REAL ANALYSIS Spring 2003 - SOLUTIONS TO SOME PROBLEMS

2003

REAL ANALYSIS Spring 2003

SOLUTIONS TO SOME PROBLEMS

Warning:These solutions may contain errors!!

PREPARED BY SULEYMAN ULUSOY

PROBLEM 1. Suppose f

: X → [0, ∞] is measurable for n = 1, 2, 3, ...;

≥ f

≥ ... ≥ 0; f

(x) → f (x) as n → ∞, for every x ∈ X .

a)Give a counterexample to show that we do not have generally the following result.

lim

n→∞

dµ =

fdµ.

b) Without changing any other assumptions just add one more assumption and prove

that the conclusion is satisﬁed in this case.

SOLUTION.

a)The standard example is the following.Let f

= 1

[n,∞)

for n = 1, 2, 3, ...,where 1

represents the characteristic function of the set X.Then one can easily show that f = 0

but lim

n→∞

dµ does not exist.

Consider f

[n,∞)

for n = 1, 2, 3, ...,where 1

represents the characteristic function

of the set X. Then one can easily checks that f = 0 but lim

n→∞

dµ does not exist.

Here is another example:In this example we give a counterexample to the case where we

have strict inequality.

Let f = 0.Let f

if −∞ < x ≤ n and f

if n < x < ∞.Thus f

strictly

decreases to f = 0 but

dµ = ∞ for all n yet

dµ = 0.

b) If we assume that f

∈ L

(X, µ) then the conclusion is satisﬁed.Here is the proof :

Consider the sequence g

= f

− f

since f

> f

> ... > 0 we have g

n+1

> g

> 0

for all n = 1, 2, 3, .....Also g

(x) → f

(x) −f(x) as n → ∞.Now we apply the Monotone

Convergence Theorem to the sequence g

and get lim

n→∞

dµ =

lim

n→∞

dµ.

But this means that

dµ−lim

n→∞

dµ =

dµ−

fdµ.Now since f

∈ L

(X, µ)

we can delete the term

dµ from both sides and get lim

n→∞

dµ =

fdµ.

PROBLEM 2. Suppose µ(X) < ∞, f

is a sequence of bounded complex measurable

functions on X, and f

→ f uniformly on X. Prove that lim

n→∞

dµ =

fdµ.

Show by a counterexample that the conclusion is not valid if we omit µ(X) < ∞.

SOLUTION. We ﬁrst establish that the function f is integrable.For this note that we

can choose N so large that |f

− f| < /2 and for any given  > 0.Then we have the

following estimate :

|f|dµ ≤

− f|dµ +

|dµ ≤ (/2)µ(X) + Mµ(X) (*) where M < ∞ is a

bound for f

(by assumption f

’s are bounded for all n).Since µ(X) < ∞ the right hand

side of (*) is ﬁnite,which shows that f is integrable.Now we have N so that |f

−f| < /2

for ﬁxed .Now consider the following estimate for large enough n:

− f|dµ ≤

|f − f

|dµ +

− f

|dµ < (/2)µ(X) + (/2)µ(X).

Since  > 0 is arbitrary and µ(X) < ∞ we have that lim

n→∞

dµ =

fdµ.

For a counterexample to the case µ(X) = ∞,consider f

= 1/n1

[

0, n) where 1

repre-

sents the charactristic function of the set X. Then one can easily checks that

dµ = 1, ∀n

but f = 0.Hence the conclusion does not necessarily hold when µ(X) = ∞.

PROBLEM 3. Suppose f

∈ L

(X, µ).Prove that to each  > 0 there exists a δ > 0

such that

|f|dµ <  whenever µ(E) < δ.

SOLUTION.Deﬁne f

= |f| ∧ n. i.e. f

= |f|, if |f| ≤ n and f

= n ,if |f| > n.Then

→ |f| as n → ∞. Therefore we can use Monotone Convergence Theorem as f

n+1

≥ f

and f

≥ 0.Now choose N so that

|f − f

| < /2 for given  > 0.Then we have the

following estimate.

|f|dµ <

|f − f

|dµ +

|dµ < (/2) + Nµ(E) < /2 + Nµ(E) <  whenever

µ(E) < δ < /(2N).Thus the assertion is proved.

PROBLEM 4. Let X be an uncountable set,let M be the collection of all sets E ⊂ X

such that either E or E

is at most countable, and deﬁne µ(E) = 0 in the ﬁrst case,

µ(E) = 1 in the second case.Prove that M is a σ-algebra in X and that µ is a measure

on M.

SOLUTION. The solution is obtained by direct applications of deﬁnitions.

M is a σ-algebra in X: Clearly ∅ and X are in M.Also suppose F is a member of M then

we have two cases.Either F is countable , in this case (F

)

= F is countable which shows

that F

is in M or F

is countable which shows that F

is in M. The most interesting

part is to show that M is closed under countable unions.Suppose E

, E

, ... is contable

collection of sets each of which is in M.Then we have to consider the following cases:

1.) Suppose each E

is countable.In this case the union will be countable and hence it

will lie in M.

2.) Suppose now that there exists E

for some k such that E is uncountable.Then E

countable.And by the De Morgan’s Law we have (∪E

)

⊂ E

and this shows that the

complement of the union is countable and so the union lies in M.

Therefore M is a σ-algebra in X.

µ is a measure on M :We need to show the following assertions.

1.)µ(∅) = 0

2.)Countable additivity.i.e.If E

, E

, ... is contable collection of sets each of which is

in M then µ(∪E

) = Σµ(E

The ﬁrst assertion is obvious as ∅ is countable.So by deﬁnition µ(∅) = 0.

The interesting part is the second assertion.Suppose ﬁrst that each of E

is countable. In

this case the union of these sets is also countable.So we have µ(∪E

) = 0 by deﬁnition.On

the other hand since each E

is countable µ(E

) = 0 for all i.Thus the second assertion

above holds in this case.

Now suppose that ∃ k such that E

is uncountable.Then E

is countable and by the De

Morgan’s Law used above we again have (∪E

)

⊂ E

which shows that µ(∪E

) = 1.If

we consider the summation Σµ(E

) we see that it is equal to 1 since the only term that

is nonzero(1) is µ(E

).So we again have the validity of the second assertion.

Thus µ deﬁned above is a measure in M.

PROBLEM 5. Let E

be a sequence of measurable sets in X, such that

∞

k=1

µ(E

) < ∞.(∗)

a.)Then show that almost all x ∈ X lie in at most ﬁnitely many sets E

b.)Is the conclusion still valid if we omit the condition (∗)?

SOLUTION.This is known as the Borel-Contelli’s Lemma.There are two ways to do

the part a.)

First Proof:If A is the set of all x which lie in inﬁnitely many E

, we need to prove that

µ(A) = 0.Put

g(x) =

∞

k=1

(x), (x ∈ X)

where 1

represents the characteristic function of the set K.Observe that for each x,

each term in this series is either 1 or 0.Hence x ∈ X if and only if g(x) = ∞.But we

know that the integral of g is equal to the sum in (∗).Thus g ∈ L

(µ),and so g(x) < ∞

a.e.

Second Proof:From set theory we see that the set we are looking for is A = ∩

∞

n=1

∪

∞

k=n

Deﬁne F

= ∪

∞

k=n

.Then clearly F

n+1

⊂ F

.Thus

µ(A) = lim

n→∞

µ(F

) = lim

n→∞

µ(∪

∞

k=n

) ≤ lim

n→∞

∞

k=n

µ(E

)(∗∗)

But the last term in (∗∗) is the limit of the remainder term of the series (∗) which is

ﬁnite.Hence it goes to 0.Thus the assertion is proved.

b.)As you guess the conclusion is not valid if we omit the ﬁniteness condition in (∗).Here

is a simple example.Take E

= (−∞, 1/n).Then the sum in (∗) is equal to ∞.And

A = (−∞, 0] which has measure ∞.

PROBLEM 6. Find a sequence (f

) of Borel measurable functions on R which decreses

uniformly to 0 on R , but

dm = ∞ for all n.Also,ﬁnd a sequence (g

) of Borel

measurable functions on [0, 1] such that g

→ 0 pointwise but

dm = 1 for all n.

(here m is the Lebesgue measure!)

SOLUTION.For the ﬁrst part f

= 1/n1

[1,∞)

works(easy Calculus exercise!!).For the

second part one sees that g

= n1

[0,1/n]

satisﬁes all the assertions(again this is easy to

verify).

PROBLEM 7. Show that Monotone Convergence Theorem can be proved as a corollary

of the Fatou’s lemma.

SOLUTION. Applly the Fatou’s lemma to the following sequences (f +f

) and (f −f

Since f

↑ f both of these sequences are non-negative. Hence, application of the Fatou’s

lemma to the sequence (f + f

) gives lim inf

≥

f. And application of Fatou’s

lemma to the sequence (f − f

) gives lim sup

≤

f.Combination of these two

inequalities proves the Monotone Convergence Theorem.

PROBLEM 8. Let f ∈ L

and

f < ∞, then show that the set

{x : f (x) > 0} is σ−ﬁnite.

SOLUTION.This follows from the following equality.

∪

∞

n=1

{x : f (x) > 1/n} = {x : f(x) > 0}.

Each of the sets on the left hand side of this equality is of ﬁnite measure , since otherwise

f would not have ﬁnite integreal.And this shows that the set in question is the union of

sets of ﬁnite measure.

PROBLEM 9.

a.)If f is nonnegative and integrable on A, then show that

µ({x : x ∈ A, f(x) ≥ c}) ≤ 1/c

f(x)dµ

b.)If

|f(x)|dµ = 0, prove that f(x) = 0 a.e.

SOLUTION.

a.)This is known as the Chebyshev’s inequality.If A

= {x : x ∈ A, f(x) ≥ c}, then

f(x)dµ =

f(x)dµ +

A−A

f(x)dµ ≥

f(x)dµ ≥ cµ(A

This proves the result.

b.)By the Chebyshev’s inequality,

µ({x : x ∈ A, f(x) ≥ 1/n}) ≤ n

f(x)dµ = 0, ∀n = 1, 2, ....

Therefore,

µ({x : x ∈ A, f(x) 6= 0}) ≤

∞

n=1

µ({x : x ∈ A, |f(x)| ≥ 1/n}) = 0.

And this clearly proves the desired result.

PROBLEM 10.

a.)Consider a measure space (X, µ) with a ﬁnite,positive,ﬁnitely additive measure µ.

Prove that µ is countably additive if and only if it satisﬁes the following condition.

If A

is a decresing sequence of sets with empty intersection then

lim

n→∞

µ(A

) = 0.

b.)Now suppose that X is locally compact Hausdorﬀ space, that Br is the Borel σ−algebra,

and that µ is ﬁnite, positive, ﬁnitely additive measure on Br.Suppose moreover that µ

is regular, that is for each B ∈ Br we have,

µ(B) = sup

{µ(K) : K ⊆ B, K − compact}

Prove that µ is countably additive.

SOLUTION.

a.)Suﬃcency:Let (B

) be countably many measurable sets which are mutually dis-

joint.Let A

= ∪

∞

i=n+1

.Then ∩

∞

n=1

= ∅.We have

µ(∪

∞

n=1

) = lim

n→∞

{µ(∪

i=1

) + µ(∪

∞

i=n+1

)}

= lim

n→∞

{

∞

i=1

µ(B

) + µ(A

)} =

∞

i=1

µ(B

Therefore µ is a measure.The necessity is obvious.

b.) If µ is not countably additive ,by a.) there is a decreasing sequence (A

) of measurable

sets with empty intersection such that

lim

n→∞

µ(A

) = inf µ(A

) > 0.

For each n there exists K

contained in A

, such that

µ(A

) < µ(K

) + 1/2

n+1

inf

µ(A

Then

µ(A

− ∩

i=1

) ≤

i=1

µ(A

− K

) < 1/2 inf

µ(A

)

which implies that µ(∩

i=1

) 6= 0 and therefore ∩

i=1

6= ∅.

Thus {∩

i=1

: n ∈ N} is a decreasing sequence of nonempty compact subsets in the

compact space K

.So ∩

∞

i=1

6= ∅,which contradicts the fact that ∩

∞

n=1

= ∅.

PROBLEM 11. Let λ be Lebesgue measure on R. Show that for any Lebesgue

measurable set E ⊂ R with λ(E) = 1, there is a Lebesgue measurable set A ⊂ E with

λ(A) = 1/2.

SOLUTION. Deﬁne the function f : R → [0, 1] by f(x) = λ(E ∩ (−∞, x]), where

x ∈ R.It is continuous by the following inequality

|f(x) − f(y)| ≤ |x − y|,

wherex, y ∈ R.Since lim

x→−∞

f(x) = 0 and lim

x→∞

f(x) = 1, there is a point x

∈ R

such that f(x

) = 1/2.Put A = E ∩ (−∞, x

PROBLEM 12. Let m be a countably additive measure deﬁned for all sets in a

σ−algebra M.

a.)If A and B are two sets in M with A ⊂ B, then show that m(A) ≤ m(B).

b.)Let (B

) be any sequence of sets in M.Then show that m(∪

∞

n=1

) ≤

∞

n=1

m(B

SOLUTION.These are almost trivial statements.

a.)We have B = A ∪ (B − A) and using countable additivity of m by taking other sets

to be empty gives, m(B) = m(A) + m(B −A).But m(B −A) ≥ 0. So the result follows.

b.)By set theory we can ﬁnd a mutually disjoint sequence (A

) such that ∪

∞

n=1

∪

∞

n=1

.So we have,

m(∪

∞

n=1

) = m(∪

∞

n=1

) =

∞

n=1

m(A

) ≤

∞

n=1

m(B

where the second equality follows from the countable additivity of m and the last in-

equlity follows from the fact that each term in the sum on the left is less than or equal

to the corresponding term on the right.i.e.A

⊆ B

, ∀n hence by part a.), m(A

) ≤

m(B

), ∀n. So the result follows.

PROBLEM 13. a.)Let (E

) be an inﬁnite decreasing sequence of Lebesgue measurable

sets, that is, a sequence with E

n+1

⊂ E

for each n. Let m(E

) be ﬁnite, where m is the

Lebesgue meausre.Then show that m(∩

∞

i=1

) = lim

n→∞

m(E

b.)Show by acounterexample that we can not omit the condition m(E

) is ﬁnite.

SOLUTION.

a.) Let E = ∩

∞

i=1

, and let F

= E

− E

i+1

.Then E

− E = ∪

∞

i=1

, and the sets F

are

pairwise disjoint.Hence,

m(E

− E) =

∞

i=1

m(F

) =

∞

i=1

m(E

− E

i+1

But m(E

) = m(E

) + m(E

−E), and m(E

) = m(E

i+1

) + m(E

−E

i+1

),since E ⊂ E

and E

i+1

⊂ E

. Since m(E

) ≤ m(E

) < ∞, we have m(E

− E) = m(E

) − m(E) and

m(E

− E

i+1

) = m(E

) −m(E

i+1

).Thus

m(E

) −m(E) =

∞

i=1

(m(E

) −m(E

i+1

)

= lim

n→∞

i=1

(m(E

) −m(E

i+1

)

= lim

n→∞

(m(E

) −m(E

)

= m(E

) − lim

n→∞

m(E

Since m(E

) < ∞, we have m(E) = lim

n→∞

m(E

)

b.)Let E

= (n, ∞).Then m(E

) = ∞ for all n,whereas ∩

∞

n=1

= ∅.

PROBLEM 14.

a.)Show that we may have strict inequality in Fatou’s Lemma.

b.)Show that Monotone Convergence Theorem need not hold for decreasing sequences

of functions.

SOLUTION.

a.)Consider the sequence (f

) deﬁned by f

(x) = 1 if n ≤ x < n + 1, with f

(x) = 0

otherwise.Then, lim inf f(x) = 0 but

(x) = 1, ∀n. Hence the strict inequality holds.

b.)Let f

(x) = 0 if x < n and f

(x) = 1 if x ≥ n.i.e.f

(x) = 1

[

n, ∞) where 1

represents the characteristic function of the set A. Now clearly f

& 0 but lim

n→∞

is undeﬁned.Hence M.C.T. does not hold in this case.

PROBLEM 15. Let (f

) be a sequence of nonnegative measurable functions that

converge to f, and suppose that f

≤ f, ∀n. Then show that

f = lim

SOLUTION.We know that f

→ f and f

≤ f, ∀n. Therefore we can choose a sub-

sequence f

such that f

% f as k → ∞. Thus an application of the Monotone

Convergence Theorem to the sequence f

gives the result.

PROBLEM 16. Suppose A ⊂ R is Lebesgue measurable and assume that

m(A ∩(a, b)) ≤ (b − a)/2

for any a, b ∈ R, a < b.Prove that m(A) = 0.

SOLUTION.If m(A) 6= 0 there is an n such that m(A ∩(n, n + 1) 6= 0.There is an open

set U in (n, n + 1) such that

A ∩(n, n + 1) ⊆ U ⊆ (n, n + 1)

and m(U) < m(A ∩ (n, n + 1)) + ,where  < m(A ∩ (n, n + 1)).

There are at most countably many disjoint intervals (a

, b

)

s such that U = ∪

, b

Then A ∩(n, n + 1) = ∪A ∩ (a

, b

).We have

m(A∩(n, n+1)) =

m(A∩(a

, b

) ≤

−a

)/2 = 1/2m(U) < 1/2(m(A∩(n, n+1))+)

which gives m(A ∩(n, n + 1)) < , a contradiction.

PROBLEM 17. Choose 0 < λ < 1 and construct the Cantor set K

as follows:Remove

from [0, 1] its middle part of length λ; we are left with two intervals L

and L

. Remove

from each of them their middle parts of length λ|I

|, i = 1, 2, etc and keep doing this ad

inﬁmum. We are left with the set K

. Prove that the set K

has Lebesgue measure 0.

SOLUTION.Calim : For any n ∈ N, the total length of intervals removed in the n’th

step is λ(1 −λ)

n−1

The claim holds for n = 1. Assume that it holds for k ≤ n. Then the total length of

intervals removed in the k + 1

th step is

λ(1 −

i=1

λ(1 −λ)

i−1

) = λ(1 − λ)

By induction the claim holds for any n ∈ N.

It follows that the Lebesgue measure of K

1 −

∞

n=1

λ(1 −λ)

n−1

= 0.

PROBLEM 18. Let A ⊂ [0, 1] measurable set of positive measure.Show that there

exist two points x

6= x

in A with x

− x

rational.

SOLUTION.Denote all rational numbers in [−1, 1] by r

, r

, ..., r

, ...Denote A

= {x+

: x ∈ A}.Then m(A

) = m(A) > 0. A

⊂ [−1, 2].Thus,

∪

∞

i=1

⊂ [−1, 2].

Suppose that A

∩ A

= ∅ if n 6= m. Then

∞

n=1

m(A

) ≤ m([−1, 2]) = 3,

which contradics m(A) > 0. Therefore there must be some n, m such that A

∩A

6= ∅.

Take z ∈ A

∩ A

. Then we can ﬁnd x

, x

∈ A such that

z = x

+ r

= x

+ r

Thus x

− x

= r

− r

PROBLEM 19. Let f : R

→ R be an arbitrary function having the property that for

each  > 0, there is an open set U with λ(U) <  such that f is continuous on R

−U(in

the relative topology).Prove that f is measurable.

SOLUTION.Let U

be an open set such that λ(U

) < 1/k and f is continuous on

− U

.Let f

= f1

−U

(where 1

represents the characteristic function of the set

A), thenf

is measurable.For any  > 0,

∗

({x : |f

− f|(x) ≥ }) = m

∗

({x ∈ U

: |f(x)| ≥ }) ≤ 1/k.

It follows that (f

) converges to f in measure.Since Lebesgue measure is complete f is

measurable.

PROBLEM 20. Prove or disprove that composition of two Lebesgue integrable func-

tions with compact support f, g : R → R is still integable.

SOLUTION.It is not true.For example, letf(x) = 1

{0}

(x) and g(x) = 1

{0,1}

(x), where

represents the characteristic function of the set A.Then f and g are integrable func-

tions with compact support. But,since g ◦ f ≡ 1, the function g ◦ f is not integrable.

PROBLEM 21. Let (X, M, µ) be a positive measure space with µ(X) < ∞. Show

that a measurable function f : X → [0, ∞) is integrable (i.e. one has

fdµ < ∞) if

and only if the series

∞

n=0

µ({x : f (x) ≥ n})

converges.

SOLUTION. Suppose f is integrable. Then

∞

n=0

µ({x : f (x) ≥ n}) =

∞

n=0

∞

m=n

µ({x : m ≤ f (x) < m + 1})

∞

m=0

n=0

µ({x : m ≤ f (x) < m + 1}) =

∞

m=0

(m + 1)µ({x : m ≤ f(x) < m + 1})

∞

m=0

mµ({x : m ≤ f(x) < m + 1}) +

∞

m=0

µ({x : m ≤ f (x) < m + 1})

∞

m=0

{x:m≤f(x)<m+1}

f(x)dµ(x) + µ(X) =

(f + 1)dµ < ∞

Conversely,

fdµ =

∞

m=0

{x:m≤f(x)<m+1}

f(x)dµ(x)

≤

∞

m=0

(m + 1)µ({x : m ≤ f(x) < m + 1}) =

∞

n=0

µ({x : f (x) ≥ n}) < ∞

which shows that f is integrable.

PROBLEM 22.

a.) Is there a Borel measure µ (positive or complex) on R with the property that

fdµ = f(0)

for all continuous f : R → C of compact support? Justify.

b.) Is there a Borel measure µ (positive or complex) on R with the property that

fdµ = f

(0)

for all continuous f : R → C of compact support? Justify.

SOLUTION.

a.) Yes. Let µ(E) = 1

(0), where 1

represents the characteristic function of the set A,

for any Borel set E.

b.) No. If there were such a Borel measure , let Φ ≥ 0 be a continuously diﬀerentiable

function of compact support, taking value 1 on [−1, 1].Then a contradiction occurs from

the following limits.

lim

n→∞

Φ(t)e

t/n

dt =

Φ(t)dt > 0

and

lim

n→∞

(Φ(t)e

t/n

)

t=0

= lim

n→∞

t/n

/n)|

t=0

= 0.

PROBLEM 23. Let f

be a sequence of real-valued functions in L

(R) and suppose

that for some f ∈ L

(R)

∞

−∞

(t) −f(t)|dt ≤ 1/n

, n ≥ 1.

Prove that f

→ f almost everywhere with respect to Lebesgue measure.

SOLUTION. Since

sup

k=1

k+1

− f

|dt ≤

∞

k=1

(1/(k + 1)

+ 1/k

) < ∞

by Levi’s theorem there is a measurable set E of measure 0 such that for any t ∈ R −E,

sup

k=1

k+1

− f

|(t) < ∞.

Therefore for any t ∈ R − E,

(t) = f

(t) +

k=2

− f

k−1

)(t)

converges. It follows that f

→ f almost everywhere.

PROBLEM 24. Let m denote the Lebesgue measure on [0, 1] and let (f

) be a sequence

in L

(m) and h a non-negative element of L

(m).Suppose that

i.)

gdm → 0 for each g ∈ C([0, 1]) and

ii.) |f

| ≤ h for all n.

Show that

dm → 0 for each Borel subset A ⊂ [0, 1].

SOLUTION.For any  > 0, there is a δ > 0 such that

hdm < , whenever m(E) < δ.

For such a δ there are a compact set K and an open set U such that (1)K ⊆ A ⊆ U

and (2)m(U − K) < δ. There is a continuous function g : [0, 1] → R such that

(3)0 ≤ g ≤ 1, (4)g = 1 on K and (5)g = 0 outside U. Then we have

lim sup

n→∞

dm = lim sup

n→∞

dm|

≤ lim sup

n→∞

gdm| + |

− g)dm|

≤ lim sup

n→∞

gdm| + |

U−K

dm| ≤ 

It follows that lim

n→∞

dm = 0.

NOT E : Here 1

represents the characteristic function of the set B.

PROBLEM 25.

a.)Prove the Lebesgue Dominated Convergence Theorem.

b.)Here is a version of Lebesgue Dominated Convergence Theorem which is some kind

of extension of it.Prove this.

Let (g

) be asequence of integrable functions which converges a.e. to an integarble

function g.Let (f

) be asequence of measurable functions such that |f

| ≤ g

and (f

)

converges to f a.e. If

g = lim

, then

f = lim

c.) Show that under hypotheses of the part b.) we have

− f| → 0 as n → ∞.

d.)Let (f

) be asequence of integrable functions such that f

→ f a.e. with f is inte-

grable.Then show that

|f − f

| → 0 as n → ∞ if and only if

| →

|f| as n → ∞.

SOLUTION.

a.) First let us state the theorem properly.

Let g be integrable over E and let (f

) be a sequence of measurable functions such that

| ≤ g on E and for almost all x in E we have f(x) = lim f

(x). Then

f = lim

P roof :The function g − f

is nonnegative so by Fatou’s Lemma we have

(g −f

) ≤ lim inf

(g −f

Since |f| ≤ g, f is integrable, and we have

g −

f ≤

g −lim sup

whence

f ≥ lim sup

Similarly, considering g + f

, we get

f ≤ lim inf

and this completes the proof.

b.) We will try to use the same idea as in the above proof. Take h

:= g

− f

, by

noting that h

≥ 0 and k

:= g

+ f

.Applying the Foatou’s lemma to these sequences

we get the follwing inequalities combination of which proves the result.

Application of Fatou’s lemma to h

gives lim sup

≤

f and application of the

Fatou’s lemma to k

gives lim inf

≥

c.)Take f = 0, g

= |f

| + |f| and g = 2|f | in part b.).Note that now our sequence is

− f| which tends to 0 as n → ∞.

d.)Suppose

|f −f

| → 0 as n → ∞.We have ||f

|−|f|| ≤ |f

−f| and this immediately

gives

| →

|f| as n → ∞.

Conversely, suppose that

| →

|f| as n → ∞.We will use part b.).Take g

2(|f

| + |f|) and note that g

→ 4|f| and note also that |f| is integrable by Fatou’s

lemma.Now the result follows from part b.) by taking f

= |f

− f| + |f

| −|f|.

PROBLEM 26. Evaluate

lim

n→∞

(1 + x/n)

−2x

dx,

justifying any interchange of limits you use.

SOLUTION. W know that lim

n→∞

(1+x/n)

= e

and (1+x/n)

≤ (1+x/(n+1))

n+1

Also we have (1 + x/n)

≤ e

. Therefore we get (1 + x/n)

% e

, which gives that

(1 + x/n)

−2x

≤ e

−x

.Therefore we can apply the Dominated Convergence Theorem to

the function (1 + x/n)

with the dominating function e

−x

.An easy computation gives,

lim

n→∞

(1 + x/n)

−2x

dx = lim

n→∞

∞

[0,n]

(x)(1 + x/n)

−2x

∞

lim

n→∞

[0,n]

(x)(1 + x/n)

−2x

dx =

∞

−x

dx = 1.

PROBLEM 27.

a.) Let (a

) be a sequence of nonnegative real numbers. Set µ(∅) = 0, and for every

nonempty subset A of N (set of natural numbers) set µ(A) =

n∈A

.Show that the

set function µ : P (N) → [0, ∞] is a measure.

b.) Let X be a nonempty set and let f : X → [0, ∞] be a function.Deﬁne µ by

µ(A) =

a∈A

f(x) if A 6= ∅ and is at most countable, µ(A) = ∞ if A is uncountable,

and µ(∅) = 0.Show that µ is a measure.

SOLUTION.

a.) If (A

) is a sequence of pairwise disjoint subsets of N and A = ∪

∞

n=1

, then note

that

µ(A) =

k∈A

∞

n=1

(

k∈A

) =

∞

n=1

µ(A

This clearly shows that µ is a measure.

b.) We need to show the σ−additivity of µ. For that let (A

) be asequence of pairwise

disjoint sequence of subsets of X. Set A = ∪

∞

n=1

. If some A

is uncountable then A is

likewise uncountable , and hence , in this case µ(A) =

∞

n=1

µ(A

) = ∞ holds. On the

other hand, if each A

is at most countable then A is also at most countable, and so

µ(A) =

x∈A

f(x) =

∞

n=1

[

a∈A

f(x)] =

∞

n=1

µ(A

)

also holds.Therefore µ is σ−additive and hence it is a measure.

PROBLEM 28. Let F be a nonempty collection of subsets of a set X and let f : F →

[0, ∞] be a function. Deﬁne µ : P (X) → [0, ∞] by µ(∅) = 0 and

µ(A) = inf{

∞

n=1

f(A

) : (A

) ⊆ F, and, A ⊆ ∪

∞

n=1

}

for each A 6= ∅, with inf ∅ = ∞. Show that µ is an outer measure.

SOLUTION.(1) By deﬁnition we have µ(∅) = 0.

(2) (Monotonicity) : Let A ⊆ B and let (A

) be a sequence in F with B ⊆ ∪

∞

n=1

Then A ⊆ ∪

∞

n=1

, and so µ(A) ≤

∞

n=1

f(A

). Therefore

µ(A) ≤ inf{

∞

n=1

f(A

) : (A

) ⊆ F, and, B ⊆ ∪

∞

n=1

} = µ(B).

If there is no sequence (A

) with B ⊆ ∪

∞

n=1

, then µ(B) = ∞, and clearly µ(A) ≤ µ(B).

(3) (Subadditivity) : Let (E

) be a sequence of subsets of X and let E = ∪

∞

n=1

.If

∞

n=1

µ(E

) = ∞, then µ(E) ≤

∞

n=1

µ(E

) is obvously ture. So, assume

∞

n=1

µ(E

) <

∞ and let  > 0. For each n pick a sequence (A

) of F with E

⊆ ∪

∞

k=1

and

∞

k=1

f(A

) < µ(E

) + /2

Clearly, E ⊆ ∪

∞

n=1

∪

∞

k=1

holds, and so

µ(E) ≤

∞

n=1

∞

k=1

f(A

) <

∞

n=1

[µ(E

) + /2

] =

∞

n=1

µ(E

) + .

Since  > 0 is arbitrary , it follows that

µ(E) = µ(∪

∞

n=1

) ≤

∞

n=1

µ(E

Therefore µ is an outermeasure.

PROBLEM 29. Let f : R → R be a Lebesgue integrable function.Show that

lim

t→∞

f(x)cos(xt)dλ(x) = lim

t→∞

f(x)sin(xt)dλ(x) = 0

SOLUTION. This is known as the Riemann-Lebesgue lemma. Since simple functions

are dense in integrable functions, it suﬃces to prove the result for the special case

f = 1

[a,b)

where 1

represents the characteristic function of the set B. So, let f = 1

[a,b)

where −∞ < a < b < ∞. In this case, for each t > 0 we have

f(x)cos(xt)dλ(x)| = |

cos(xt)dx| = |sin(xt)/t|

x=b

x=a

| = |{sin(bt)−sin(at)}/t| ≤ 2/t,

and so lim

t→∞

f(x)cos(xt)dλ(x) = 0 holds. In a similar fashion, we can show that

lim

t→∞

f(x)sin(xt)dλ(x) = 0.

PROBLEM 30. For a sequence (A

) of subsets of a set X deﬁne

lim inf A

= ∪

∞

n=1

∩

∞

i=n

and lim sup A

= ∩

∞

n=1

∪

∞

i=n

Now let (X, S, µ) be a measure space and let (E

) be a sequence of measurable sets.Show

the following:

a.) µ(lim inf E

) ≤ lim inf µ(E

)

b.) If µ(∪

∞

n=1

) < ∞, then µ(lim sup E

) ≥ lim inf µ(E

)

SOLUTION.

a.) Note that ∩

∞

i=n

% lim inf E

and ∩

∞

i=n

⊆ E

holds for each n.Thus,

µ(lim inf E

) = lim

n→∞

µ(∩

∞

i=n

) ≤ lim inf µ(E

b.) Note that ∪

∞

i=n

& lim sup E

. Hence,since µ(∪

∞

n=1

) < ∞

µ(lim sup E

) = lim

n→∞

µ(∪

∞

i=n

) ≥ lim sup µ(E

PROBLEM 31.

a.) Let X be a nonempty set and let δ be the Dirac measure on X with respect to a point.

Show that every function f : X → R is integrable and that

fdδ = f(a)δ(a) = f(a).

b.) Let µ be the counting measure on N(set of natural numbers). Show that a function

f : N → R is integrable if and only if

∞

n=1

|f(n)| < ∞. Also, show that in this case

fdµ =

∞

n=1

f(n).

SOLUTION.

a.) Note that f = f (a)1

{a}

a.e. holds. Therefore, the function f is integrable and

fdδ = f(a)δ({a}) = f(a).

b.) Let f : N → R. Since every function is measurable, f is integrable if and only if

both f

and f

−

are integrable. So, we can assume that f(k) ≥ 0 holds for each k.

If φ

k=1

f(k)1

{k}

, then (φ

) is a sequence of step functions such that φ

% f(k) as

n → ∞ for each k, and

dµ =

k=1

f(k) %

∞

k=1

f(k)

as n → ∞. This shows that f is integrable if and only if

∞

k=1

f(k) < ∞, and in this

case

fgµ =

∞

k=1

f(k).

PROBLEM 32. Let (X, S, µ) be a measure space and let f

, f

, ... be nonnegative

integrable functions such that f

→ f a.e. and lim

dµ =

fdµ. If E is a measurable

set, then show that lim

n→∞

dµ =

fdµ.

SOLUTION. By assumptions the functions f

, f

, ...are nonnegative and

integrable (because 0 ≤ f

≤ f

) and f

→ f1

holds. Using Fatou’s lemma we

get

fdµ =

lim inf f

dµ ≤ lim inf

dµ = lim inf

dµ.(∗)

Similarly, we have

fdµ ≤ lim inf

dµ.(∗∗)

Therefore,

fdµ =

fdµ +

fdµ ≤ lim inf

dµ + lim inf

dµ

≤ lim inf(

dµ +

dµ) = lim inf(

dµ) =

fdµ,

It follows that

fdµ +

fdµ = lim inf

dµ + lim inf

dµ,

and from (*) and (**), we see that

lim inf

dµ =

fdµ.

Now let (g

) be a subsequence of (f

). Then ,

→ f a.e. and lim

n→∞

dµ =

fdµ. By the above result, we infer that

lim inf

dµ =

fdµ,

and so there exists a subsequence (g

) of the sequence (g

) such that lim

dµ =

fdµ. In other words, we have shown that every subsequence of a sequence of real

numbers (

dµ) has a convergent subsequence converging to

fdµ.This means that

lim

n→∞

dµ =

fdµ holds.

PROBLEM 33. Let f : [0, ∞) → R be a continuous function such that f(x+1) = f(x)

holds for all x ≥ 0. If g : [0, 1] → R is an arbitrary continuous function, then show that

lim

n→∞

g(x)f(nx)dx = (

g(x)dx)(

f(x)dx).

SOLUTION. Observe that by induction f(x + k) = f(x) holds for all x ≥ 0 and for

all nonnegative integers k.

The change of variables u = nx yields

g(x)f(nx)dx = 1/n

g(u/n)f(u)du = 1/n

i=1

i−1

g(u/n)f(u)du.

Letting t = u − i + 1, we get

i−1

g(u/n)f(u)du =

g((t + i −1)/n)f (t + i − 1)dt =

g((t + i −1)/n)f (t)dt.

Consequently,

g(x)f(nx)dx =

[

i=1

1/ng((t + i −1)/n)]f (t)dt =

(t)dt(∗)

where h

(t) = [

i=1

1/ng((t + i − 1)/n)]f(t) . Clearly, h

is a continuous function

deﬁned on [0, 1]. In addition, note that if |g(x)| ≤ K and |f(x)| ≤ K hold for each

x ∈ [0, 1], then h

(t) ≤ K

for all t ∈ [0, 1].i.e. the sequence (h

) is uniformly bounded

on [0, 1]. Now, note that if 0 ≤ t ≤ 1 then (i − 1)/n ≤ (t + i − 1)/n ≤ i/n.Thus, if m

and M

denote the minimum and maximum values of g, respectively, on the interval

[(i −1)/n, i/n], then m

≤ g((t + i − 1)/n) ≤ M

holds for each 0 ≤ t ≤ 1. Let

i=1

1/nm

and S

i=1

1/nM

and note that R

and S

are two Riemann sums(the smallest and the largest ones)

for the function g corresponding to the partition {0, 1/n, 2/n, ..., (n − 1)/n, 1}. Hence,

lim

n→∞

= lim

n→∞

g(x)dx.From,

(t) −R

.f(t)| = |[

i=1

1/ng((t + i −1)/n)]f (t) −R

.f(t)|

= |([

i=1

1/ng((t + i −1)/n)] − R

).f(t)| ≤ (S

− R

)|f(t)|,

we see that lim

n→∞

(t) = f(t)

g(x)dx and in fact h

converges uniformly. Now, by

(*) and the Lebesgue Dominated Convergence Theorem we obtain

lim

n→∞

g(x)f(nx)dx = lim

n→∞

(t)dt

[ lim

n→∞

(t)]dt =

[f(t)

g(x)dx]dt = (

g(x)dx)(

f(x)dx).

PROBLEM 34. Show that

∞

sin

(x)

dx =

SOLUTION.Consider the function f(x) = 1, ifx = 0, f (x) = sin

(x)/x

, if0 < x ≤

1, f(x) = 1/x

, ifx > 1.Note that f is Lebesgue integrable over [0, ∞).By the inequality

0 ≤

sin

(x)

≤ f(x), we see that the function

sin

(x)

is Lebesgue integrable over [0, ∞).

Now for each r,  > 0, we have



sin

(x)

dx = −



sin

(x)d(

)

−sin

(x)



2sinx

cosx

sin



−

sin

2

sinx

dx.

Thus, we see that

∞

sin(x)

dx = lim

r→∞

→0



sin

(x)

dx =

∞

sin(x)

dx =

PROBLEM 35.

a.)Let (f

) be a sequence of measurable functions and let f : X → R. Assume that

lim

n→∞

µ({x : |f

(x) −f(x)| ≥ }) = 0(∗)

holds for every  > 0. Show that f is measurable.

b.) Assume that (f

) ⊆ M satisﬁes f

↑ and f

→

f(i.e. f

goes to f in measure).

Show that

lim

n→∞

dµ =

fdµ.

c.) Assume that (f

) ⊆ M satisﬁes f

≥ 0 a.e. and f

→

f(i.e. f

goes to f in

measure). Show that f ≥ 0 a.e.

SOLUTION.

a.)We will show that there is a subsequence (f

) of (f

) that converges to f a.e. as the

given condition (∗) is equivalent to the condition that f

→ f in measure.

Pick a sequence (k

) of strictly increasing positive integers such that µ({x : |f

(x) −

f(x)| ≥ 1/n}) < 2

−n

for all k > k

. Set E

:= {x : |f

(x) −f (x)| ≥ 1/n} for each n and

let E := ∩

∞

n=1

∪

∞

k=n

.Then,

µ(E) ≤ µ(∪

∞

k=n

) ≤

∞

k=n

µ(E

) ≤ 2

1−n

holds for all n, and this shows that µ(E) = 0. Also, if x is not in E, then there exists some

n such that x is not in ∪

∞

k=n

, and so |f

−f| ≤ 1/m holds for each m ≥ n.Therefore,

lim f

(x) = f(x) for each x ∈ E

, and so f

→ f a.e. holds. Thus, f is measurable as

limit of a sequence of measurable functions is itself measurable.

b.)By part a.) there exists a subsequence f

which converges to f a.e. Since f

↑, it

easily follows that f

↑ f.Now apply the Monotone Convergence Theorem to deduce the

result.

c.) Again by part a.) there exists a subsequence f

which converges to f a.e. Thus, we

must have f ≥ 0 a.e.

PROBLEM 36. Let g be an integrable function and let (f

) be a sequence of integrable

functions such that |f

| ≤ g a.e. holds for all n. Suppose that f

→

f (i.e.f

goes to f

in measure), then show that f is an integrable function and lim

− f|dµ = 0.

SOLUTION. By the above exercise we know that if f

converges to f in measure then

there exists a subsequence f

of f

which converges to f a.e.Thus, |f| ≤ g a.e. And

application of the Lebesgue Dminated Convergence Theorem gives that f is integrable.

Now, assume on the contrary that lim

− f|dµ 6= 0.Thus, assume that for some

 > 0 there exists a subsequence (g

) of (f

) such that

− f|dµ ≥ . But we know

that there exists a subsequence (h

) of (g

) with h

→ f a.e. Now Lebesgue Dominated

Convergence Theorem implies ) <  ≤

− f|dµ → 0, which is a contradiction.

Therefore, we must have lim

− f|dµ = 0.

PROBLEM 37. Let f be a.e. positive measurable function and let

= µ({x ∈ X : 2

i−1

< f(x) ≤ 2

})

for each integer i. Show that f is integrable if and only if

∞

−∞

< ∞.

SOLUTION. Let E

:= {x ∈ X : 2

i−1

< f(x) ≤ 2

}, i = 0, +

−

1, +

−

2, ...Set φ

i=−n

for n = 1, 2, 3, ... Then there exists some function g with φ

↑ g a.e.Clearly,

g is a measurable function and 0 ≤ f ≤ g a.e.

Assume that f is integrable. Then, each φ

is a step function, and by φ

≤ 2f, it follows

that

∞

i=−∞

= lim

n→∞

dµ ≤ 2

fdµ < ∞

On the other hand, if

∞

−∞

< ∞, then each φ

is a step function, and so g is

integrable. Since 0 ≤ f ≤ g, f it follows that f is also integrable.

PROBLEM 38.

a.) Let f ∈ L

(µ)(i.e. f is integrable) and let  > 0. Show that

µ({x ∈ X : |f(x)| ≥ }) ≤ 

−1

|f|dµ

b.) If f

→ f in L

(µ) then show that f

→ f in measure.

SOLUTION.

a.) Consider the measurable set E = {x ∈ X : |f(x)| ≥ }.Then,the follwing estimate

gives the result.

|f|dµ ≥

|f|1

dµ ≥

1

dµ = µ(E).

b.) From part a.) we have the follwing inequality,

µ({x ∈ X : |f

(x) −f(x)| ≥ }) ≤ 

−1

− f|dµ.

But if f

→ f in L

(µ) holds then the right hand side of this inequality goes to 0 and

that shows that f

→ f in measure.

PROBLEM 39. Suppose f is integrable on a set A. Then, show that given  > 0 there

exists a δ > 0 such that

f(x)dµ| < 

for every measurable set E ⊂ A of measure less than δ.

SOLUTION. There are various ways to do this. Here is our proof :

The result is obvious when f is bounded, since then

f(x)dx| ≤

|f|dµ ≤ (sup

x∈E

|f(x)|)µ(E)

In the general case, let

:= {x ∈ A : n ≤ f(x) ≤ n + 1}

:= ∪

n=0

:= A − B

Then,

|f(x)|dx =

∞

n=0

|f(x)|dµ.Let N be such that

∞

n=N+1

|f(x)|dµ =

|f(x)|dµ <



and let 0 < δ <



2(N+1)

. Then µ(E) < δ implies

f(x)dµ| =

|f(x)|dµ =

E∩B

|f(x)|dµ +

E∩C

|f(x)|dµ

≤ (N + 1)µ(E) +

|f(x)dµ <



= .

PROBLEM 40. Suppose f is integrable function on R(⇔ f ∈ L

(R)).Then, show

that

lim

t→0

|f(t + x) − f(x)|dx = 0.

SOLUTION. Note that the result is immediate when f is a step function. Now let f

be an arbitrary integrable function and let  > 0. If f ≈ f

+ f

+ ...(approximate

f by a step function which can be done as step functions are dense in L

), then there

exists n

∈ N such that

∞

n=n

|dx <



We have

|f(x + t) − f(x)|dx ≤

n=1

(x + t) −

n=1

(x)|dx

∞

n=n

(x + t)|dx +

∞

n=n

(x)|dx

n=1

(x + t) −

n=1

(x)|dx

∞

n=n

(x)|dx

n=1

(x + t) −

n=1

(x)|dx + 2



Since

n=1

(x) is a step function, we have

lim

t→0

n=1

(x + t) −

n=1

(x)|dx = 0.

Consequently,

|f(t + x) − f(x)|dx <  for suﬃciently small t.This proves the result.

PROBLEM 41. Show that every extended real valued measurable function f is the

limit of a sequence (f

) of simple functions.

SOLUTION. Suppose ﬁrst that f ≥ 0. For every n = 1, 2, 3, ..., and for every x ∈ X,

write

(x) = (i − 1)/2 if (i −1)/2 ≤ f(x) < i/2, for i = 1, 2, ...2

(x) = n if f(x) ≥ n.

Clearly f

is a nonnegative simple function, and the sequence (f

) is increasing. If

f(x) < ∞, then, for some n,

0 ≤ f (x) −f

(x) ≤ 2

−n

;

if f(x) = ∞, then f

(x) = n for every n.Recalling that the diﬀerence of two simple

functions is a simple function, application of the procedure above to f

and f

−

separately

proves the result for arbitrary f.

PROBLEM 42. Suppose µ is a probability measure on X i.e.µ(X) = 1.Let A

, A

, ... ∈

U be sets in the σ−algebra U such that

i=1

µ(A

) > n −1. Show that µ(∩

k=1

) > 0.

SOLUTION. Since µ(A

) = 1 − µ(A

), we have

i=1

µ(A

) = n −

i=1

µ(A

) < 1.By

the semi-additivity of the measure we have,

µ(

[

i=1

) ≤

i=1

µ(A

) < 1.

Therefore,

µ(

i=1

) = 1 − µ((

i=1

)

) = 1 − µ(

[

i=1

) > 0.

PROBLEM 43. Suppose f is an integrable function on X = R

i) Show that ∀ > 0, there exists a measurable set with ﬁnite measure such that f is

bounded on A and

(X−A)

|f|dµ < 

ii) From this deduce that

lim

µ(E)→0

|f|dµ = 0.

SOLUTION.

i)We can assume that f ≥ 0. Consider the following sets

:= {x : f(x) = 0}, A

:= {x : 1/n ≤ f(x) ≤ n}, A

∞

:= {x : f(x) = ∞}.

Clearly, A

⊆ A

⊆ ... and also

∞

n=1

(X − A

) = A

∪ A

∞

. Note that µ(A

∞

) = 0

as f is integragle. Thus, we have

lim

n→∞

(X−A

)

fdµ =

fdµ +

∞

fdµ = 0

Therefore ∃n

such that letting A = A

(X−A)

fdµ < . Furthermore, f is bounded

on A by n

and also A has ﬁnite measure as

µ(A) ≤

fdµ < ∞

ii) Let A be as in part i) with



instead of . Let M = sup

f. Then if E is a measurable

set and µ(E) <



we have

fdµ ≤

(X−A)

fdµ +

(E∩A)

fdµ

≤

(X−A)

fdµ +

(E∩A)

fdµ <



+ Mµ(E) <



+ M



= .

PROBLEM 44.

i) Show that an algebra A is a σ−algebra if and only if it is closed under countable

increasing unions(i.e. If (E

)

∞

j=1

⊂ A and E

⊂ E

⊂ ..., then

∞

j=1

∈ A.)

ii) Suppose µ

, µ

, ... are measures on (X, M) and a

, a

, ... ∈ [0, ∞). Show that

is a measure on (X, M).

SOLUTION.

i) We just need to show that A is closed under all countable unions since the other

direction is obvious. For that, let (E

) be any sequence of sets in A. Let F

= E

and

= ∪

j=1

for n > 1. Since A is an algebra, F

∈ A. Note that F

⊆ F

n+1

, so

∞

[

j=1

∞

[

j=1

∈ A.

ii) Let m(E) =

(E). It is clear that m(∅) = 0 as m(∅) =

(∅) = 0.It

is also obvious that m(E) ≥ 0.For the countable additivity, recall the fact that series

of nonnegative terms can be added and multiplied termwise and rearrenged arbitrarily

without changing the sum. Thus, m is a measure.

PROBLEM 45.

i) Let (X, M, µ) be a measure space and E, F ∈ M. Show that

µ(E) + µ(F) = µ(E ∪ F ) + µ(E ∩ F).

ii) Given a measure space (X, M, µ) and E ∈ M, deﬁne µ

(F ) = µ(A ∩ E). Show that

is a measure.

SOLUTION.

i) Write the set as a disjoint union of two sets as follows E = (E − F ) ∪ (E ∩ F ).Now

write F as the disjoint union F = (F − E) ∪ (E ∩ F ) and E ∪ F as the disjoint union

(E − F ) ∪ (F − E) ∪ (E ∩ F ). Then we have,

µ(E ∪ F ) + µ(E ∩ F) = µ(E − F ) + µ(F − E) + 2µ(E ∩ F )

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

= µ(E) + µ(F)

ii) Clearly, µ

(A) ≥ 0 and µ

(∅) = 0. Countable additivity is also very easy to verify.

Suppose A

∩ A

= ∅ when m 6= n, then

(

∞

[

n=1

) = µ(E ∩

∞

[

n=1

) =

∞

n=1

µ(E ∩ A

) =

∞

n=1

PROBLEM 46. We know that if µ is a measure(in the sense that it is count-

ably additive) then it is continuous from below. {i.e. If A

⊆ A

⊆ ...then

µ(

) = lim µ(A

)}.Similarly, we know that if µ(X) is ﬁnite then µ is continuous

from above.{i.e. If ... ⊆ A

⊆ A

then lim

n→∞

µ(A

) = µ(

).}. Show that

a ﬁnitely additive measure is a measure if and only if it is continuous from below.Now

suppose that µ(X) < ∞. Show that µ is a measure if and only if it is continuous from

above.Give a counterexample to show that if µ is a measure with µ(X) = ∞ then it is

not necessarily continuous from above.

SOLUTION. Suppose that µ is a ﬁnitely additive and continuous from below. Let

)

∞

j=1

be a disjoint sequence of measurable sets. Then for each n let F

= ∪

j=1

, so

that we have F

⊆ F

⊆ .... By ﬁnite additivity, µ(F

) =

j=1

µ(E

), and by continuity

from below,

µ(

∞

[

j=1

) = µ(

∞

[

n=1

) = lim

n→∞

µ(F

) = lim

j=1

µ(E

) =

∞

j=1

µ(E

This clearly shows that µ is countably additive, and hence is a measure.

We will try to use previous part. Suppose µ(X) < ∞ and µ is continuous from above.

We will show that under these conditions µ is also continuous from below which, together

with previous part, will prove that µ is countably additive and hence is a measure. Take

a decresing sequence of measurable sets E

⊆ E

⊆ ..., then ... ⊆ E

⊆ E

, and so,

µ(

∞

[

j=1

) = µ(X −

∞

j=1

) = µ(X) −µ(

∞

j=1

)

= µ(X) − lim

j→∞

µ(E

) = lim

j→∞

[µ(X) − µ(E

)] = lim

j→∞

µ(E

Hence µ is continuous from below and by the previous part it is therefore countaly

additive.

PROBLEM 47. Let µ

∗

be an outer measure on X and let (A

)

∞

j=1

be a sequence of

disjoint µ

∗

− measurable sets(in the sense of Carethedory). Show that µ

∗

(E∩(

∞

)) =

∞

∗

(E ∩ A

) for any E ⊂ X.

SOLUTION. First we prove that for each ﬁnite n,

∗

(E ∩

[

j=1

) =

j=1

∗

(E ∩ A

)

For this we use induction on n. This statement is obvious when n = 1. Suppose that it

is true for n = k.Let n = k + 1 and note that

∗

(E ∩

k+1

[

j=1

) ≥ µ

∗



(E ∩

k+1

[

j=1

) ∩A

k+1



+ µ

∗



(E ∩

k+1

[

j=1

) ∩A

k+1



= µ

∗

(E ∩ A

k+1

) + µ

∗

(E ∩

[

j=1

) = µ

∗

(E ∩ A

k+1

) +

j=1

∗

(E ∩ A

)

By the monotonicity of the outer measure , we then have,

∗

(E ∩

∞

[

j=1

) ≥ µ

∗

(E ∩

[

j=1

) =

j=1

∗

(E ∩ A

Since n is arbitrary, it follows that

∗

(E ∩

∞

[

j=1

) ≥

∞

j=1

∗

(E ∩ A

)

The other inequality follows from the countable subadditivity.Hence equality holds.

PROBLEM 48. Let f : R

→ R be a Lebesgue measurable function such that

m({x : |f (x)| > t}) ≤

, t > 0.

Prove that there exists a constant C

such that for any Borel set E ⊂ R

of ﬁnite and

positive measure

|f(x)|dx ≤ C

m(E).

SOLUTION.

|f|dm =

∞

m({x : |f | > t} ∩E)dt

√

|E|

m({x : |f | > t} ∩E)dt +

∞

√

|E|

m({x : |f | > t} ∩E)dt

≤ |E|

|E|

∞

√

|E|

dt =

√

|E|+



−c



∞

√

|E|

√

|E|+

|E|

= 2

√

|E| = C

|E|.

Thus C

= 2

√

c and therefore the assertion is proved.

PROBLEM 49. Let B(m, 1) be m−dimensional ball of radius 1 centered at the origin

in R

a.) Show that there exists a function f : R → [0, 1] such that

m(B(n + 1, 1)) = m(B(n, 1))

[f(t)]

dt.

Here m denotes the Lebesgue measure.

b.) Show that

[f(t)]

dt → 0 as n → ∞.

c.) Show that for any positive number A, A

m(B(n, 1)) → 0 as n → ∞.

SOLUTION.

a.) Let B(n + 1, r) =



x ∈ R

: x

+ ...x

n+1

= r



.Integrate over t = x

n+1

to determine

the volume of B(n + 1, 1). Then we get,

m(B(n + 1, 1)) =

−1

m(B(n,

√

1 −t

))dt

= 2

−1

m(B(n, 1))



√

1 −t



dt = m(B(n, 1))

∞

−∞

f(t)dt

where f(t) = 1

[−1,1]

√

1 −t

, here 1

represents the characteristic function of the set A.

The ﬁrst equality follows from the F ubini

s theorem, and the second equality follows

from the fact that m(B(n, r)) = r

m(B(n, 1)). This fcat follows from a linear change of

variables.

b.) Note that [f(t)]

→ 0 pointwise if t 6= 0 and deﬁne f

:= f

, then f

≥ f

≥

... ≥ 0, f

is integrable and f

→ 0 almost everywhere, so by the ﬁrst question we have

[f(t)]

dt → 0.

c.)

m(B(n, 1)) = A

m(B(n − 1, 1)

[f(t)

n−1

= A

m(B(n − 2, 1))

[f(t)]

n−2

[f(t)]

n−1

= ... = A

m(B(0, 1))

[f(t)]

dt...

[f(t)]

n−1



[f(t)]



...



[f(t)]

n−1



Here we used the fact that m(B(0, 1)) = 1. By part b.), for k suﬃciently large,

[f(t)]

dt < s < 1 for some ﬁxes s with 0 ≤ s < 1. But this shows A

m(B(n, 1)) & 0.

PROBLEM 50. Let f be an integrable function. Show that

i) µ(



x : |f (x)| ≥ a



) ≤

|f|dµ.

ii) µ(



x : |f (x)| ≥ a



) = o(

) as a → ∞.

SOLUTION.

µ(



x : |f (x)| ≥ a



) =



x:|f(x)|≥a



dµ ≤



x:|f(x)|≥a



|f|

dµ ≤

|f|dµ.

ii) From part i) if a → ∞ then µ(



x : |f (x)| ≥ a



) → 0. This clearly proves the result.

PROBLEM 51. Let



X, M



be a measure space. If f ∈ L

, let λ(E) =

fdµ, for

E ∈ M. Show that λ is a measure on M, and for any g ∈ L

gdλ =

fgdµ.

SOLUTION. Since f ≥ 0, λ(E) =

fdµ ≥ 0, and λ(∅) = 0. If A is a disjoint union

of (A

)

∞

n=1

, then

λ(A) =

fdµ =

.fdµ =

fdµ =

λ(A

Therefore, λ is a measure on M.

If g ∈ L

is simple, and g =

, then

gdλ =

.λ(E

) =

fdµ =

.fdµ =

gfdµ.

If g ∈ L

is arbitrary, we can ﬁnd a sequence (φ

)

∞

of nonnegative simple functions that

increases pointwise to the function g. Then the sequence





increases pointwise to

the function g.f. Thus, by the Monotone Convergence Theorem,

gdλ = lim

n→∞

dλ = lim

n→∞

fdµ =

gfdµ.

PROBLEM 52. Let f(x) = x

−1/2

if 0 < x < 1, f(x) = 0 otherwise. Let (r

)

∞

be an

enumeration of rationals, and set g(x) =

∞

−n

f(x − r

).Show that

a.) g ∈ L

(m) and in particular g < ∞ a.e.



Here m is the Lebesgue measure



b.) g

< ∞ a.e. but g

is not integrable on any interval.

c.) g is discontinuous at every point and unbounded on every interval, and it remains

so after any modiﬁcation on a Lebesgue null set.

SOLUTION.

a.) Let f

:= 1

[1/n,1]

f, then f

≥ 0 for all n and f

% f pintwise. We have,

dm =

1/n,1]

fdm =

1/n

−1/2

dx = 2 − 2.(

)

1/2

Thus, since f

% f pointwise

fdm = lim

n→∞

dm = lim

n→∞

2 −2.(

)

1/2

= 2.

Therefore, we have

|g|dm ≤

∞

n=1

−n

|f(x − r

)|dm =

∞

n=1

−n

.2 = 2 < ∞.

and this clearly shows that g ∈ L

(m), and g < ∞ a.e.

b.) Since g < ∞ a.e., it is also true that g

< ∞ a.e. Fix any interval



a, b



for some

a < b, there is r

∈



a, b



∩ Q since Q is dense in R. There exists M ∈ N such that

when m ≥ M, r

∈ (a, b). Then

(a,b)

(x) ≥ 2

−2n

]

f(x − r

)

for m ≥ M. Therefore,

(a,b)

dm ≥ 2

−2n

]

f(x − r

)

= 2

−2n

]

(x −r

)

−1

dm = 2

−2n



ln(m) −ln(M)



for all m ≥ M, so

a,b)

dm =

dm can only be ∞, and this clearly shows that g

can not be integrable.

c.) If g = h a.e., we also have

(a,b)

|h|

dm = ∞ for any interval (a, b). Thus, h can not

be bounded on (a, b). Moreover, h is discontinuous at any point x, since otherwise h

would be bounded on some interval containing x.

PROBLEM 53. Compute the following limits and justify the calculations:

a.) lim

n→∞

∞



1 +



−n

sin(

)dx ;

b.) lim

n→∞



1 + nx



1 + x



−n

dx;

c.) lim

n→∞

∞

nsin(

)



x(1 + x

)]

−1

dx;

d.) lim

n→∞

∞

n(1 + n

)

−1

dx , (a ∈ R)

SOLUTION.

a.) For n ≥ 2 we have





1 +



−n

sin(

)



≤



1 +



−2

Note also that the function



1 +



−2

is integrable over [0, ∞). Thus, by the Dominated

Convergence Theorem ,we have

lim

n→∞

∞



1 +



−n

sin(

)dx =

∞

lim

n→∞



1 +



−n

sin(

)dx = 0.

b.)





1 + nx



1 + x



−n



≤ 1, and

1dx = 1. Thus by the Dominated Convergence

Theorem, we have

lim

n→∞



1 + nx



1 + x



−n

dx =

lim

n→∞



1 + nx



1 + x



−n

dx = 0.

c.)



nsin(

)



x(1 + x

)]

−1



≤ (1 + x

)

−1

and note that (1 + x

)

−1

is integrable over

[0, ∞).Hence, again by the Dominated Convergence Theorem, we have,

lim

n→∞

∞

nsin(

)



x(1+x

)]

−1

dx =

∞

lim

n→∞

nsin(

)



x(1+x

)]

−1

dx =

∞

(1+x

)

−1

d.) lim

n→∞

∞

n(1 + n

)

−1

dx = lim

n→∞

∞

(1 + y

)

−1

dy = lim

n→∞

arctan(y)|

∞

= 0

if a > 0; =

if a = 0; = π if a < 0.

PROBLEM 54. Suppose f

and f are measurable complex-valued functions and φ :

C → C.

a.) If φ is continuous and f

→ f a.e., then show that φ ◦f

→ φ ◦ f a.e.

b.) Show that if φ is uniformly continuous and f

→ f uniformly, almost uniformly, or

in measure, then φ◦f

→ φ◦f, uniformly, almost uniformly, or in measure, respectively.

c.) Give counterexamples when the continuity assumptions on φ are not satisﬁed.

SOLUTION.

a.) Since φ is continuous by assumption, f

→ f implies that φ ◦ f

→ φ ◦ f , and

so {x : φ ◦ f

9 φ ◦ f} ⊂ {x : f

9 f}. Thus, µ({x : φ ◦ f

9 φ ◦ f}) = 0 as

µ({x : f

9 f}) = 0.Therefore, φ ◦ f

→ φ ◦ f a.e.

b.) Since φ is uniformly continuous, for any  > 0, there exists a δ() > 0 such that

|x − y| < δ() implies that |φ(x) − φ(y)| < . Now, if f

→ f uniformly, ∀ > 0, there

is M ∈ N such that when n ≥ M, for all x ∈ X, |f

(x) − f(x)| < δ(), and so that

|φ ◦f

(x) −φ ◦f| < . But this shows that φ ◦f

→ φ ◦ f, uniformly.

If f

→ f almost uniformly, then for any 

, 

> 0, there is a set E ∈ F (σ − algebra)

and a natural number M ∈ N, such that µ(E) < 

, and when n

≥ M for x ∈

X − E, |f

(x) − f(x)| < δ(

), and so |φ ◦ f

− φ ◦ f| < 

. This clearly shows that

φ ◦f

→ φ ◦ f almost uniformly.

If f

→ f in measure, then ∀ > 0, µ({x : |f

(x) −f(x)| > δ()}) →). Since

{x : |φ ◦ f

− φ ◦f| > } ⊂ {x : |f

− f| > δ()},

we have µ({x : |φ ◦f

−φ◦f| > }) → 0. But this means that φ◦f

→ φ ◦f , in measure.

c.) A counterexample for a.) is f

(x) =

, f(x) = 0, and φ = 1

{0}

A counterexample for b.) is X = R, f

(x) = x +

, f(x) = x and φ(x) = x

PROBLEM 55. Suppose f

→ f almost uniformly, then show that f

→ f a.e. and

in measure.

SOLUTION.First let us recall what it means to converge almost uniformly: It means,

for all 

, 

> 0, there is a set E such that µ(E) < 

and x ∈ (X − E) implies

(x) −f(x)| < 

Since f

→ f almost uniformly, for any n ∈ N, there is E

∈ M(σ-algebra such that

µ(E

) <

and f

→ f on E

. Let E =

∞

, then µ(E) = 0 and f

→ f on

∞

= E

. Thus, f

→ f a.e.

Since f

→ f almost uniformly, for every 

, 

> 0, there is E ∈ M and n

∈ N

such that µ(E) < 

and when n > n

, |f

(x) − f (x)| < 

for x is not in E, and so

µ({x : |f

(x) −f(x)| ≥ 

}) ≤ µ(E) < 

. Thus,

µ({x : |f

(x) −f(x)| ≥ 

}) → 0.

Therefore, f

→ f in measure as 

and 

are arbitrary.

PROBLEM 56. Show that if f : [a, b] → C is Lebesgue measurable and  > 0, then

there is a set E ⊂ [a, b] such that m(E

) <  and f|

is continuous. Moreover, E may

be taken to be compact.

SOLUTION. Since

∞

n=1

{x : |f (x)| > n} = ∅, there is M ∈ N such that m({x :

|f(x)| > M}) <



. Let E

= {x : |f(x)| ≤ M}, and deﬁne h(x) = 1

f(x). Now

h ∈ L

[a, b], so we can ﬁnd a subsequence of a sequence of continuous functions (g

)

which tends to f a.e.Without loss of generality we assume that g

→ h a.e. Applying

Egoroﬀ’s Theorem, we have E

∈ M such that m(E

) <



and g

→ h uniformly on

. Then, we have that h is continuous on E

and so f is continuous on E

∩ E

for f

diﬀers from h only on E

. Now, m((E

∩ E

)

) = m(E

∪ E

) < , there is an open set

O containing E

∪ E

and m(O) < . Let E = O

, then E is compact, m(E

) <  and

E ⊂ E

∩ E

, so f is continuous on E.

PROBLEM 57. A measure µ is called semi-ﬁnite if every set of inﬁnite measure

contains a subset of ﬁnite, nonzero measure. Show that every σ−ﬁnite measure is semi-

ﬁnite.

SOLUTION. Since µ is σ−ﬁnite, we know that

X =

∞

[

j=1

, µ(E

) < ∞.

Without loss of generality we may assume that E

are disjoint. Let A be an arbitrary

set with µ(A) = ∞. Then,

µ(A) =

∞

j=1

µ(A ∩E

Each A ∩E

has ﬁnite measure as it is a subset of E

. Since the sum is ∞, at least some

of the sets A ∩ E

must have nonzero measure(actually, inﬁnitely many). Pick any of

them.

PROBLEM 58. Let µ be the counting measure on N. Prove that f

→ f in measure

if and only if f

→ f uniformly.

SOLUTION. Assume f

→ f in measure. This means that for any  > 0

µ({x : |f

(x) − f (x)| > }) → 0 as n → ∞. Since µ only takes integer values, this

is equivalent to : ∃N so that µ({x : |f

(x) − f(x)| > }) = 0 for n > N. This says,

(x)−f(x)| ≤  for n > N and for all x. But this says f

→ f uniformly. The converse

is obvious.

PROBLEM 59. Prove that for a > 0,

∞

−∞

−x

cos(ax)dx =

√

πe

−

SOLUTION. Deﬁne,

(x) = e

−x

j=0

(−1)

(ax)

(2j)!

and

g(x) = e

−x

∞

j=0

(−1)

(ax)

(2j)!

= e

−x

cosh(ax) = e

−x

+ e

−ax

Then, g ∈ L

, and |f

(x)| ≤ g(x). Thus, we can use Dominated Convergence Theorem

and integrate the series for f term by term. A simple calculation yields,

∞

−∞

−x

(ax)

(2n)!

dx = a

√

Thus,

∞

−∞

−x

cos(ax)dx =

∞

n=0

(−1)

√

∞

n=0

(−a

/4)

√

πe

−a

PROBLEM 60. Let (q

) be an enumeration of rationals in [0, 1]. Deﬁne the function

f on [0, 1] by,

f(x) =

n,so,q

−n

(a.) Where is this function continuous/discontinuous?

(b.) Is this function Riemann integrable?

(c.) Is this function Lebesgue integrable?

SOLUTION.

(a.) If x is rational, then x = q

for some n. and f (x+) − f(x−) = 2

−n

, so f is not

continuous at x.

Let x be irrational. Claim: f is continuous at x. To prove the claim ﬁx  > 0.

Choose N so large that

∞

n=N

−n

< .

Now, choose δ > 0 so small that the interval (x −δ, x + δ) does not contain any of

the q

with n < N. Then for y ∈ (x −δ, x + δ),

|f(x) − f(y)| <

n>N

−n

< .

(b.) The answer is yes. There is a theorem saying, if a function is bounded on a

bounded interval and it has at most countably many points of discontinuity, then

it is Riemann integrable. Our function satisﬁes the conditions so it is Riemann

integrable.

(c.) Yes. It is Lebesgue integrable. For this we can use a theorem which says if f is

properly Riemann integrable, then it is Lebesgue integrable.

Or we can try to do it directly. The partial sums

(x) =

n≤N,q

−n

are non-negative, increasing and bounded above. So either by the Monotone Con-

vergence Theorem or by the Dominated Convergence Theorem f is Lebesgue in-

tegrable.

PROBLEM 61. If f and g are two continuous functions on a common open set in R

that agree everywhere on the complement of a set of zero Lebesgue measure,then, show

that in fact f and g agree everywhere.

SOLUTION. Let f and g be two continuous functions such that f(x) = g(x), for

all x ∈ A

and m(A) = 0. Consider any point a ∈ A. Consider also the open ball

B(a, r) = {y : |y −a| < r}. Since m(A) = 0 and m(B(a, r)) > 0 it is not possible to have

B(a, r) ⊂ A for any r > 0. Therefore, for all r > 0, B(a, r) contains points in A

. Thus,

there exists a sequence of points each lying in A

and converging to a.i.e. There exists

)

∞

n=1

such that a

∈ A

for all n and a

→ a. But, then lim

n→∞

f(a

) = lim

n→∞

g(a

)

since f(a

) = g(a

) for all n. Since f and g are continuous on A

the above equality

gives,

f( lim

n→∞

) = g( lim

n→∞

)

⇒ f(a) = f( lim

n→∞

) = g( lim

n→∞

) = g(a).

Since, a ∈ A is arbitrary we have g(a) = f(a) for all a ∈ A. Since also f (b) = g(b) for

all b ∈ A

we have f(x) = g(x) for any x ∈ R