UCI Real Analysis Qualifying Solutions

Sang TruongDepartment of Mathematics, UC Irvine

June 16, 2017

Acknowledgement

This work cannot be done without great help fromProfessor Lana, Alec Fox, Fernando Quintino, and Kai-wei Zhao.

1

Contents

1 Fall 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Spring 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Fall 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Spring 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 Fall 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Spring 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

7 Fall 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

8 Spring 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

9 Fall 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

10 Fall 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

11 Spring 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

12 Fall 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

13 Spring 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

14 Spring 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

15 Fall 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

16 Fall 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

17 Fall 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

18 Spring 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2

1 Fall 2016

Problem 1. Let (X,A, µ) be a σ-finite measure space, and let A0 be a sub-σ-algebra of A.Given an a nonnegative A-measurable function f on X, show that there is a nonnegativeA0-measurable function f0 on X, such that

ˆX

fg dµ =

ˆX

f0g dµ

for every nonnegative A0-measurable function g. In what sense is f0 unique?

Proof. Note that this problem is not true if we do not require (X,A0, µ) to be σ-finite. Asa counter-example, take the usual Lebesgue measure on R, and A0 = {∅,R}. Let g = 1and f = χ[0,1]. Then LHS is 1 while RHS is either 0 and ∞ (contradiction �). Thus we willassume (X,A0, µ) is σ-finite in this proof.

We first construct f0 that holds for g to be simple functions, i.e. we want f0 satisfies

ˆE

f dµ =

ˆE

f0 dµ, for any E ∈ A0. (1.1)

Consider the σ-finite measure space (X,A0, µ). Let ν be a measure defined on this space by

ν(E) =

ˆE

f dµ for any E ∈ A0.

It is clear that ν is a σ-finite measure and ν � µ. Radon-Nikodym theorem allows us toconstruct an A0-measurable function f0 = dνdµ ≥ 0. Then (1.1) follows by the definition of f0.

In the case when g ≥ 0 is A0-measurable, we can find an increasing sequence of A0-simplefunctions ψn converges to g pointwise. Then fψn and f0ψn monotonically converges to fgand f0g respectively. By previous case and the Monotone convergence theorem

ˆX

fg dµ = lim

ˆX

fψn dµ = lim

ˆX

f0ψn dµ =

ˆX

f0g dµ.

Therefore, our f0 satisfies all the hypotheses. By Radon-Nikodym’s construction, f0 is uniqueup to µ �A0 .

Problem 2. Let (X,A, µ) be a measure space. Suppose there exists an extended real valuedA-measurable function f on X such that f > 0 µ-a.e. and f is µ-integrable. Prove that(X,A, µ) is a σ-finite measure space.

3

Proof. By Chebyshev’s inequality,

µ{f > 1/n} ≤ nˆX

f < +∞.

Therefore, X = {f = 0} ∪ (⋃∞n=1{f > 1/n}) is σ-finite.

Problem 3. Let 0 ≤ f ∈ L1(R). Set f1 = f and fn+1 = f ∗ fn for n = 1, 2, . . .

(i) Show that the series∞∑n=1

fn converges in L1(R) if and only if

ˆRf dx < 1.

(ii) Suppose that f(x) = 0 for x < 0 and that f is bounded for x < A, for some A > 0.

Show that∞∑n=1

fn(x) converges uniformly for x < A.

Proof. (i) Let P =´R f . Since fn ≥ 0, we can apply Tonelli’s theorem.

‖fn+1‖1 =ˆR

ˆRf(t)fn(x− t) dt dx

=

ˆR

ˆRf(t)fn(x− t) dx dt

=

ˆRf(t)

ˆRfn(x− t) dx dt

=

ˆRf(t)

ˆRfn(x) dx dt

= P‖fn‖1

The fourth equality is obtained thanks to the translational invariance of Lebesgueintegrals over R. By induction, we see that ‖fn‖ = P n. Therefore, if P < 1, then∑‖fn‖1 converges. Hence

∑fn converges (since L

1 is complete). Conversely, if∑fn

converges, then ‖fn‖1 → 0. This implies P < 1.

(ii) Let f be bounded by M on (0, A). It is clear that by induction, fn(x) = 0 when x < 0for every natural number n. So we only need to consider x in the interval (0, A). For

4

x in this interval, we have

fn+1(x) =

ˆRfn(t)f(x− t) dt

=

ˆ x0

fn(t)f(x− t) dt

≤Mˆ x

0

fn(t1) dt1

≤M2ˆ x

0

(ˆ t10

fn−1(t2) dt2

)dt1

...

≤Mnˆ x

0

ˆ t10

. . .

ˆ tn−10

f1(tn) dtn . . . dt1

≤Mn+1ˆ x

0

ˆ t10

. . .

ˆ tn−10

dtn . . . dt1

= Mn+1 · xn

n!

≤M · (MA)n

n!.

The series∑M · (MA)

n

n!is absolutely summable (to M ·eMA). Therefore, by Weierstrass

M-test, the series∑fn(x) converges uniformly.

Problem 4. Construct a nonnegative measurable function f on [0, 1] such that

(i) f ∈ Lm[0, 1] for any m > 0;

(ii) ess supIf = +∞ for any interval I ⊂ [0, 1].

Proof. Let g = − log x · χ(0,1] ≥ 0, then we see that ‖g‖m =´ 1

0gm < +∞ for every m > 0.

Enumerate all the rationals as {q1, q2, . . .} and define

f(x) =∞∑n=1

1

2ng(x− qn) dx.

For m ≥ 1, by Minkowski’s inequality,

‖f‖m ≤∞∑n=1

1

2n‖g(x− qn)‖m ≤ ‖g‖m

∞∑n=1

1

2n< +∞.

5

For 0 < m < 1, by the subadditivity of the function xm, we get

‖f‖mm ≤∞∑n=1

1

2mn‖g(x− qn)‖mm ≤ ‖g‖mm

∞∑n=1

1

2mn< +∞.

Therefore, f ∈ Lm[0, 1] for any m > 0. The fact that ess supIf = +∞ is clear by the densityof rational numbers and log x goes unbounded as x goes to 0.

Problem 5. Assume that f is a nonnegative measurable function on [0, 1] and

ˆ[0,1]

f dx =

ˆ[0,1]

f 2 dx =

ˆ[0,1]

f 3 dx.

Show that f(1− f) = 0 a.e.

Proof. By Cauchy-Schwarz inequality,(ˆ[0,1]

f 2)2

=

(ˆ[0,1]

f√f ·√f

)2≤ˆ

[0,1]

f 3ˆ

[0,1]

f.

From the hypothesis, the equality must occur. Therefore, f√f = k

√f a.e. for some constant

k 6= 0. Hence, f(x) = 0 or f(x) = k for almost every x ∈ [0, 1]. Let A = {x ∈ [0, 1] : f(x) =k}. We see that ˆ

A

k =

ˆA

k2 =

ˆA

k3.

So kµA = k2µA = k3µA. If µA = 0, then f = 0 a.e. Otherwise, k = 1 whence f = 0 orf = 1 a.e.

Another proof: Notice that´ 1

0f(1 − f)2 = 0. Thus f(1 − f)2 = 0 a.e. Hence f(1 − f) = 0

a.e. x ∈ [0, 1].

Problem 6. Let (fn) be a sequence of measurable functions on Rd such that |fn(x)| ≤ 1 forall x and all n and assume that

fn → f a.e., as n→∞.

Show that g ∗ fn → g ∗ f uniformly on each compact set in Rd, if g ∈ L1(Rd).

6

Proof. Let K be a compact set, we need to show g ∗ fn ⇒ g ∗ f on K. Fix an ε > 0. Sinceg is integrable, we can find a compact set M ⊃ K such that

´Mc|g| ≤ ε. Since f ≤ 1 a.e.,∣∣∣∣ˆ

Mcg(y) (fn(x− y)− f(x− y)) dy

∣∣∣∣ ≤ 2ε, for any x ∈ K. (1.2)It is clear that the set K −M = {k −m : k ∈ K, m ∈ M} is bounded. Hence, there is acompact set X that contains (K−M)∪M . Since g is integrable, we can find δ > 0 such that´A|g| ≤ ε for any µA ≤ δ. By Egoroff’s theorem, there exists X0 ⊆ X such that µ(X0) < δ

and fn ⇒ f on X\X0. Therefore, there is an integer N such that for every n ≥ N andx ∈ K,∣∣∣∣ˆ

M∩(X\X0)g(y) (fn(x− y)− f(x− y)) dy

∣∣∣∣ ≤ ε ˆM∩(X\X0)

|g| ≤ εˆRd|g|⇒ 0 (1.3)

∣∣∣∣ˆM∩X0

g(y) (fn(x− y)− f(x− y)) dy∣∣∣∣ ≤ 2ˆ

X0

|g| ≤ 2ε. (1.4)

From (1.2), (1.3), (1.4), we see that g ∗ fn ⇒ g ∗ f on M , hence on K.

2 Spring 2016

Problem 1. Assume that f ∈ L1([0, 1]). Compute

limk→+∞

ˆ[0,1]

|f |1k dx.

Proof. Note that f is finite a.e. Let A = {|f | ≤ 1} ∩ [0, 1], B = {1 < |f |

for some f ∈ L1([0, 1]) and any g ∈ C([0, 1]). Prove that 0 ≤ f ≤ 1 a.e.

Proof. We will show that the hypothesis still holds when g is replaced by h = χE, whereE is a bounded interval. Giving such h, there exists a sequence of continuous functions gj

such that gjL1−→ h and 0 ≤ gj ≤ 1. By Riesz’s lemma, we can extract a subsequence that

converges pointwise to h. WLOG, let gja.e.−−→ h.

Claim 1 :

limj→∞

ˆ[0,1]

fgj =

ˆ[0,1]

fh.

This is true by Dominated convergence theorem because fgja.e.−−→ fh and

∣∣fgj∣∣ ≤ ∣∣f ∣∣ whichis integrable.

Claim 2 :

limj→∞

ˆ[0,1]

fngj =

ˆ[0,1]

fnh uniformly in n.

Indeed, since∣∣fn∣∣ ≤ 1, we have∣∣∣∣ˆ

[0,1]

fn(gj − h)∣∣∣∣ ≤ ‖gj − h‖1 −→ 0.

This completes the proof of Claim 2. From the two claims, we get

limn→∞

ˆfnh = lim

n→∞limj→∞

ˆfngj

= limj→∞

limn→∞

ˆfngj

= limj→∞

ˆfgj =

ˆfh.

The second equality is obtained since the inner limit is uniformly in n by Claim 2. Plugh = χE into above equation, we get

1

µ(E)

ˆE

f = limn→∞

1

µ(E)

ˆE

fn ∈ [0, 1] for µ(E) > 0.

By Lebesgue differentiation theorem, we can conclude that 0 ≤ f ≤ 1 for a.e.

Remark : There is a cleaner way to write out the convergence without interchanging twolimits. ˆ

[0,1]

|fnh− fh| ≤ˆ

[0,1]

|fnh− fngj|+ˆ

[0,1]

|fngj − fgj|+ˆ

[0,1]

|fgj − fh|

goes to zero for some fixed large j and for any n ≥ N for some N > 0. The argument is stillthe same as above.

8

Problem 3. Let f, g ∈ L2(R,ML, µL). Show that f ∗ g is a continuous function on Rvanishing at infinity, that is, f ∗ g ∈ C(R) and

lim|x|→+∞

(f ∗ g)(x) = 0.

Proof. If g is continuous with compact support, then f ∗ g is clearly uniformly continuous.For g ∈ L2, we can find a continuous function with compact support h such that ‖g− h‖2 issmall. By triangle inequality

|f ∗ g(x+ δ)− f ∗ g(x)| ≤ |f ∗ g(x+ δ)− f ∗ h(x+ δ)|+ |f ∗ h(x+ δ) + f ∗ h(x)|+ |f ∗ h(x)− f ∗ g(x)|.

The first and last terms can be bounded by ‖f‖2‖g − h‖2. The second term is small sinceh is continuous with compact support. Note that the choice of δ does not depend on x.Therefore, f ∗ g is uniformly continuous.

There are sequences of continuous functions with compact support such that fnL2−→ f and

gnL2−→ g. Therefore,

‖fn ∗ gn − f ∗ g‖∞ ≤ |fn ∗ gn − fn ∗ g|+ |fn ∗ g − f ∗ g|≤ ‖fn‖2‖gn − g‖2 + ‖fn − f‖2‖g‖2→ 0 as n→∞.

Therefore, fn ∗ gn converges to f ∗ g uniformly. This implies f ∗ g → 0 as |x| → +∞.

Problem 4. Let (X,A, µ) be a finite measure space, and let p1 ∈ (1,∞]. Let {fn}n∈Nbe a uniformly bounded sequence in Lp1(X,A, µ), that is, {fn}n∈N ⊂ Lp1(X,A, µ) andsupn∈N ‖fn‖Lp1 < ∞. Suppose f = limn→∞ fn exists µ-a.e. Prove that f ∈ Lp(X,A, µ) forall p ∈ [1, p1] and fn → f in Lp(X,A, µ) for all p ∈ [1, p1).

Proof. We have |fn|p1a.e.−−→ |f |p1 . By Fatou’s lemma,ˆ

X

|f |p1 ≤ lim infn→∞

ˆX

|fn|p1 ≤M.

This shows f ∈ Lp1 . Since the measure space is finite, fn and f also belong to Lp for anyp ∈ [1, p1). To show fn

Lp−→ f , by Vitali convergence theorem, it suffices to show that {|fn|p}is uniformly integrable over X (tightness is automatic since the measure space is finite).

Fix an ε > 0, we need to find δ > 0 such that´A|fn|p ≤ ε for any n and any measur-

able set A ⊂ X with µ(A) < δ. Let R > 0, since p1 − p > 0, we have this inequality

|fn|p · χ{|fn|≥R} ·Rp1−p ≤ |fn|p1 .

9

Integrate both sides over X, we see thatˆA∩{|fn|≥R}

|fn|p ≤M

Rp1−p. (2.1)

On the other hand, ˆA∩{|fn| 0} is σ-finite. Then we can apply Tonelli’s theorem for the non-negativemeasurable function χE in the product space:

(µ× µL)(E) =ˆX×R

χE(x, y) d(µ× µL)

=

ˆX

ˆRχE(x, y) dµL dµ

=

ˆX

ˆRχ[0,f(x))(y) dµL dµ

=

ˆX

f dµ.

Case 2: {f > 0} is not σ-finite. Then´R f dµ = +∞. We need to show µ× µL(E) = +∞.

If b > a > 0, then {x : a < f(x) < b} × [0, a] ⊆ E, so a · µ{x : a < f(x) < b} ≤ µ× µL(E).So it suffices to find a, b > 0 such that µ{x : a < f(x) < b} = +∞. Indeed, let An = {x :

1n+1

< f ≤ 1n} and Bn = {x : n < f ≤ n+ 1}, then

{x : f > 0} =∞⋃n=1

An ∪∞⋃n=1

Bn ∪ {f =∞}.

Thus, at least one of An’s or Bn’s must have infinite measure.

10

Problem 6. Let f ∈ L1(R), and let a1, . . . , ak ∈ R and b1, . . . , bk ∈ R\{0}. Assume thatthe vectors

ajbj

are all distinct. Determine

limt→∞

ˆR

∣∣∣∣∣k∑j=1

f(bjx+ taj)

∣∣∣∣∣ dx.Proof. We claim that the limit is

∑kj=1

1|bj |‖f‖1. It is clear that

ˆR

∣∣∣∣∣k∑j=1

f(bjx+ taj)

∣∣∣∣∣ dx ≤k∑j=1

ˆR

∣∣f(bjx+ taj)∣∣ dx = k∑j=1

1

|bj|‖f‖1.

Let ε > 0 be very small. Since f is integrable, there exists R > 0 such that´

(−R,R)c |f | ≤ ε.

Note that the condition |bjx + taj| ≤ R is equivalent to x ∈ B R|bj |

(tajbj

). Since

ajbj

’s are all

distinct, we can find N such that for every t > N , all the balls B R|bj |

(tajbj

)are disjoint.

ˆR

∣∣∣∣∣k∑j=1

f(bjx+ taj)

∣∣∣∣∣ dx ≥k∑i=1

ˆB R|bi|

(taibi

)∣∣∣∣∣k∑j=1

f(bjx+ taj)

∣∣∣∣∣ dx≥

k∑i=1

ˆB R|bi|

(taibi

)(∣∣f(bix+ tai)∣∣− k∑

j 6=i

∣∣f(bjx+ taj)∣∣) dx≥

k∑i=1

1

|bj|

ˆ(−R,R)

∣∣f(x)∣∣ dx− k(k − 1)ε≥

k∑i=1

1

|bj|(‖f‖1 − ε

)− k(k − 1)ε.

Let ε → 0, and combining with the beginning inequality, we can conclude the limit is∑kj=1

1|bj |‖f‖1.

3 Fall 2015

Problem 1. Let E be a measurable subset of [0, 2π]. Assume that f ∈ C(R) is 1-periodic,i.e. f(x+ 1) = f(x). Compute

limn→+∞

ˆE

f(nx) dx.

Justify your answer.

11

Proof. We will show a more general problem: if f ∈ L1(R) and g ∈ C(R) is continuous withperiod T > 0, then

limn→∞

ˆRf(x)g(nx) dx =

(ˆR

f

)(1

T

ˆ T0

g

). (3.1)

Case 1 : f = χ(a,b). We need to show

limn→∞

ˆ ba

g(nx) dx =b− an

ˆ T0

g. (3.2)

For any n ∈ N large, there are unique integers k, l (depending on n) such that{(l − 1)T ≤ nb ≤ lT,kT ≤ na ≤ (k + 1)T.

Simplifying this givesa− bT− 2n≤ k − l

n≤ a− b

T.

Therefore, limn→∞k−ln

= a−bT. So we have

LHS of (3.2) =1

n

ˆ nbna

g(x) dx

=1

n

ˆ lTna

g +1

n

ˆ kTlT

g +1

n

ˆ nbkT

g.

The first and third terms are bounded by 1n

´ T0|g| that goes to 0. The second term goes to

a−bT

´ T0g by our previous observation. Hence (3.2) is proved. It follows that (3.1) also holds

for step functions.

Case 2 : f ∈ L1(R). Note that f can be approximated in L1 norm by continuous functionswith compact supports, and continuous functions with compact supports can be approxi-mated in L1 norm by step functions. Therefore we can approximate f in L1 norm by a stepfunction ϕ. We see that ∣∣∣∣ˆ

Rf(x)g(nx) dx−

(ˆRf

)1

T

(ˆ T0

g

)∣∣∣∣≤∣∣∣∣ˆ

R(f(x)− ϕ(x))g(nx) dx

∣∣∣∣+∣∣∣∣ˆRϕ(x)g(nx) dx−

(ˆRϕ

)1

T

(ˆ T0

g

)∣∣∣∣+∣∣∣∣ˆR(ϕ(x)− f(x)) 1

T

ˆ T0

g

∣∣∣∣ .Since g is continuous and periodic, it is bounded. So the first term is small. The last termis also small by our approximation. The second term is small thanks to the previous case.So (3.1) is proved.

12

Problem 2. Suppose f ∈ L1[0, 1] and assume that there exists C > 0 such that for allmeasurable subsets E ⊂ [0, 1], we have

ˆE

|f(x)| dx ≤ Cµ(E)12 .

Show that f ∈ Lp[0, 1] for 1 ≤ p < 2. Show that the statement fails for p = 2 by giving acounter-example.

Proof. Lemma: If µ(X)

Problem 3. Show that a function f : Rn → R+ is measurable if and only if A = {(x, y) :0 ≤ y ≤ f(x)} is a measurable subset of Rn+1.

Proof. (⇒) Assume f is measurable. Consider g : (x, y) 7→ y, f̃ : (x, y) 7→ f(x), andh : (x, y) 7→ y− f̃(x). For any open set E ⊆ R, f̃−1(E) = f−1(E)×R is measurable, so f̃ ismeasurable. Similarly, g−1(E) = Rn×E is measurable, so g is measurable. Then h is measur-able since it is the sum of two measurable functions. Therefore, A = g−1[0,∞) ∩ h−1(−∞, 0]is measurable.

(⇐) Suppose A is measurable. Then for any α ≥ 0, f̃−1[α,∞) = A ∩ g−1{α} is mea-surable. So f̃ is measurable. This implies that f is also measurable.

Problem 4. Let f ∈ L1(R) and set

fh(x) =1

2h

ˆ x+hx−h

f(t) dt, h > 0.

Show that fh ∈ L1(R) and fh → f in L1(R) as h→ 0.

Proof. We use Tonelli’s theorem to show fh ∈ L1(R).ˆR|fh| ≤

ˆR

ˆR

1

2h|f(t)| · χ(x−h,x+h)(t) dt dx

=

ˆR

ˆR

1

2h|f(t)| · χ(t−h,t+h)(x) dx dt

=

ˆR|f | < +∞.

(3.3)

Let F (x) =´ x

0f(t) dt, then F is absolutely continuous. By Lebesgue differentiation theorem,

F (x+ h)− F (x− h)2h

−→ F ′(x) for a.e. x ∈ R.

This means fha.e.−−→ f as h → 0. Passing to sequence, we have fn

a.e.−−→ f as n → ∞. ByFatou’s lemma: ˆ

R|f | ≤ lim inf

n→∞

ˆR|fn|. (3.4)

From (3.3) and (3.4) we get ‖fn‖1 → ‖f‖1. We see that

|fn|+ |f | − |fn − f | ≥ 0. (3.5)

14

LHS of (3.5) is in L1(R) and converges pointwise a.e. to 2|f |. By Fatou’s lemma

2

ˆR|f | ≤ lim inf

n→∞

ˆR(|fn|+ |f | − |fn − f |)

This implies

lim supn→∞

ˆR|fn − f | ≤ 0.

Therefore, limn→∞´R |fn − f | = 0, i.e. fn converges to f in L

1(R).

Problem 5. Let (X,A, µ) be a measure space and let fk : X → R be a sequence ofmeasurable functions satisfying the following:

ˆX

|fk|2 dµ ≤ 2015, for all k,

and ˆX

fjfk = 0 for all j 6= k.

Prove that for all β > 3/2,

limn→+∞

1

nβ

n2∑k=1

fk(x) = 0, for a.a. x ∈ X.

Proof. First observe that

ˆX

(1

nβ

n2∑k=1

fk(x)

)2dµ =

ˆX

∑n2k=1 |fk|2 + 2

∑1≤i 0. By Chebyshev’s inequality and our observation

µ

x ∈ X :(

1

nβ

n2∑k=1

fk(x)

)2> ε2

︸ ︷︷ ︸En

≤ 2015ε2n2β−2

.

Since 2β − 2 > 1, we have∑∞

n=1 µ(En) < +∞. By Borel-Cantelli’s lemma, a.e. x ∈ Xbelongs to at most finitely many En’s. Hence we are done.

15

Problem 6. Let A,B ⊂ Rn be Lebesgue measurable sets and assume that for every x ∈ Qn,there exists a null set Nx such that

A+ x ⊂ B ∪Nx.

Here A + x = {a + x : a ∈ A}. Show that if A is not a null set, then the complement of Bin Rn is a null set.

Proof. Suppose m(A) > 0, it is enough to show that there is a zero measure set V such thatV ∪ (A+Q) = R. Indeed, if this is not the case, then there is a positive finite measure set Esuch that E ∩ (A+Q) = ∅. We show that this is a contradiction. Consider the convolution

ϕ(y) =

ˆRχA(x+ y) · χE(x) dx.

We have ϕ(q) = m(E ∩ (A+Q)

)= 0 for every q ∈ Q. Since ϕ is continuous and Q is dense,

ϕ(y) = 0 on R. We have a contradiction because´R ϕ = m(A)m(E) > 0.

4 Spring 2015

Problem 1. Show that if f ∈ L4(R) then

limc→1

ˆR|f(cx)− f(x)|4 dx = 0.

Proof. Fix an ε > 0. First, the conclusion is clear if f is continuous with compact support.For a general case, approximate f by a continuous function g with compact support, i.e.‖f − g‖4 ≤ ε. We use a well-known lemma

If p > 0, then |a+ b|p ≤ 2p(|a|p + |b|p).

in the case p = 4.

ˆR|f(cx)− f(x)|4 ≤ 16

ˆR|g(cx)− g(x)|4 + 16

ˆR|f(cx)− f(x)− g(cx) + g(x)|4

≤ 16ˆR|g(cx)− g(x)|4 + 162

ˆR|f(cx)− g(cx)|4 + 162

ˆR|f(x)− g(x)|4

= 16

ˆR|g(cx)− g(x)|4 + 2 · 162‖f − g‖4

≤ 16ˆR|g(cx)− g(x)|4 + 2 · 162ε.

Let ε→ 0 and c→ 1, then RHS goes to 0, so does LHS.

16

Problem 2. Let fn : (0,∞) → R, n = 1, 2, . . . be a sequence of Lebesgue measurablefunctions such that fn → f a.e. as n → ∞. Assume that there exists g : (0,∞) → R suchthat |fn| ≤ g, n = 1, 2, . . ., and g ∈ L1(0, a) for all 0 < a 0. Show that fn ∈ L1(0,∞), f ∈ L1(0,∞) and fn → f in L1(0,∞)as n→∞.

Proof. By changing of variables,

ˆ ∞a

∣∣fn(y)∣∣ dy = ˆ ∞a2

∣∣fn(√x)∣∣2√x

dx

≤ C2a2−→ 0 as a→∞.

This also implies fn ∈ L1(0,∞) and ‖f‖1 is uniformly bounded. So f ∈ L1(0,∞) by Fatou’slemma. By Lebesgue dominated convergence theorem, fn

L1−→ f on any (0, a). Thus (fn)is uniformly integrable. The above inequality implies (fn) is tight. Therefore, fn

L1−→ f on(0,∞) by Vitali’s convergence theorem.

Problem 3. Assume that f ∈ C1(0, 1) and

C =

ˆ(0,1)

x|f ′|p dx < +∞ for some p > 2.

Show that limx→0+ f(x) exists.

Proof. Let (an)∞n=1 ⊂ (0, 1) such that an → 0. We need to show limn→∞ f(an) exists. It is

suffices to show {f(an)}∞n=1 is Cauchy. Let q be the conjugate of p, then q < 2 < p.∣∣f(an)− f(am)∣∣ = ∣∣∣∣ˆ anam

f ′∣∣∣∣ ≤ ˆ 1

0

(x1/p

∣∣f ′∣∣)(x−1/pχ(min(am,an),max(am,an)

))≤ C1/p

∣∣∣∣ˆ anam

x−q/p∣∣∣∣

=C

1p

1− q/p

∣∣∣a1− qpn − a1− qpm ∣∣∣−→ 0 as m,n→∞

because a1− q

pn → 0, so it is Cauchy. Thus we are done.

17

Problem 4. Suppose that E ⊂ [0, 1]2 is measurable. Denote

Ex = {y ∈ [0, 1] : (x, y) ∈ E} and Ey = {x ∈ [0, 1] : (x, y) ∈ E}.

Show that if m(Ex) = 0 for a.e x ∈ [0, 1/2], then

m({y ∈ [0, 1] : m(Ey) = 1}) ≤1

2.

Proof. By Tonelli’s theorem,

ˆ 10

m(Ey) dy =

ˆ 10

m(Ex) dx =

ˆ 12

0

m(Ex) dx ≤1

2.

Therefore we have:

m{y ∈ [0, 1] : m(Ey) = 1} =ˆ{y∈[0,1]:m(Ey)=1}

m(Ey) dy ≤ˆ 1

0

m(Ey) dy ≤1

2.

Problem 5. Let f ∈ Lp(R), 1 < p 1− 1p

. Show that the series

∞∑n=1

ˆ n+n−αn

|f(x+ y)| dy

converges for a.e. x ∈ R.

Proof. Let g(x) =∑∞

n=1

´ n+n−αn

|f(x+ y)| dy. We will show that g ∈ L1loc(R). Then it mustbe finite a.e. It suffices to show that g ∈ L1(k, k + 1).ˆ k+1k

∞∑n=1

ˆ n+n−αn

∣∣f(x+ y)∣∣ dy dx = ∞∑n=1

ˆ k+1k

ˆ n+n−α+xn+x

∣∣f(y)∣∣ dy dx=∞∑n=1

ˆR

∣∣f(y)∣∣ ˆ k+1k

χ(y−n−n−α,y−n)(x) dx dy

=∞∑n=1

ˆ k+n+1+n−αk+n

∣∣f(y)∣∣ˆ χ(y−n−n−α,y−n)(x) dx dy=∞∑n=1

n−αˆ k+n+1+n−αk+n

∣∣f(y)∣∣ dy.The third equality is obtained by solving the condition (y−n−n−α, y−n)∩ (k, k+1) 6= ∅. Let(an) = {n−α}∞n=1 and bn =

{´ k+n+1+n−αk+n

∣∣f(y)∣∣ dy}∞n=1

. Then (an) ∈ `q (q is the conjugate

18

number of p), and (bn) ∈ lp since

∞∑n=1

ˆ k+n+1+n−αk+n

∣∣f ∣∣ = ∞∑n=1

(ˆ k+n+1k+1

∣∣f ∣∣+ ˆ k+n+1+n−αk+n+1

∣∣f ∣∣) ≤ 2‖f‖p.Thus, the conclusion is followed by Holder’s inequality for sum

∞∑n=1

anbn ≤ ‖an‖`q · ‖bn‖`p < +∞.

Problem 6. Suppose E ⊂ R is measurable and E = E+ 1n

for every natural number n ≥ 1.Show that either m(E) = 0 or m(Ec) = 0.

Proof. By induction we have E + Q≥0 = E. This implies E + Q

Problem 3. Denote

E =

{x ∈ [0, 1] : there exists infinitely many p, q ∈ N such that

∣∣∣∣x− pq∣∣∣∣ ≤ 1q3

}.

Show that m(E) = 0.

Proof. Rewrite E as

E =∞⋂n=1

⋃q≥n

∞⋃p=1

{x ∈ [0, 1] :

∣∣∣∣x− pq∣∣∣∣ ≤ 1q3

}︸ ︷︷ ︸

En

.

Since (En) is a nested sequence, m(E) = limn→∞m(En). We see that for each p, q, m{∣∣∣x− pq ∣∣∣ ≤ 1q3} ≤

2q3. And for each q ≥ n large enough, there are less than q solutions for p such that∣∣∣x− pq ∣∣∣ ≤ 1q3 . Therefore,

m(E) ≤ limn→∞

∞∑q≥n

m

(∞⋃p=1

{x ∈ [0, 1] :

∣∣∣∣x− pq∣∣∣∣ ≤ 1q3

})≤ lim

n→∞

∑q≥n

q

q3

= limn→∞

∞∑q≥n

1

q2= 0.

Problem 4. Assume that η ∈ L1(R) is a non-negative function satisfying´R η dx = 1. Show

that for any f ∈ L1(R),‖f ∗ η‖1 ≤ ‖f‖1.

Proof. This is a special case for Young’s inequality. We can use Tonelli’s theorem to prove.

‖f ∗ η‖1 =ˆR

∣∣∣∣ˆRf(x+ y)η(y) dy

∣∣∣∣ dx≤ˆR

ˆR|f(x+ y)|η(y) dx dy

=

ˆR

ˆR|f(x)|η(y) dx dy

=

ˆR|f(x)| dx,

where the third equality is obtained because integral over R is translational invariant.

20

Problem 5. Let f : R→ R be continuous and periodic with period one. Prove that

limn→∞

ˆ 10

f(nx) cos2(2πx) dx =1

2

ˆ 10

f(x) dx.

Proof. See a simlar proof at (3).

Problem 6. Let x = 0.n1n2 . . . be the decimal representation of x ∈ [0, 1]. Compute withjustification

´[0,1]

f(x) dx, where

(a) f(x) = maxi∈N |ni − ni+1|

(b) f(x) = maxi∈N

∞∑k=i

nk2k−i

Proof. (a) We show f = 9 a.e. Let E = {f < 9} and Ek = {|nk − nk+1| < 9}, thenE =

⋂∞k=1Ek. Divide the interval E0 = [0, 1] into 10

2 equal subintervals. Then E1is E0 removed [0.09, 0.1] and [0.90, 0.91]. Thus m(E1) =

98100m(E0). To find m(E2),

divide E0 into 104 subintervals and apply the same technique. Inductively, we have

m(Ek) =(

98100

)km(E0) =

(98100

)k. Thus m(E) = 0.

(b) We have f(x) ≤∑∞

k=i9

2k−i= 18. We show f = 18 a.e. Let Ep = {x : @ i : ni+1 = . . . =

ni+p = 9}. The same argument as before shows that m(Ep) = 0 for each p. Note thatfor each x ∈ Ep, we have f(x) ≥

∑pk=0

92k> 18− εp. This shows that f = 18 a.e.

6 Spring 2014

Problem 1. Let A be a subset of R of positive Lebesgue measure. Prove that there existk, n ∈ N and x, y ∈ A with |x− y| = k

2n.

Proof. We will use the following lemmas:

Lemma 1 : If m(A) > 0, then A− A contains an interval (centered at 0).

WLOG, assume m(A) < +∞. Consider f(x) = χA ∗ χ−A(x), which is continuous. Wesee that f(0) = m(A) > 0. Therefore, f−1(0,∞) is an non-empty open set. So thereis δ > 0 such that (−δ, δ) ⊆ f−1(0,∞). We claim that f−1(0,∞) ⊆ A − A. Indeed, iff(x) > 0, then by the definition of f , there is y such that y ∈ A and x − y ∈ −A. Hence,x = y − (y − x) ∈ A− A. Therefore, (−δ, δ) ⊆ A− A.

21

Lemma 2 : The set M =

{k

2n: k, n ∈ Z, n > 0

}is dense in R.

Indeed, let x ∈ R. Fix an ε > 0. Choose n > 0 such that 12n

< �. There is an integerk such that k ≤ x · 2n ≤ k + 1. With this choice, we have

∣∣ k2n− x∣∣ ≤ ε.

The conclusion of this problem follows directly from the two lemmas.

Remark : There is another way without using convolution. WLOG, we can assume A ⊆ [0, 1].Enumerate all the elements of M in [0, 1] as {q1, q2, . . .}. Since M is closed under addition,we are done if A+ qi ∩ A+ qj 6= ∅ for some qi 6= qj. If A+ qi ∩ A+ qj = ∅ for every i 6= j,we get

m

(∞⊔j=1

(A+ qj)

)=∞∑j=1

m(A+ qj) =∞∑j=1

m(A) = +∞.

This is a contradiction since⊔∞j=1(A+ qj) ⊆ [0, 2].

Problem 2. Either prove or give a counterexample: If a sequence of functions fn on ameasure space (X,µ) satisfies

´X|fn| dµ ≤ 1n2 , then fn → 0 µ-a.e.

Proof. From the hypothesis,

∞∑n=1

ˆX

∣∣fn∣∣ ≤ ∞∑n=1

1

n2< +∞.

By Monotone convergence theorem, we have´X

∑∞n=1

∣∣fn∣∣ < +∞. Therefore ∑∞n=1 ∣∣fn∣∣ isfinite a.e. x ∈ X. Thus, fn

a.e.−−→ 0.

Problem 3. Let f ∈ L4[a, b], and let F (x) =´ xaf(x) dx. Show that

limh→0

F (x+ h)− F (x)h3/4

= 0 for all x ∈ (a, b).

Proof. Let x ∈ (a, b). Fix an ε > 0. Since f ∈ L4[a, b], there exists δ > 0 such that

22

B = (x− δ, x+ δ) ⊂ (a, b) and ‖f · χB‖4 ≤ ε. Then for any |h| < δ, we have∣∣F (x+ h)− F (x)∣∣h3/4

≤ h−3/4ˆR

∣∣f(t)∣∣ · χ(x, x+h) dt=

ˆB

∣∣f ∣∣ · h−3/4χ(x, x+h) + ˆBC

∣∣f ∣∣ · h−3/4χ(x, x+h)≤ ‖f · χB‖4 · ‖h−3/4χ(x,x+h)‖4/3 + 0= ‖f · χB‖4 ≤ ε.

The second inequality is because BC ∩ (x, x+h) = ∅. Hence we have limh→0 F (x+h)−F (x)h3/4 = 0for every x ∈ (a, b).

Problem 4. Assume f, g ∈ L2(R). Define

A(x) =

ˆRf(x− y)g(y) dy.

Show that A(x) ∈ C(R) andlim|x|→+∞

A(x) = 0.

Proof. See problem 3 here.

Problem 5. Is it possible for a continuous function f : [0, 1]→ R to have

(i) Infinitely many strict local minima?

(ii) Uncountably many strict local minima?

Proof. (i) Let f(x) = x2 sin 1x·χ(0,1]. Then f ′(0) = limh→0

h2 sin 1h

h= 0, so f is differentiable

on [0, 1]. Moreover, f ′(x) = 2x sin 1x− cos 1

xfor x ∈ (0, 1]. Solving f ′(x) = 0 is

equivalent to solving 2 tan z = z for z ≥ 1. By the graph of tan, we see that the liney = x intersects y = 2 tanx at infinitely many points and g(z) = 2 tan z − z changessign from negative to positive at each of those points. Therefore, f has infinitely manystrict local minima.

(ii) Suppose x0 is a point where f admits a strict local minimum. Since it is strict, thereis a neighborhood of x0 where x0 is the only point that gives a local minimum. Assignx0 to a rational in this neighborhood. Hence there is a 1-1 correspondence betweenS = {x ∈ [0, 1] : f(x) is a strict local minimum} and a subset of rational numbers.Thus S must be countable.

23

Problem 6. Let A be the collection of functions f ∈ L1(X,µ) such that ‖f‖1 = 1 and´Xf dµ = 0. Prove that for every g ∈ L∞(X,µ),

supf∈A

ˆX

fg dµ =1

2(ess sup g − ess inf g).

Proof. From the hypothesis, it is clear thatˆ{f≥0}

f dµ = −ˆ{f 0small enough such that µ(U) and µ(V ) are finite, and µ

(U ∩ V

)= 0 (indeed, we can assume

U ∩ V = ∅). In this case, define

f(x) =

1

2µ(U)if x ∈ U,

− 12µ(V )

if x ∈ V ,

0 otherwise.

It is clear that this choice of f satisfies the hypotheses. Finally,ˆX

fg dµ+ ε =

ˆU

fg dµ+

ˆV

fg dµ+ ε

≥ 12µ(U)

(M − ε)µ(U)− 12µ(V )

(N + ε)µ(V ) + ε

=1

2(M −N).

That is, (6.1) is satisfied. Thus, we are done.

24

7 Fall 2013

Problem 1. Let f ∈ Lp[0, 1] for every p ∈ [1,∞]. Show that

‖f‖∞ = limp→∞‖fp‖.

Proof. It is clear that ‖f‖p ≤ ‖f‖∞ for every p ≥ 1. Fix an ε > 0, let Aε = {x ∈ [0, 1] :∣∣f ∣∣ > ‖f‖∞ − ε}. We claim that µ(A) > 0. Indeed, if µ(A) = 0, then ∣∣f ∣∣ ≤ ‖f‖∞ − ε fora.e. (contradiction to the definition of ‖f‖∞). Therefore, we have(ˆ

[0,1]

∣∣f ∣∣p) 1p ≥ (ˆAε

∣∣f ∣∣p) 1p ≥ µ(A) 1p (‖f‖∞ − ε)Combine with the first observation, we get

µ(A)1p (‖f‖∞ − ε) ≤ ‖f‖p ≤ ‖f‖∞.

Since µ(A) > 0, let ε→ 0 and p→∞, we have limp→∞ ‖f‖p = ‖f‖∞.

Problem 2. Assume that E is a subset of R2 and the distance between any two points inE is a rational number. Show that E is a countable set. (Note that this is actually true inany dimension.)

Proof. Suppose E is uncountable. Pick any x ∈ E. Since E has uncountably many elements(pigeons) and there are countably many possible distances from elements of E to x (holes),by the Pigeonhole principle, at least one hole has uncountably many pigeons. In anotherword, there is a uncountable set E1 lie on the circle (C1) centered at x. Pick any y ∈ (C1).By a similar argument, there is a uncountable set E2 lie on E1 ∩ (C2), where (C2) is a circlecenter at y. This is a contradiction since E1 ∩ (C2) contains at most 2 points.

Problem 3. Let f be a non-negative measurable function on R. Prove that if∑+∞

n=−∞ f(x+n) is integrable, then f = 0 a.e.

Proof. We have

+∞ >ˆR

+∞∑n=−∞

f(x+ n) =+∞∑

n=−∞

ˆRf(x+ n) dx

=+∞∑

n=−∞

ˆRf(x) dx.

25

Therefore,´R f = 0. By Chebyshev’s inequality, µ{f >

1n} = 0 for every positive integer n.

Thus, f = 0 a.e.

Problem 4. For each n ≥ 2, we define fn : [0, 1]→ R as follows:

fn(x) =

{n2 for x ∈

[in, in

+ 1n3

]and i = 0, 1, . . . , n− 1

0 otherwise.

(i) Show thatlimn→∞

fn(x) = 0 for a.e x ∈ [0, 1].

(ii) For g ∈ C[0, 1], what islimn→∞

ˆ[0,1]

fn(x)g(x) dx ?

(iii) (bonus problem): Is your answer to question (ii) still valid for all g ∈ L∞(0, 1)?

Proof. (i) Let x ∈ (0, 1). It suffices to show that there is an integer N > 0 such that foreach n > N , there is an integer i ∈ [0, n−1] satisfies i

n+ 1

n3< x < i+1

n. This inequality

is equivalent to

n

(x− 1

n

)< i < n

(x− 1

n3

).

The existence of i is guaranteed if

n

(x− 1

n3

)≤ n− 1 and n

(x− 1

n

)≥ 0.

The later inequality holds for every n > M , where M =⌊

1x

⌋+ 1. The first inequality

is equivalent to

x ≤ 1− 1n

+1

n3.

Since the right hand side goes to 1, there exists K > 0 such that the inequality holdsfor every n > K. Choose N = max

(M,K

); this is the constant we need to find.

(ii) ˆ[0,1]

fn(x)g(x) dx =

ˆ[0,1]

n2χtn−1i=0 [ in ,in

+ 1n3

] · g(x) dx

=1

n

n−1∑i=0

n3ˆ i

n+ 1n3

in

g(x) dx

=1

n

n−1∑i=0

g(ξi) for some ξi ∈(i

n,i

n+

1

n3

).

26

The last equality is obtained by Mean value theorem for integral. The last term is justa Riemann sum for g. Therefore,

limn→∞

ˆ[0,1]

fn(x)g(x) dx =

ˆ 10

g(x) dx.

Problem 5. Let A = {(x, y) ∈ [0, 1]2 : x + y /∈ Q, xy /∈ Q}. Find´Ay−1/2 sinx dm2, where

m2 is the Lebesgue measure in R2.

Proof. Lemma: If f : R2 → R be measurable, then its graph Γf has zero measure.

m2(Γf ) =

ˆR2χΓf (x, y) dm2

=

ˆR

ˆRχ{y=f(x)}(y) dy dx (by Tonelli’s theorem)

=

ˆR

0 dx = 0.

Let {q1, q2, . . .} be an enumeration of all rationals in [0, 1], then AC is the union of all thecurves y = qi − x and xy = qi. Thus m2(AC) = 0 by the lemma. Therefore,ˆ

A

y−1/2 sinx dm2 =

ˆ[0,1]2

y−1/2 sinx dm2.

We show that the integral exists. Indeed,

ˆ 10

ˆ 10

∣∣y−1/2 sinx∣∣ dx dy ≤ ˆ 10

y−1/2 dy = 2.

So the integrand is integrable by Tonelli’s theorem. Therefore, by Fubini’s theorem,

ˆ[0,1]2

y−1/2 sinx dx =

ˆ 10

ˆ 10

y−1/2 sinx dx = 4 sin2(

1

2

).

Problem 6. Suppose that fn ∈ L4(X,µ) with ‖fn‖4 ≤ 1 and fn → 0 µ-a.e. on X. Showthat for every g ∈ L4/3(X,µ),

limn→∞

ˆX

fn(x)g(x) dµ(x) = 0.

Hint: First consider the case when X is of finite measure.

27

Proof. First, we consider the case when µ(X) < ∞. Fix an ε > 0. Since g ∈ L4/3(X),there is δ > 0 such that ‖gχE‖4/3 ≤ ε for every µ(E) < δ. Since fn

a.e.−−→ 0, by Egoroff’stheorem, there is F ⊆ X such that µ(FC) < δ and fn ⇒ 0 on F . So there is N > 0 suchthat ‖fn‖∞ ≤ ε for every n > N . Thus,∣∣∣∣ˆ

X

fn(x)g(x) dµ

∣∣∣∣ ≤ ˆF

∣∣fn(x)∣∣∣∣g(x)∣∣+ ˆFC

∣∣fn(x)∣∣∣∣g(x)∣∣≤ εˆX

∣∣g(x)∣∣ dµ+ ‖fn‖4 · ‖gχFC‖4/3≤ εˆX

∣∣g(x)∣∣ dµ+ ε.Since X has finite measure, we have g ∈ L1(X). So the first term goes to 0 as ε→ 0.

For the case X has infinite measure, since µ{g 6= 0} is σ-finite, we can assume X is σ-finite. Let X =

⊔∞j=1Xj, where each Xj has finite measures. By Holder’s inequality,∣∣∣∣∣

ˆX

fn(x)g(x)−ˆ⋃Nj=1Xj

fn(x)g(x)

∣∣∣∣∣ ≤ ‖fn‖4 · ‖g · χ⋃∞j=N Xj‖4/3 ≤ ‖g · χ⋃∞j=N Xj‖4/3.The last term goes to 0 uniformly in n as N goes to ∞. Therefore,

limN→∞

ˆ⋃Nj=1Xj

fn(x)g(x) =

ˆX

fn(x)g(x) uniformly in n as N →∞.

Thus, we can interchange the limits

limn→∞

ˆX

fn(x)g(x) = limn→∞

limN→∞

ˆ⋃Nj=1Xj

fn(x)g(x)

= limN→∞

limn→∞

ˆ⋃Nj=1Xj

fn(x)g(x)

= limN→∞

0 = 0.

8 Spring 2013

Problem 1. For each of (a) and (b) either prove or give a counter-example. Suppose thatfor every n ∈ N, fn : X → [0, 1] is a µ-measurable function, and

limn→∞

ˆX

fn dµ = 0.

Then

28

(a) fn → 0 in measure.

(b) for µ-almost all x ∈ X, limn→∞ fn(x) = 0.

Proof. (a) This is true since fn ≥ 0. For any η > 0, by Chebyshev’s inequality

µ{|fn| > η} ≤´fn dµ

η−→ 0 as n→∞.

(b) This is false. Typewriter is a counter-example.

Problem 2. Let fn : R→ R be Borel measurable for every n ∈ N. Define E to be the set ofpoints in R such that limn→∞ fn(x) exists and is finite. Show that E is a Borel measurableset.

Proof. Since sequence in R converges if and only if it is Cauchy, we can write

E =∞⋂k=1

∞⋃N=1

⋂m,n≥N

{x ∈ R :

∣∣fn(x)− fm(x)∣∣ < 1k

}

=∞⋂k=1

∞⋃N=1

⋂m,n≥N

(fn − fm

)−1(−1k,

1

k

).

Because fn − fm is Borel measurable,(fn − fm

)−1(−1k,

1

k

)is a Borel measurable set. It

follows that E is a Borel measurable set.

Problem 3. Does there exist a nowhere dense subset of [0, 1]2 ⊂ R2

(a) of Lebesgue measure greater than 9/10?

(b) of Lebesgue measure 1 ?

If yes, construct such a set, if no, prove why not.

Proof. (a) First, we construct a fat Cantor set C ⊂ [0, 1]. Suppose we remove the middlethird of length 0 < a < 1

2. Then the total length removed is

∞∑n=0

2nan+1 = a∞∑n=0

(2a)n =a

1− 2a.

29

Thus, we want a such that

1− a1− 2a

>9

10.

So if we choose any 0 < a < 112

, then we will get m1(C) >910

. Let E = C × [0, 1]then m2(E) = m1(C) >

910

. It is clear that E is compact (since C is compact) and Edoes not contains any ball (if so, then C also contains an interval �). Therefore, E isnowhere dense.

(b) Suppose we can construct a set E nowhere dense and has full measure. Then E also hasfull measure. Thus E

chas zero measure. Unless E

cis empty, since it is an open set, it

contains some ball which has positive measure. However, Ec

= ∅ implies E = [0, 1]2.This is a contradiction since E is nowhere dense (�).

Problem 4. Let λ ∈ (0, 1), and f ∈ L1[0, 1]. Show that the integral

F (x) =

ˆ x0

1

(x− t)λf(t) dt.

exists a.e. x ∈ [0, 1] and that F ∈ L1[0, 1].

Proof. We only need to show F ∈ L1[0, 1], then F has to be finite (exists) for a.e. x ∈ [0, 1].By Tonelli’s theorem for non-negative integrand,

ˆ 10

∣∣F (x)∣∣ dx = ˆ 10

ˆ 10

1

(x− t)λ∣∣f(t)∣∣χ(0,x)(t) dt dx

=

ˆ 10

ˆ 10

1

(x− t)λ∣∣f(t)∣∣χ(t,∞)(x) dx dt

=

ˆ 10

ˆ 1t

1

(x− t)λ∣∣f(t)∣∣ dx dt

=

ˆ 10

∣∣f(t)∣∣ [(x− t)1−λ1− λ

]1t

dt

=

ˆ 10

∣∣f(t)∣∣ · (1− t)1−λ1− λ

dt

≤ 11− λ

ˆ 10

∣∣f(t)∣∣ dt < +∞.The last inequality holds since 0 < λ < 1, so 0 < (1− t)1−λ < 1.

30

Problem 5. Show that for g ∈ L3[0, 1], f ∈ L3/2[0, 1]ˆ 1

0

cos(2πnx)g(x)f(x) dx −→ 0.

Proof. By Holder’s inequality, we know that fg ∈ L1[0, 1]. Then the conclusion followsdirectly from (3)

ˆRf(x)g(x)χ[0,1] cos(2πnx) dx −→

ˆRf(x)g(x)χ[0,1] dx

ˆ 10

cos(2πx) dx = 0.

Problem 6. Assume that f, f ′ ∈ L1(R). Let

g(x) =∞∑k=0

|f(x+ k)|.

Show that g ∈ L∞(I) for any bounded interval I. (Hint: first prove for intervals with length≤ 1

2).

Proof. This is not true when f is not absolutely continuous. Let f(x) =∑∞

n=0 n1/2χ( 1n ,

1n+1)

on R; that is, the graph of f consists of horizontal segments below the curve y = x−1/2 in[0, 1]. Then f is clearly not continuous. It has zero derivative a.e. It is integrable because

ˆRf ≤ˆ

[0,1]

x−1/2 dx < +∞.

Finally, for any x ∈ (0, 1), we have g(x) = f(x), which is not bounded a.e.

Therefore, we will assume f is absolutely continuous.

First, we show that g is finite almost everywhere. Let α be a real number, we have

∞∑k=0

ˆ α+1α

∣∣f(x+ k)∣∣ = ∞∑k=0

ˆ α+1+kα+k

∣∣f ∣∣ = ˆ(α,∞)

∣∣f ∣∣ < +∞.Switching the sum and integral, we see that g ∈ L1(α, α+1). Since this is true for all α ∈ R,we can conclude g ∈ L1loc(R), thus g is finite a.e. on R.

We prove the conclusion holds for any interval J with length ≤ 12. There exists x0 ∈ J

31

such that g(x0) x0, we have

|g(x)− g(x0)| =

∣∣∣∣∣∞∑k=0

∣∣f(x+ k)∣∣− ∣∣f(x0 + k)∣∣∣∣∣∣∣

≤∞∑k=0

∣∣f(x+ k)− f(x0 + k)∣∣=∞∑k=0

∣∣∣∣ˆ x+kx0+k

f ′∣∣∣∣

≤∞∑k=0

ˆ x+kx0+k

∣∣f ′∣∣ ≤ ˆR

∣∣f ′∣∣.The last inequality is obtained because the pairs (x0+k, x+k) are pairwise disjoint. Similarly,we get the same inequality for a.e. x < x0. Therefore, g(x) ≤ g(x0) +

´R

∣∣f ′∣∣ < ∞ in J .This shows that g ∈ L∞(J). Since any bounded interval I is the finite union of intervals oflength ≤ 1

2, it is clear that g ∈ L∞(I).

9 Fall 2012

Problem 1. Find

(a) limn→∞

ˆ ∞0

e−x2+3x

n2−2x dx

(b) limn→∞

ˆ π0

x2 cos3x

ndx.

Proof. (a) The limit is´∞

0e−2x dx by Monotone convergence theorem.

(b) The limit is´ π

0x2 dx by Dominated convergence theorem.

Problem 2. Find a sequence of pointwise convergent measurable functions fn : [0, 1] → Rsuch that fn doesn’t converge uniformly on any set X with m(X) = 1.

Proof. Let fn = nχ[0, 1n ], then fn

a.e.−−→ 0. If f ⇒ 0 on X, then

1 =

ˆX

fn −→ˆX

0 = 0 (contradiction �)

32

Problem 3. Suppose f is Borel measurable andˆ 1

0

x−1/2∣∣f(x)∣∣3 dx 0. There exists δ > 0 such that´ δ

0x−1/2

∣∣f(x)∣∣3 dx ≤ ε. For every 0 < t < δ,t−5/6

ˆ t0

∣∣f(x)∣∣ dx = t−5/6 ˆ 10

(x−1/6

∣∣f(x)∣∣χ(0,δ)) (x1/6χ(0,t)) dx≤ t−5/6

(ˆ δ0

x−1/2∣∣f(x)∣∣3 dx)1/3(ˆ t

0

x1/4)2/3

≤ 45ε1/3 −→ 0 as t→ 0+.

Problem 4. If (X,µ) is a finite measure space, 1 < p1 < p2 0

λp2µ(x : f(x) > λ) 0 λ

p2µ(x : f(x) > λ), then

∞∑n=1

µ{fp1 > n} =∞∑n=1

µ{f > n1/p1}

≤∞∑n=1

M

np2/p1< +∞.

Problem 5. Let x = 0.n1n2 . . . be a decimal representation of x ∈ [0, 1]. Let f(x) = mini ni.Prove that f(x) is measurable and a.e. constant.

33

Proof. It suffices to show that f = 0 a.e. Let A = {x = 0.n1n2 . . .∣∣ nj 6= 0 for all j}

and Ak = {x = 0.n1n2 . . .∣∣ nj 6= 0 for 1 ≤ j ≤ k}. Then A ⊂ Ak for every k. Thus

m(A) ≤ m(Ak) =(

910

)kfor all k. Therefore m(A) = 0, i.e. f = 0 a.e.

Problem 6. Let fn be a sequence of measurable functions on (X,µ) with fn ≥ 0 and´Xfn dµ = 1. Show that

lim supn→+∞

f1nn (x) ≤ 1 for µ a.e. x.

Proof. We have

µ

{lim supn→∞

f 1/nn (x) > 1

}≤ µ

∞⋃m=1

{lim supn→∞

f 1/nn (x) ≥ 1 +1

m

}= µ

∞⋃m=1

∞⋂k=1

⋃n≥k

{f 1/nn (x) ≥ 1 +

1

m

}= lim

m→∞limk→∞

µ⋃n≥k

{fn(x) ≥

(1 +

1

m

)n}

≤ limm→∞

limk→∞

∞∑n≥k

(1 +

1

m

)−n ˆX

fn = 0.

The last inequality is due to Chebyshev’s inequality.

10 Fall 2011

Problem 1. Let a measurable bounded set X ⊂ Rn have the property that every continuousmap f : X → R is uniformly continuous. Show that X is compact.

Proof. Since X is bounded, we only need to show that X is closed. Suppose not, then thereis a sequence xn → x0 such that x0 /∈ X. Consider f(x) = 1‖x−x0‖ , then f is continuous butnot uniformly continuous (contradiction �). Thus, X must be compact.

Problem 2.

(a) For any ε > 0 construct an open set Uε ⊂ (0, 1) so that Uε ⊃ Q ∩ (0, 1) and m(Uε) < ε.Here and later m stands for the Lebesgue measure.

34

(b) Let A =⋃∞n=1 U

c1/n. Find m(A). Show that A

c ∩ (1/2, 5/16) 6= ∅.

Proof. (a) Enumerate all rationals as {q1, q2, . . .}. It is clear that the following Uε satisfiesall the hypotheses.

Uε =∞⋃n=0

(qn −

ε

2n+2, qn +

ε

2n+2

)∩ (0, 1).

(b) 1 ≥ m(A) ≥ m(UC1/n

)> 1− 1

nfor every n. Thus m(A) = 1. Finally, the intersection

is non-empty because AC contains all the rationals in (0, 1).

Problem 3. Show that either a σ-algebra of subsets of a set X is finite, or else it hasuncountably many elements.Hint: if the σ-algebra is not finite, how many pairwise disjoint elements does it necessarilycontain?

Proof. Let Ω be the σ-algebra. Suppose that Ω is infinite. Let us define

Bx =⋂A∈Ωx∈A

A, x ∈ X.

We claim that for x 6= y, then Bx and By are either disjoint or the same. Case 1: x ∈ Bcy.Then Bx ⊆ Bcy, so Bx ∩ By = ∅. Otherwise, if x ∈ By, then Bx ⊆ By. Similarly, we alsoget By ⊆ Bx, so Bx = By. Hence, the claim is proved. Note that we have

{Bx}x∈X ⊆ Ω.

If the family{Bx}x∈X is finite, it is clear that Ω is also finite. Otherwise, we can extract a

countable disjoint subsequence {B1, B2, . . .}. The sigma-algebra generated by this sequencecorresponds the the power set P(N) in the sense that

{Bn1 , Bn2 , . . .} ←→ {n1, n2, . . .}.

The correspondence is clearly onto. It is injective due to that fact that if⊔Ni=1 Bi =

⊔Mj=1Bj,

then M = N and all the Bi’s are equal to Bj’s. So the correspondence is one-to-one. Thusthe sigma-algebra generated by {Bn1 , Bn2 , . . .} has the same cardinality of P(N), henceuncountable. In this case, Ω is uncountable.

Problem 4. Let f : R → R be absolutely continuous and assume that f ′ ∈ L2[0, 1] andthat f(0) = 0. Show that the following limit exists and compute its value.

limx↘0

x−1/2f(x).

35

Proof. Fix an ε > 0. Since f ′ ∈ L2[0, 1], there is δ > 0 such that´ δ

0

∣∣f ′∣∣2 ≤ ε. Since f isabsolutely continuous, for every 0 < x < δ,

∣∣x−1/2f(x)∣∣ = ∣∣∣∣x−1/2 ˆ x0

f ′∣∣∣∣ ≤ x−1/2 ˆ x

0

∣∣f ′∣∣= x−1/2

ˆ δ0

∣∣f ′∣∣ · χ(0,x)≤ x−1/2

(ˆ δ0

∣∣f ′∣∣2)1/2(ˆ δ0

χ(0,x)

)1/2=

(ˆ δ0

χ(0,x)

)1/2≤ ε1/2.

Thus, the limit exists and equals to 0.

Problem 5. Let S ⊂ R be closed and let f ∈ L1[0, 1]. Assume that for all measurableE ⊂ [0, 1] with m(E) > 0 we have 1

m(E)

´Ef ∈ S. Prove that f(x) ∈ S for a.e. x ∈ [0, 1].

Proof. This follows directly from Lebesgue differentiation theorem.

Problem 6. If g is a Lebesgue measurable real-valued function on [0, 1] such that thefunction f(x, y) = 5g(x) − 7g(y) is Lebesgue integrable over the square [0, 1] × [0, 1], showthat g is Lebesgue integrable over [0, 1].

Proof. By Fubini’s theorem: since f is integrable in the product space, for a.e. y ∈ [0, 1],5g(x)− 7g(y) ∈ L1[0, 1] (F), and for a.e. x ∈ [0, 1], 5g(x)− 7g(y) ∈ L1[0, 1]. The later canbe rewritten as for a.e. y ∈ [0, 1], 5g(y)− 7g(x) ∈ L1[0, 1] (FF). Since L1 is a vector space,from (F) and (FF), we see that g(x) ∈ L1[0, 1].

11 Spring 2011

Problem 1. Suppose f and g are real-valued µL-measurable functions on R, such that (1)f is µL-integrable, and (2) g belongs to C0(R). For c > 0 define gc(t) = g(ct). Prove that(a) limc→∞

´R fgc dµL = 0, and (b) limc→0

´R fgc dµL = g(0)

´R f dµL.

36

Proof. (a) Suppose g(x) vanishes outside [a, b]. Fix ε > 0. Since f is integrable, there is

M > 0 such that´ b/ca/c

∣∣f ∣∣ ≤ ε for every c > M . Therefore,∣∣∣∣ˆRf(t)g(ct) dt

∣∣∣∣ ≤ ˆ b/ca/c

∣∣f(t)g(ct)∣∣ dt ≤ ε‖g‖∞.(b) Rewrite gc as sequence gn(t) = g (tan), where an → 0. Then fgn

a.e.−−→ g(0)f , which isintegrable. Therefore, the wanted limit is obtained by Dominated convergence theorem.

Problem 2. Suppose µ and ν are σ-finite measures on a measurable space (X,A), suchthat ν ≤ µ , and ν � µ− ν. Prove that

µ

{x ∈ X : dν

dµ= 1

}= 0.

Proof. By definition of Radon-Nikodym derivative,

ν(E) =

ˆE

dν

dµdµ.

Let E =

{x ∈ X : dν

dµ= 1

}then ν(E) = µ(E). So µ(E)− ν(E) = 0 (we need ν to be finite

here). Since ν � µ− ν, we have ν(E) = 0. Thus µ(E) = ν(E) = 0.

Problem 3. Prove that the Gamma function

Γ(x) =

ˆ(0,∞)

tx−1e−t µL(dt)

is well-defined and continuous for x > 0.

Proof. Since tx−1e−t ≤ tx−1, the integral converges at 0. And for large t, we have tx−1e−t ≤et/2 · e−t = e−t/2, i.e. integral converges at infinity. Thus, the function Γ(x) is well-defined.

Let x ∈ (a, b), by above argument, the integral converges uniformly on (a,b). Since theintegrand is continuous in x and t, we see that Γ is continuous on (a, b). Thus we see thatΓ is continuous at every x > 0.

37

Problem 4. Let (X,A, µ) and (Y,B, ν) be the measure spaces given by:

• X = Y = [0, 1].

• A = B = B[0,1], the Borel σ-algebra of [0, 1].

• µ = µL, and ν is the counting measure.

Consider the product measurable space (X × Y, σ(A × B)), and its subset E = {(x, y) ∈X × Y : x = y} ⊂ X × Y .

(a) Show that E ∈ σ(A× B).

(b) Show that´X

(´YχE dν

)dµ 6=

´Y

(´XχE dµ

)dν.

(c) Explain why Tonelli’s theorem is not applicable.

Proof. (a) The function f : (x, y) 7→ x− y is Borel measurable because it is the differencebetween two Borel measurable functions. Therefore, E = f−1({0}) is Borel measurable.

(b) LHS is 1 while RHS is 0.

(c) Tonelli’s theorem is not applicable since counting measure is not σ-finite.

Problem 5. Suppose f ∈ C1[0, 1] (that is, f is continuous, and continuously differentiableon [0, 1]), f(0) = f(1), and f > f ′ everywhere.

(a) Prove that f > 0 everywhere.

(b) Prove that ˆ(0,1)

f 2

f − f ′dµL ≥

ˆ(0,1)

f dµL.

Proof. (a) Suppose there exists x0 ∈ [0, 1) such that f(x0) ≤ 0. Since f > f ′, f cannot beconstant on any neighborhood of x0. Thus there is δ > 0 such that f < 0 on (x0, x0+δ).We claim that f ′ < 0 on (x0, 1]. Indeed, if there is z = inf{x ∈ (x0, 1] : f ′(x) ≥ 0}, thenf ′(z) ≥ 0 > f(z) (contradiction �). Therefore f < 0 on (x0, 1]. So f(0) = f(1) < 0.By the same argument as before, f ′ < 0 on [0, 1]. This implies f is strictly decreasingon [0, 1] (contradiction to f(0) = f(1)).

38

(b) By Cauchy-Schwarz’s inequality,(ˆ 10

f

)2=

(ˆ 10

f√f − f ′

·√f − f ′

)2≤ˆ 1

0

f 2

f − f ′·ˆ 1

0

f − f ′

=

ˆ 10

f 2

f − f ′·ˆ 1

0

f.

Thus, we get the desired inequality.

Problem 6. Suppose (fn)∞n=1 is a sequence of measurable functions on [0, 1]. For x ∈ [0, 1],

define h(x) = #{n : fn(x) = 0} (the number of indices n for which fn(x) = 0). Assumingthat h

ε > 0 there exists a set E of measure µ(E) ≤ ε such that (fn) converges uniformly to foutside the set E.

Proof. This is a statement Egoroff’s theorem.

Problem 3. Suppose that f ∈ Lp[0, 1] for some p > 2. Prove that g(x) = f(x2) ∈ L1[0, 1].

Proof. By changing of variable,´ 1

0|f(x2)| dx = 1

2

´ 10|f(x)| · x−1/2 dx. Since p > 2, we have

q < 2. So x−1/2 ∈ Lq[0, 1]. Thus, by Holder’s inequality, f · x−1/2 ∈ L1[0, 1]. Therefore,g(x) = f(x2) ∈ L1[0, 1].

Problem 4. Assume that E ⊂ [0, 1] is measurable and for any (a, b) ⊂ [0, 1],

µ(E ∩ [a, b]

)≥ 1

2(b− a).

Show that µ(E) = 1.

Proof. Lebesgue’s differentiation theorem states that

limB→x

1

µ(B)

ˆB

χE = χE(x) for a.e. x ∈ [0, 1].

This can be rewritten as

limB→x

µ(E ∩ B)µ(B)

= χE(x) for a.e. x ∈ [0, 1].

From the hypothesis, we get χE(x) = 1 for a.e. This implies µ(E) = 1.

Problem 5. Let f be a real-valued uniformly continuous function on [0,∞). Show that iff is Lebesgue integrable on [0,∞), then limx→∞ f(x) = 0.

Proof. Suppose limx→∞ f(x) 6= 0, then there is an ε > 0 and an increasing sequence (xn)∞n=1of positive numbers such that xn → +∞ and |f(xn)| ≥ ε for every n ∈ N. There exists η > 0and a subsequence of (xn) (which we rename (xn) for simplicity) such that xn+1 − xn < ηfor every n. Since f is uniformly continuous, there is 0 < δ < η such that

∣∣f(xn + δ)∣∣ ≥ εfor every n ∈ N. Thus

´ xn+δxn

|f | ≥ δε. This is a contradiction to f ∈ L1[0,∞) sinceˆ ∞

0

|f | ≥∞∑n=1

ˆ xn+δxn

|f | ≥∞∑n=1

δε = +∞ (contradiction �).

40

Therefore, we must have limx→∞ f(x) = 0.

Problem 6. Consider the Lebesgue measure space (R,ML, µL) on R . Let f be an extendedreal-valued ML-measurable function on R. For x ∈ R and r > 0 let Br(x) = {y ∈ R :|y − x| < r}.With r > 0 fixed, defne a function g on R by setting

g(x) =

ˆBr(x)

f(y)µL(dy) for x ∈ R.

(a) Suppose f is locally µL-integrable on R. Show that g is a real-valued continuousfunction on R.

(b) Show that if f is µL-integrable on R, then g is uniformly continuous on R.

Proof. (a) Fix x ∈ R. Since f is locally integrable in R, there is δ = δ(x) > 0 such that´E|f | ≤ ε whenever µ(E) < δ. For y sufficiently close to x, we get µ(Br(x)\Br(y)) =

µ(Br(y)\Br(x)) < δ. Therefore,

|g(x)− g(y)| =∣∣∣∣ˆBr(x)

f −ˆBr(y)

f

∣∣∣∣≤ˆBr(x)\Br(y)

|f |+ˆBr(y)\Br(x)

|f |

≤ ε+ ε = 2ε.

(b) Similar to part (a), but if f is integrable on R, we can choose δ uniformly over x ∈ R.It follows that g is indeed uniformly continuous.

13 Spring 2010

Problem 1. Suppose f is Lipschitz continuous in [0, 1], that is,∣∣f(x) − f(y)∣∣ ≤ L|x − y|

for some constant L. Show that

(a) m(f(E)

)= 0 if m(E) = 0.

(b) If E is measurable, then f(E) is also measurable.

Proof. (a) Lemma: If f is Lipschitz with constant L, then m(f(α, β)

)≤ L(β−α). Indeed,

since f is continuous, it maps the interval (α, β) to some interval (f(c), f(d)), wherec, d ∈ [α, β]. By the Lipschitz’s condition,

m(f(α, β)

)= f(d)− f(c) ≤ L|c− d| ≤ L(β − α).

41

Fix an ε > 0. Since E is measurable, there is an open set G =⊔∞j=1(aj, bj) such that

E ⊆ G and m(G) ≤ m(E) + ε = ε. We have

m(f(G)

)=∞∑j=1

m(f(aj, bj)

)≤

∞∑j=1

L(bj − aj) = m(G) ≤ ε.

Let ε→ 0, we have m(f(G)

)= 0. Thus m

(f(E)

)= 0.

(b) Since f maps open interval onto (open) interval, the image of an open interval ismeasurable. Since f is injective, image of intersection is intersection of images. Thusimage of a Gδ set is measurable. For any measurable set E, we can find a Gδ setG ⊇ E such that m(G) = m(E). Then f(G) is measurable. By previous part, wehave m

(f(G\E)

)= 0. Thus f(G\E) is measurable. Finally, f(E) = f(G) \ f(G\E)

is measurable.

Problem 2. Let {qk} be all the rational numbers in [0, 1]. Show that

∞∑k=1

1

k21√|x− qk|

converges a.e. in [0, 1].

Proof. For any x ∈ QC ∩ [0, 1],

∞∑k=1

ˆ 10

1

k21√|x− qk|

dx =∞∑k=1

1

k2

(ˆ qk0

1√qk − x

dx+

ˆ 1qk

1√x− qk

dx

)=∞∑k=1

2

k2(√

qk +√

1− qk)

≤∞∑k=1

2√

2

k2< +∞.

Thus, the series converges a.e. x ∈ [0, 1].

Problem 3. Suppose that f(t) =´ t

0g(s) ds where g(s) is integrable over [0, 1]. Show that

limn→∞

2n−1∑k=1

∣∣∣∣f (k + 12n)− f

(k

2n

)∣∣∣∣2 = 0.

42

Proof. We need to show that

limn→∞

2n−1∑k=0

(ˆ k+12n

k2n

|g|

)2= 0.

Let 0 < ε < 1. Since g is integrable, there is δ > 0 such that´E|g| ≤ ε whenever µ(E) < δ.

Choose N large enough such that for every n > N , we have 12n< δ.

2n−1∑k=0

(ˆ k+12n

k2n

|g|

)2≤

2n−1∑k=0

ε

(ˆ k+12n

k2n

|g|

)= ε

ˆ 10

|g|.

Since ε can be chosen arbitrarily small, the summation goes to 0.

Problem 4. Let f be a real-valued uniformly continuous function on [0,∞). Show that iff is Lebesgue integrable on [0,∞), then limx→∞ f(x) = 0.

Proof. See a proof here.

Problem 5. Let (X,M, µ) be a measure space and let f be an extended real-valued M-measurable function on X such that

´X|f |p dµ 0. Since´X|f |p < ∞, there is δ > 0 such that

´E|f |p < ε whenever

µ(E) < δ. By Chebyshev’s inequality, µ{|f | ≥ λ} ≤´X |f |

p

λp→ 0, so there is M > 0 such

that µ{|f | ≥ λ} ≤ δ for every λ > M . So´{|f |≥λ} |f |

p < ε for every λ > M . Again, by

Chebyshev’s inequality,

λpµ{|f | ≥ λ} ≤ˆ{|f |≥λ}

|f |p < ε for every λ > M.

Therefore, we have limλ→∞ λpµ{x ∈ X : |f | ≥ λ} = 0.

Problem 6. Consider the Lebesgue measure space (R,ML, µL) on R. Let f be a µL-integrable extended real-valued ML-measurable function on R. Show that

limh→0

ˆR

∣∣f(x+ h)− f(x)∣∣µL(dx) = 0.43

Proof. The statement holds trivially when f is continuous with compact support. Fix anε > 0. For any f integrable, we can find a continuous function g with compact support suchthat ‖f − g‖1 ≤ ε. Then we haveˆR

∣∣f(x+ h)− f(x)∣∣ ≤ ˆR

∣∣f(x+ h)− g(x+ h)∣∣+ ∣∣g(x+ h)− g(x)∣∣+ ∣∣g(x)− f(x)∣∣= 2

ˆR

∣∣f(x)− g(x)∣∣+ ˆR

∣∣g(x+ h)− g(x)∣∣The first integral is bounded by ε and the second integral goes to 0. Hence we are done.

14 Spring 2009

Problem 1. Let (X,A, µ) be a measure space. Let {fn : n ∈ N} and f be real-valuedA-measurable functions on a set D ∈ A. Suppose fn

µ−→ f on D, that is, {fn : n ∈ N}converges to f in measure µ on D. Let F be a real-valued uniformly continuous function onR. Show that F ◦ fn

µ−→ F ◦ f on D.

Proof. Fix ε > 0. Since F is uniformly continuous, there is δ > 0 such that |F (x)−F (y)| ≤ εwhenever |x− y| ≤ δ. Therefore {

∣∣F ◦ fn − F ◦ f ∣∣ > ε} ⊆ {∣∣fn − f ∣∣ > δ}. Since fn µ−→ f , wehave

µ{∣∣F ◦ fn − F ◦ f ∣∣ > ε} ≤ µ{∣∣fn − f ∣∣ > δ} −→ 0.

Problem 2.

(a) Let f ∈ Lp[0, 10], p ≥ 1. Prove that limt↘1(t− 1)1p−1 ´ t

1f(s) ds = 0.

(b) Suppose´∞

0x−2|f |5 dx 0. Choose δ > 1 such that ‖fχ(1,δ)‖p ≤ ε. For any t ∈ (1, δ), we have

(t− 1)1p−1ˆ t

1

f(s) ds = (t− 1)1p−1ˆ 10

0

(fχ(1,δ)

)(χ(1,t)

)ds

≤ (t− 1)1p−1‖fχ(1,δ)‖p · ‖χ(1,t)‖q

t = ‖fχ(1,δ)‖p ≤ ε.

(b) See a proof here.

44

Problem 3. Let fn(x) =n2χ[− 1n ,

1n ]

. Prove that for g ∈ L1(R),ˆ ∣∣∣∣ˆ fn(y − x)g(x) dx− g(y)∣∣∣∣ dy → 0 as n→∞.

Proof. We want to show that

limn→∞

ˆR

∣∣∣∣∣n2ˆ 1

n

− 1n

g(y − x)− g(y) dx

∣∣∣∣∣ dy = 0.First, let’s show this holds for g continuous with compact support. In this case, g(y−x)−g(x)also has a compact support K for x ∈

[− 1n, 1n

]. Thus we only need to show that

limn→∞

ˆK

∣∣∣∣∣n2ˆ 1

n

− 1n

g(y − x)− g(y) dx

∣∣∣∣∣ dy = 0.By Lebesgue differentiation theorem,

n

2

ˆ 1n

− 1n

g(y − x) dx = n2

ˆ y+ 1n

y− 1n

g(x) dxa.e.−−→ g(y).

Thus the integrand goes to zero and uniformly bounded by 2‖g‖∞. Since µ(K) < +∞, thelimit goes to 0 by Bounded convergence theorem.

Next, for any integrable g, we can approximate g by a continuous function ϕ with com-pact support, i.e. ‖g − ϕ‖1 < ε. Then we have

ˆR

∣∣∣∣∣n2ˆ 1

n

− 1n

g(y − x)− g(y) dx

∣∣∣∣∣ dy≤ˆR

∣∣∣∣∣n2ˆ 1

n

− 1n

ϕ(y − x)− ϕ(y) dx

∣∣∣∣∣ dy +ˆR

∣∣∣∣∣n2ˆ 1

n

− 1n

ϕ(y − x)− g(y − x) dx

∣∣∣∣∣ dy+

ˆR

∣∣∣∣∣n2ˆ 1

n

− 1n

ϕ(y)− g(y) dx

∣∣∣∣∣ dy.The first integral goes to 0 by previous case. The second and third integrals are bounded by‖ϕ− g‖1 ≤ ε. Hence, the limit goes to 0.

Problem 4. Suppose f is a bounded nonnegative function on (X,µ) with µ(X) =∞. Showthat f is integrable if and only if

∞∑n=1

1

2nµ{x ∈ X : f(x) > 1

2n}

Proof. (⇐) This follows directly from the inequality∞∑n=1

χ{f> 12n} ≤ 2f.

(⇒) Suppose the series converges and f ≤M , thenˆX

f dµ =∞∑n=1

ˆ{ M

2n−1≥f>M

2n}f dµ

≤∞∑n=1

M

2n−1µ

{M

2n−1≥ f > M

2n

}=∞∑n=1

M

2n−1

(µ

{f >

M

2n

}− µ

{f >

M

2n−1

})< +∞.

Problem 5. Let f be an element and {fn : n ∈ N} be a sequence in Lp(X,A, µ) wherep ∈ [1,∞) such that limn→∞ ‖fn − f‖p = 0. Show that for every ε > 0, there exists δ > 0such that for all n ∈ N we haveˆ

E

|fn|p dµ < ε for every E ∈ A such that µ(E) < δ.

Proof. By Minkowski’s inequality,

‖fn‖p ≤ ‖f − fn‖p + ‖f‖p.

There is N > 0 such that ‖f − fn‖ is small for all n > N . We can find δ0, δ1, . . . , δN for theintegrability of f, f1, . . . , fN . Choose δ = min

(δ0, δ1, . . . , δN

). This choice of δ satisfies the

conclusion.

Problem 6. Let (X,Σ, µ) be a measure space with µ(X)

15 Fall 2008

Problem 1. Let f be Lebesgue integrable function of real line. Then

limn→∞

ˆ ∞−∞

f(x) sin(nx) dx = 0.

Proof. See Fall 2015 Problem 1.

Problem 2. Let g be an absolutely continuous monotone function on [0, 1]. Prove that, ifE ⊂ [0, 1] has Lebesgue measure zero, then the set g(E) also has Lebesgue measure zero.

Proof. Fix ε > 0. Let δ respond to the ε challenge regarding the criterion for the absolutecontinuity of g. Since m(E) = 0, E ⊆

⊔∞i=1(xi, yi), where

∑∞i=1

∣∣yi − xi∣∣ < δ. Theng(E) ⊆ g (

⊔∞i=1(xi, yi)) =

⊔∞i=1 g(xi, yi). Therefore,

m(g(E)

)≤

∞∑i=1

∣∣g(yi)− g(xi)∣∣ < ε,where the first inequality is obtained by the monotonicity of g. Since ε can be arbitrarilysmall, we get m(g(E)) = 0.

Problem 3. Let ν be a finite Borel measure on the real line, and set F (x) = ν(−∞, x].Prove that ν is absolutely continuous with respect to Lebesgue measure µL if and only if Fis an absolute continuous function. In this case, show that the Radon-Nikodym derivative isthe derivative of F , that is, dν

dµL= F ′ a.e.

Proof. The problem follows directly from this lemma: if ν is a finite measure space, thenν � µ if and only if for each ε > 0, there exists δ > 0 such that µ(E) < δ implies ν(E) < εfor every measurable set E. Indeed, suppose the ε − δ condition does not hold, then thereis ε0 > 0 and a sequence of measurable sets (En) such that µ(En) <

12n

and ν(En) ≥ ε0.Let An =

⋃∞k=nEk, then ν(Ak) ≥ ε0 and µ(An) ≤

12n−1

. Finally, let A∞ =⋂∞n=1An, then

µ(A∞) = 0 by continuity of measure. On the other hand, ν(A1) ≤ ν(X) < ∞, so bycontinuity of measure, ν(A∞) = lim ν(An) ≥ ε0 (contradiction � to ν � µ).

Problem 4. Let µ be a measure and λ, λ1, λ2 be signed measures on a measurable space(X,A). Prove

47

(a) If λ ⊥ µ and λ� µ, then λ = 0.

(b) If λ1 ⊥ µ and λ2 ⊥ µ, then, if we set λ = c1λ1 + c2λ2 with c1, c2 real numbers suchthat λ is a signed measure, we have λ ⊥ µ.

(c) If λ1 � µ and λ2 � µ, then, if we set λ = c1λ1 + c2λ2 with c1, c2 real numbers suchthat λ is a signed measure, we have λ� µ.

Proof. (a) There exists A such that λ(A) = µ(Ac) = 0. Since λ� µ, we have λ(Ac) = 0.Thus λ(X) = λ(A) + λ(Ac) = 0.

(b) By definition, there exists A and B such that |λ1(A)| = µ(Ac) = 0 and |λ2(B)| =µ(Bc) = 0. Then |λ| ≤ (c1||λ1|+ |c2||λ2|)(A ∩ B) = 0, and µ(Ac ∪ Bc) = 0.

(c) µ(A) = 0 implies λ1(A) = λ2(A) = 0. Hence, λ(A) = 0.

Problem 5. Let (X,A, µ) be a measure space. Let {fn}n∈N and f be extended real-valuedA-measurable functions on a set D ∈ A sucht that limn→∞ fn = f on D. Then for everyα ∈ R, we have:

(a) µ{D : f > α} ≤ lim infn→∞ µ{D : fn ≥ α}

(b) µ{D : f < α} ≤ lim infn→∞ µ{D : fn ≤ α}

Proof. (a) If f(x) > α, then fn(x) ≥ α for n > N . Thus χ{f>α} ≤ χ{fn≥α} for n > N .Take integral both sides, we get the conclusion.

(b) Similar to (a).

Problem 6. Let (X,A, µ) be a measure space. Let {fn}n∈N and f be extended real-valuedA-measurable functions on a set D ∈ A with µ(D) < ∞. Show that fn converges to 0 inmeasure on D if and only if

limn→∞

ˆD

|fn|1 + |fn|

dµ = 0.

Proof. (⇐) We want to show µ {|fn| < ε} ≤ η for large n. Indeed, there is N > 0 such thatfor all n > N , we have

η

(ε

ε+ 1

)≥ˆ{|fn|>ε}

|fn||fn|+ 1

dµ ≥ εε+ 1

µ {|fn| < ε} .

48

(⇒) Now suppose fnµ−→ 0. Then there is N > 0 such that µ

{|fn| > ε2µ(D)

}≤ ε

2for all

n > N . Then, for every n > N , we have

ˆD

|fn|1 + |fn|

=

ˆ{|fn|> ε2µ(D)}

|fn|1 + |fn|

+

ˆ{|fn|≤ ε2µ(D)}

|fn|1 + |fn|

≤ 1 · µ{|fn| >

ε

2µ(D)

}+

ε

2µ(D)· µ(D)

= ε −→ 0.

16 Fall 2007

Problem 1. Let 1 ≤ p < q < ∞. Which of these are true or false? (all in Lebesguemeasures)

(a) Lp(R) ⊂ Lq(R)

(b) Lq(R) ⊂ Lp(R)

(c) Lp[2, 5] ⊂ Lq[2, 5]

(d) Lq[2, 5] ⊂ Lp[2, 5].

Proof. (a) False. Any f(x) = x−αχ(0,1), where1q< α < 1

p, is a counter-example.

(b) False. Any f(x) = x−αχ(1,∞), where1q< α < 1

p, is a counter-example.

(c) False. f(x) = x−1qχ(0,1) is a counter-example.

(d) True by Holder’s inequality.

Problem 2. Does there exist a Lebesgue measurable subset A of R such that for everyinterval (a, b), we have µL(A ∩ (a, b)) = b−a2 ? Either construct the set or prove it does notexist.

Proof. By Lebesgue differentiation theorem,

limb−a→0

µ(A ∩ (a, b)

)b− a

a.e.−−→ χA.

LHS is always 1/2 while RHS is either 0 or 1 (contradiction �).

49

Problem 3. Let f be a real valued measurable function on the finite measure space (X,Σ, µ).Prove that the function F (x, y) = f(x) − 5f(y) + 4 is measurable in the product measurespace (X ×X, σ(Σ× Σ), µ× µ), and that F is integrable if and only if f is integrable.

Proof. F is a sum of measurable functions, so it is measurable. If F is integrable, then F (x, ·)is integrable for a.e. x ∈ X by Fubini’s theorem. It is clear that if F (x, ·) is integrable forone x, then f is integrable (since the measure space is finite). To show the other direction,we use Tonelli’s theorem:ˆ

X×X

∣∣F (x, y)∣∣ dµ dµ ≤ ˆX

ˆX

|f(x)|+ 5|f(y)|+ 4 dµ dµ

=

ˆX

‖f‖+ 5(|f(y)|+ 4

)µ(X) dµ

= 6‖f‖µ(X) + 4µ(X)2 < +∞.

Problem 4. Let f ∈ L3/2([0, 5],ML, µL

). Prove that

limt↘0

1

t13

ˆ t0

f(s) ds = 0.

Proof. Fix an ε > 0. Since f ∈ L3/2[0, 5], there exists δ > 0 such that ‖fχ(0,δ)‖3/2 < ε. Forevery t ∈ (0, δ), by Holder’s inequality:

t−1/3ˆ t

0

|f | ds ≤ t−1/3‖fχ(0,δ)‖3/2 · ‖χ(0,t)‖3 < ε.

Thus, the limit is zero.

Problem 5. Let {fn : n ∈ N} be a sequence of real-valued functions and f be a real-valuedfunction on [a, b] such that limn→∞ fn(x) = f(x) for x ∈ [a, b]. Let V ba f be the total variationof f on [a, b]. Show that

V ba f ≤ lim infn→∞

V ba fn.

Proof. For every partition P , we see that V (fn, P )→ V (f, P ). If V ba f = +∞, then for anylarge M > 0, there is a partition P0 such that V (f, P0) > M . Thus V (fn, P ) > M − ε forlarge n. This show that lim infn→∞ V

ba fn = limn→∞ V

ba fn = +∞.

Now suppose V ba f < +∞. Assume that V ba f > lim infn→∞ V ba fn, then there is an ε > 0and a subsequence V ba fnk + 2ε < V

ba f . By definition of total variation, there is a partition

P ∗ such that V (f, P ∗) > V ba f − ε. Therefore, V (f, P ∗) > V ba fnk + ε ≥ V (fnk , P ) + ε. Thiscontradicts to the fact that V (fnk , P

∗)→ V (f, P ∗).

50

Problem 6. Let f ∈ L1(R,ML, µL). With h > 0 fixed, define a function φh on R by setting

φh(x) =1

2h

ˆ x+hx−h

f(t)µL(dt) for x ∈ R.

(a) Show that φh is ML-measurable on R.

(b) Show that φh ∈ L1(R,ML, µL) and ‖φh‖1 ≤ ‖f‖1.

Proof. See Fall 2015 Problem 4.

17 Fall 2006

Problem 1. Given a measure space (X,A, µ). let f be a non-negative extended real-valuedA-measurable function on a set D ∈ A with µ(D) < ∞. Let Dn = {x ∈ D : f(x) ≥ n} forn ∈ N. Show that

´Df du

(a) The set {f 6= 0} is of σ-finite measure.

(b) If f ≥ 0, then f = limn φn pointwise for some increasing sequence of simple functions,each of which vanishes outside a set of finite measure.

(c) For every ε > 0, there is a simple function φ such that

ˆX

∣∣f − φ∣∣ < ε.Proof. Typical approximate theorems.

Problem 4. Let (fn) and (gn) be sequences of measurable functions on a measurable set E,and let f and g are measurable functions on E. Suppose fn → f and gn → g in measure. Isit true that f 3n + gn → f 3 + g in measure if (a) µ(E) = 2; (b) µ(E) =∞.

Proof. For the case of finite measure, it is enough to show that if fnµ−→ f and gn

µ−→ g, thenfngn

µ−→ fg. We use the subsequence principle: a sequence (xn) converges to x in R if andonly if every subsequence (xnk) has a further subsequence (xnkj ) that converges to x. Back

to the problem, since fnµ−→ f , there is a subsequence fnk

a.e−→ f . Since gnkµ−→ g, there is

a subsubsequence gnkja.e.−−→ g. Thus, we have fnkj gnkj

a.e.−−→ fg. Since the measure is finite,fnkj gnkj

µ−→ fg. Hence the claim is proved. For infinite measure, consider a counter-example:gn = 0 and fn = x+

1n.

Problem 5. Suppose f ∈ L1[0, 1]. Let F (x) =´ x

0f(t) dt. Let φ be a Lipschitz function.

Show that there exists g ∈ L1[0, 1] such that

φ(F (x)

)=

ˆ x0

g(t) dt.

Proof. It is enough to show φ ◦ F is an absolutely continuous function. Fix an ε > 0. SinceF is absolutely continuous, there exists δ > 0 corresponding to the ε-challenge for F . Then,for any

∑∞n=1 |yn − xn| < δ, we have

∞∑n=1

∣∣φ ◦ F (xn)− φ ◦ F (yn)∣∣ ≤ ∞∑n=1

M∣∣F (xn)− F (yn)∣∣ < Mε.

Therefore, φ ◦ F is absolutely continuous.

52

Problem 6.

(a) Suppose f ∈ Lp[0, 1], p > 1. Show that limt→0 t−1+1p´ t

0f(x) dx = 0.

(b) Suppose´ 1

0x−1|f |3 dx

Show that fn converges to f a.e. on D.

Proof. We have∑∞

n=1

´D|fn−f |p <

∑∞n=1 εn < +∞. So

´D

∑∞n=1 |fn−f |p

Proof. Since {f > 0} is σ-finite, we can assume µ is σ-finite. By Tonelli’s theorem,ˆ ∞

0

b(t) dt =

ˆ ∞0

ˆX

χ{f(x)≥t}(x) dµ dt

=

ˆX

ˆ ∞0

χ{t≤f(x)}(t) dt dµ

=

ˆX

f(x) dµ.

55

Fall 2016Spring 2016Fall 2015Spring 2015Fall 2014Spring 2014Fall 2013Spring 2013Fall 2012Fall 2011Spring 2011Fall 2010Spring 2010Spring 2009Fall 2008Fall 2007Fall 2006Spring 2006