thenm(f(E)) ≤ m(f(G)) ≤

∑

|f(di) − f(ci)| < ε.

So m(f(E)) = 0.

By the conclusion of the Problem 2, we can know that if E ⊂ R is Lebesgue measurable,then f(E) is Lebesgue measurable.

4 Lebesgue measure

Question: Let f ∈ AC[a, b] for all a < b where a, b ∈ R. Suppose that f is strictlyincreasing. Show that if A is Lebesgue measurable then

m(f(A)) =

∫

A

f ′(t)dt. (4.1)

Proof: Case 1: A is an open interval A = (a0, b0).

On one hand, since f is strictly increasing, m(f(A)) = f(b0)−f(a0); on the other hand,∫

Af ′(t)dt =

∫ b0a0

f ′(t)dt = f(b0) − f(a0). Thus, (4.1) holds.

Case 2: If A is an open set, then A =⊔

In where Ii ∩ Ij = ∅, i 6= j, In = (an, bn).

II


On one hand,

m(f(A)) =

∞∑

n=1

(f(bn) − f(an)) = f(b∞) − f(a1).

On the other hand,

∫

A

f ′(t)dt =

∞∑

n=1

∫ bn

an

f ′(t)dt =

∞∑

n=1

(f(bn) − f(an)) = f(b∞) − f(a1).

It implies the (4.1).

Case 3: If A is a Gσ-set, it suffices to show that∫

A

f ′(x)dx ≤ m∗(f(A)). (4.2)

Choose open intervals In such that f(A) ⊂ ∪In, then E ⊂ ∪Jn, where Jn = f−1(In).

Choose sequences {α(n)k } and {β

(n)k } such that

limk→∞

α(n)k = inf

x∈Jn

{x},

limk→∞

β(n)k = sup

x∈Jn

{x}.

Then we get∫

Jn

f ′(x)dx = limk→∞

∫ β(n)k

α(n)k

f ′(x)dx ≤ m(In).

As a sequence,∫

A

f ′(x)dx ≤

∫

∪Jn

f ′(x)dx ≤∑

∫

Jn

f ′(x)dx ≤ m(In).

By the definition of outer measure, we can get (4.2).

5 Signed measure

Question: Let µ = µ+ −µ− be the Jordan decomposition of a signed measure µ defined ona measurable space (X,A). Define the total variation of µ to be the measure |µ| = µ++µ−.Show that for each A ∈ A,

|µ|(A) = sup

{

n∑

i=1

|µ(Ai)| : (Ai)ni is a partition of A with Ai ∈ A, n ≥ 1

}

. (5.1)

III


Proof: Let E,F be a Hahn decomposition for µ. That is to say X = E ∪F , E ∩F = ∅,and E is positive while F is negative respect to µ.

On one hand, we know that

µ+A = µ(A ∩ E),

µ−A = −µ(A ∩ F ),

⇒ |µ|(A) = µ(A ∩ E) − µ(A ∩ F ). (5.2)

On the other hand,

∫

A

(χA∩E − χA∩F ) dµ =

∫

A∩E

1dµ −

∫

A∩F

1dµ = µ(A ∩ E) − µ(A ∩ F ). (5.3)

By Eqn.(5.2) and Eqn.(5.3), we know that

|µ|(A) =

∫

A

(χA∩E − χA∩F ) dµ

=

n∑

i=1

∫

Ai

(χAi∩E − χAi∩F ) dµ,

where (Ai)ni=1 is a partition of A with Ai ∈ A.

Since for each Ai,∫

Ai(χAi∩E − χAi∩F ) dµ ≥ |µ(Ai)|, so we have

|µ|(A) =

n∑

i=1

∫

Ai

(χAi∩E − χAi∩F ) dµ ≥ sup

{

n∑

i=1

|µ(Ai)|

}

, (5.4)

for any partition (Ai)ni=1.

Besides, if we choose A1 = A ∩ E and A2 = A ∩ F , then

|µ|(A) = µ(A1) − µ(A2) = |µ(A1)| + |µ(A2)|,

which means

sup

{

n∑

i=1

|µ(Ai)|

}

≥ |µ|(A). (5.5)

By Eqn.(5.4) and (5.5), we can get the conclusion (5.1).

IV

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