Combine Solution

8/6/2019 Combine Solution

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Real Analysis Homework 10, due 2007-11-28 in class

1. (10 points) Given t he funct ion

f(x, y) = x2 y

2

(x2 + y2)2, (x, y) I = (0, 1) (0, 1)

comput e t he following it erated integrals (hint: use t rigonomet ric subst it uti on) :

Z1

0

Z1

0

f(x, y) dx

dy and

Z1

0

Z1

0

f (x, y) dy

dx.

Is f(x, y) L (I) or not? Give your reasons.

Solution:

For fixed x we haveZ

1

0

f(x, y) dy =

Z1

0

x2 y2

(x2 + y2)2

dy =

Zt an1

1

x

0

x2 x2 t an2

(x2 + x2 t an2 )2

d (x t an )

=1

x

Zt an1

1

x

0

cos2d =1

x sin

t an1

1

x

cos

t an1

1

x

=

1

1 + x2.

Hence Z1

0

Z1

0

x2 y2

(x2 + y2)2dy

dx =

4.

Similarly (by symmet ry) we have

Z 1

0

x2 y2

(x2 + y2)2

dx =1

1 + y2

and so Z1

0

Z1

0

x2 y2

(x2 + y2)2

dx

dy =

4.

Since t he two it erated integrals are different, by Fubini theorem, f(x, y) 6 L (I) .

2. (10 points) Do Exercise 1 in p. 96.

Solution:

(a) . We set Ex = {y R : (x, y) E}and Ey = {x R : (x, y) E} . By Tonell i s T heorem, wehave ZZ

E

E (x, y) dxdy =

Z

R

Z

Ex

E (x, y) dy

dx =

Z

R

Z

Ey

E (x, y) dx

!

dy.

By assumpt ion we knowREx

E (x, y) dy = 0 a.e. in x R and soRR

EE (x, y) dxdy = |E| = 0. By

Tonell i s T heorem again, we have |Ey| = 0 a.e. in y R.

(b) . Let E =

(x, y) R2 : f(x, y) =

. It is a measurable set i nR2. Set Ex = {y R : f(x, y) = }{x R : f (x, y) = }. By (a) we have

|E| = 0 in R2 |Ex| = 0 i n R for a.e. x R

|Ey| = 0 in R for a.e. y R.


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Solution:

Let h1 (x, y) = f (x) . As a function on R2n, it is measurable since f (x) : Rn R

S{} is

measurable. More precisely, for any a R we have

(x, y) R2n : f(x, y) > a

= {x Rn : f(x) > a} Rn

where by repeat ed applicat ion of Lemma 5.2, we know t hat t he RHS i s a measurable set in

R2n. Simil arly, t he funct ion h2 (x, y) = g (y) is also a measurable funct ion onR

2n. Then by Theorem

4.10, we know t hat

h1 (x, y) h2 (x, y) = f(x) g (y) : R2n R

[{}

is also a measurable funct ion on R2n.

Given E1 Rn, E2 R

n, both are measurable in Rn. By E1 (x) E2 (y) = E1E2 (x, y) , weknow that E1E2 (x, y) 0 is a measurable funct ion on R

2n. Hence t he set E1 E2 is measurablein R2n.

By Tonell is T heorem

|E1 E2| =

ZZ

E1E2

E1E2 (x, y) dxdy =

Z

Rn

Z

Ex

E1E2 (x, y) dy

dx

=

Z

E1

Z

E2

E1E2 (x, y) dy

dx =

Z

E1

Z

E2

[E1 (x) E2 (y)] dy

dx

=

Z

E1

E1 (x) dx

Z

E2

E2 (y) dy = |E1| |E2| .


Solution:

Wefirst know that f(x)f(y) i s measurable on (0, 1)(0, 1) . By Fubini Theorem, if F (x, y) =f(x)f(y) is integrable on (0, 1) (0, 1) , then for a.e. y (0, 1) , F (x, y) L1 (0, 1) ( as a funct ionof x). Hence f(x) L1 (0, 1) .


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Solut ion: Choose a = x, b = x and int egrate over [0, 1] t o get

Z10

Z10 |f(t + x) f(t x)| dtdx c.

Hence t he funct ion F (t, x) = |f(t + x) f(t x)| is integrable on E = [0, 1] [0, 1] . Consider thelinear transformation = t + x, = t x. We haveZZ

E

F (t, x) dtdx =1

2

ZZE

F

+

2,

2

dd =

1

2

ZZE

|f() f()| dd

where E is t he diamond-shaped region i n (, ) space wit h vert ices (0, 0) , (1, 1) , (1,1) , (2, 0) . Inpart icular we know t hat |f() f()| L (E) . By Fubini T heorem, t here exist s 0

1

2, 32

such

t hat as a funct ion of we have

|f(0)

f()|

L E0 , E0 = { : (0, )

E} , E0 > 1.By |f()| |f(0) f()| + |f(0)| , we see t hat |f()| L

E0

. Since f() is periodic with

period 1 andE0

> 1, we know that f L (0, 1) .


Solut ion: By definit ion, we have (assume f L1 (R))

f(x) =

Z

f(t) cosxtdt iZ

f(t) sin xtdt, x R.

Weclaim that if f, g L1 (R) , then thefunction f(t y) g (y) L1 R2 (a a funct ion of (t, y)). Tosee t his, not e t hat

[f(t y) g (y)]+ |f(t y)| |g (y)| , (t, y) R2and by Tonelli s Theorem we haveZZ

R2

[f(t y) g (y)]+ dtdy ZZ

R2

|f(t y)| |g (y)| dtdy =ZR

|g (y)| dy

ZR

|f()| d < .

SimilarlyRR

R2[f(t y) g (y)] dtdy < . Hence t he claim is t rue. Now

\(f g) (x) =Z

(f g) (t) cosxtdt iZ

(f g) (t) sin xtdt

= Z

Z

f(t

y) g (y) dy cosxtdt iZ

Z

f(t

y) g (y) dy sin xtdt= I1 + iI2

where by Fubini T heorem

I1 =

Z

g (y)

Z

f(t y) cosxtdt

dy =

Z

g (y)

Z

f() cosx (y + ) d

dy

=

Z

g (y)

Z

f() [cosxy cosx sin xy sin x] d

dy

=

Z

f() cosxd

Z

g (y) cosxydy

Z

f() sin xd

Z

g (y) sin xydy

= Reh

f(x) g (x)i

.


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Simil arl y we have

I2 = Z

g (y)

Z

f(t y) sin xtdt

dy

= Z

g (y)

Z

f() [sin xy cosx + cosxy sin x] d

dy

= Z

f() cosxd

Z

g (y) sin xydyZ

f() sin xd

Z

g (y) cosxydy

= Imh

f(x) g (x)i

.

We conclude t he identi ty\(f g) (x) = f(x) g (x) , x R.


Solut ion: Let Vn (r) bethevolumeof theball in Rn with radius r > 0. For convenience, denot e

Vn (1) = Vn. We have

Lem m a 0.1 There holds the formula

Vn =

Z1

1

Vn1

p1 2

d.

P roof. Let S =

(x1, , xn) : x21

+ + x2n 1

. Then

Vn =

ZS

dx1 dxn =

Z1

1

"ZExn

dx1 dxn1

#dxn =

Z1

1

|Exn | dxn

where

Exn = {(x1, , xn

1) : (x1, , xn

1, xn) S}=

(x1, , xn1) : x21 + + x

2n1 1 x2n

and so

Vn =

Z1

1

Vn1

p1 x2n

dxn.

Rem ark 0.2 Similarly we have

Vn (r) =

Zrr

Vn1

pr2 2

d for any r > 0. (0.1)

Lem m a 0.3 We haveVn (r) = r

nVn for any r > 0. (0.2)

P roof. We can prove (0.2) using (0.1) and inducti on. Assume (0.2) holds for all dimensions less

than or equal to n 1. Then

Vn (r) =

Zrr

Vn1

pr2 2

d =

Zrr

Vn1

r

s1

r

2d

=

Z1

1

Vn1

rp

1 2

rd =

Z1

1

rn1Vn1

p1 2

rd (by induction hypothesis)

= rnZ11

Vn1p

1 2 d = rnVn.


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By t he above, we have

Vn =

Z1

1

Vn1

p1 2

d =

Z1

1

1 2n12 Vn1d

= 2Vn1 Z1

0

1

2n12d.

T he proof is done.


Solut ion: We already know that (from calculus)Z

ex2

dx =

.

By Tonell i s T heorem we can easil y getZRn

e|x |2

dx =

n

.


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1. (10 point s) Do Exercise 1 in p. 123.

Solution: f is defined and measurable on Rn. Assume |f| > 0 on a set E with |E| > 0. Weknow t hat t here exist s N N such that

|f| > 1N

> 0. Denote the set

|f| > 1

N

by EN. We

have 1NEN (x) |f(x)| on EN (and on Rn also) so

1

NEN (x) f

(x) on Rn.

Hence by est imate (7.7) in p. 104, t here exist s a large number d such that

c1 |EN|

N

1

|x|n

1

NEN (x) f

(x) for all |x| d.

On t he compact set S = {x Rn : 1 |x| d} , since f (x) is lower semicontinuousonRn with f (x) >0 everywhere, it attains it s posit ive minimum on S (see exercise 7, p. 61). Hence t here exist s a small

posit ive const ant c2 such that f (x) c2|x|n on S. Let c = min

c1 |EN|N , c2

. We have

f (x) c

|x|nfor all |x| 1.

2. (10 point s) Do Exercise 2 in p. 123.

Solution: Assume || M on Rn and x is in t he Lebesgue set of f. We have

|(f ) (x) f(x)|

Z

Rn

|f(x y) f(x)| | (y)| dy

=

Z

B(O)

|f(x y) f(x)| | (y)| dy M

n

Z

B(O)

|f(x y) f(x)| dy

where B (O) = {|y| } has measure C(n) n. Hence

|(f ) (x) f(x)| C(n) M 1

|B (O)|

Z

B(O)

|f(x y) f(x)| dy

and we know t hatli m0

1

|B (O)|

Z

B(O)

|f(x y) f(x)| dy = 0

due t o T heorem 7.16.

3. (10 points) Let C0 (Rn) be t he space of all continuous funct ions on Rn with compact support.We know t hat it is dense in t he space L1 (Rn) (Lemma 7.3 of the book). It is also clear thateach g (x) C0 (R

n) is uniformly continuous on Rn. Use this dense propert y t o show t hat iff L1 (Rn) , then we have the following property called " Continuity of Translation inL1" :

li my0

Z

Rn

|f(x + y) f(x)| dx = 0.


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Solution: For any > 0, choose a function g (x) C0 (Rn) withRRn

|f(x) g (x)| dx < 3

.Then for any y Rn we also have

Z

Rn

|f(x + y) g (x + y)| dx 0 such t hat it contains t hesupport of g. Then g is uniformly continuous on Br (O) and so there exists 0 < < 1 such t hat if|y| , then

|g (x + y) g (x)| 3

|Br+ 1 (O)|for all x Rn.

Now if |y| < < 1, we have

Z

Rn

|f(x + y) f(x)| dx

Z

Rn

|f(x + y) g (x + y)| dx +

Z

Rn

|g (x + y) g (x)| dx +

Z

Rn

|g (x) f(x)| dx

3+

Z

Br+ 1 (O)

|g (x + y) g (x)| dx +

3 .

4. (10 point s) T here are many applicat ions of t he use of convolut ion in analysis. One easy

example is t he following. Let

h (t) =

0 if t 0

e1

t if t > 0.

It is known that h (t) is a C function on R. Next let g (x) = h 1 |x|2 , x Rn, then

g (x) C0 (Rn) . One can divide it by its integral over Rn so that the new function (x)

C0 (Rn) satisfies

RRn

(x) dx = 1. For any number > 0, let (x) =1n

x

. Then it

satisfies (x) C

0 (Rn) , (x) 0, (x) > 0 |x| < ,

RRn

(x) dx = 1. Showthat:

(a) If f C(Rn) , then (f ) (x) converges uniformly to f(x) on compact subset s of Rn

as 0+ . (I t is easy t o see t hat (f ) (x) C (Rn) . You do not have t o show t his.)

(b) If f C0 (Rn) , then for any > 0, (f ) (x) also has compact support .

Solution: (a) . We know that f is uni formly cont inuous on compact subset s of Rn

. Let S bea compact subset of Rn. For any > 0 t here exist s > 0 such that

|f(x y) f(x)|

for all x S, |y| .For x S we have

|(f ) (x) f(x)|

=

Z

Rn

[f(x y) f(x)] (y) dy

=

Z

B(O)

[f(x y) f(x)] (y) dy

Z

B(O)

|f(x y) f(x)| (y) dy

Z

B(O)

(y) dy =


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(b) . Assume S is t he compact support of f. Then

(f ) (x) =

Z

Rn

f(x y) (y) dy =

Z

B(O)

f(x y) (y) dy.

From above wesee t hat if x / Swith dist (x, S) > , t hen we also have xy / S for any y B (O) .For such x, we have

(f ) (x) =

Z

B(O)

f(x y) (y) dy = 0.

Hence (f ) (x) also has compact support .


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1. (10 point s) Do Exercise 21 in P. 144.

Solut ion:

Let f Lp (Rn) , where 0 < p < is a constant. We first recall t he following elementarycalculus inequality:

L e m m a 1 For any 0 < p < , there exist positi ve constants c (p) and C(p) depending only onp such that

c (p) (ap + bp) (a + b)p C(p) (ap + bp) (1)for all a > 0, b > 0.

Let {rk}k= 1 be the set of all rati onal numbers. For any k and any 0 < p < , by (1) the

function |f(y) rk|p

is clearly locally integrable on Rn

. By T heorem 7.11, for each k we have

l imQ&x

1

|Q|

ZQ

|f(y) rk|p dy = |f(x) rk|p (2)

for a.e. x Rn. Let Zk be the set such that (2) is not valid, |Zk| = 0, and set Z =Sk= 1 Zk,

|Z| = 0. If x / Z, t hen by t he inequality ( in below, Q is centered at x and k is arbitr ary)1

|Q|

ZQ

|f(y) f(x)|p dy 1|Q|

ZQ

{|f(y) rk| + |rk f(x)|}p dy

C(p)|Q| ZQ |f(y) rk|

p dy +C(p)

|Q| ZQ |rk f(x)|p dy

=C(p)

|Q|

ZQ

|f(y) rk|p dy + C(p) |rk f(x)|p .

Hence

limsupQ&x

1

|Q|

ZQ

|f(y) rk|p dy 2C(p) |rk f(x)|p for all x / Z.

By choosing rk approximating f(x) (not e t hat f(x) is finit e almost everywhere in Rn; without loss

of generali ty, we can assume it is finit e everywhere), we obtain

lim supQ&x

1

|Q| ZQ|f(y) f(x)|p dy = 0 for all x / Z

for any 0 < p < .

2. (10 points) Prove the following more general version of t he T chebyshev inequality: Assume

f 0 is measurable on E satisfyingRE

fpdx < , where 0 < p < is a constant. Then forany > 0 we have

|{x E : f(x) > }| 1p

ZE

fpdx.

Solut ion:

Over t he set S := |{x E : f(x) > }| we have fp p and soZE

fpdx ZS

fpdx p |S| .


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3. (10 points) Let H(x) be t he Heaviside funct ion given by

H(x) =

1 if x > 0

1

2if x = 0

0 if x < 0.

Find t he set of t hose x R such that

li mh0+

1

2h

Zx+ hxh

f() d = f(x) . (3)

Det ermine if t he point x = 0 is in t he Lebesgue set of f or not.

Solut ion:

Clearly for any xR with x 6= 0 we have (3). For x = 0, we have

li mh0+

1

2h

Zx+ hxh

f() d = limh0+

1

2h

Zhh

f() d = limh0+

1

2h

Zh0

f() d =1

2= f(0) .

Hence t he set of t hose x R such t hat (3) holds is t he collect ion of all real numbers.To det ermine if x = 0 is in t he Lebesgue set of f or not , we comput e

li mh0+

1

2h

Zhh

|f() f(0)| d = limh0+

1

2h

Zhh

f() 12

d = 12 6= 0.

Hence x = 0 is n ot in t he Lebesgue set of f.

4. (10 points) Let g (x) be t he funct ion given by

g (x) =

(sin 1

xif x 6= 0

0 if x = 0.

Determine if the point x = 0 is in t he Lebesgue set of g or not. Also let G (x) =Rx

0g (s) ds,

x R. Do we have G0 (0) = g (0) or not.

Solut ion:

(T his solut ion is provided by TA Yu-Chu Lin) We claim t hat x = 0 is not in t he Lebesgue set

of g. Compute

li mh0+

1

2h

Zhh

|g () g (0)| d = limh0+

1

2h

Zhh

sin 1

d = lim

h0+1

h

Zh0

sin 1

d

= li mq

q

Zq

|sin y|

y2dy

.

We look at q = 2 p, p N. Over t he intervals

I1 =

2p +

4, 2p +

3

4

, I2 =

2p + 2 +

4, 2p + 2 +

3

4

I3 =

2p + 4 + 4

, 2p + 4 + 34

,


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we have |sin y|

2

2, which implies

Z2p

|sin y|

y2dy

2

2

ZI1I2I3

1

y2dy

= 22

(1

2p + 4

12p + 3

4

!+

12p + 2 +

4

12p + 2 + 3

4

!+

)

2

2

2

(1

2p + 34

2 + 12 (p + 1) + 3

4

2 + 12 (p + 2) + 3

4

2 + )

.

To est imat e t he sum of t he series (denot e t he series by S) we note that (by comparing areas)

SZp

12x + 3

4

2 dx = 12 2p + 34

and so

2p

Z2p

|sin y|y2

dy 22

2

2p

2

2p + 34

28

as p

which implies t hat

li mq

q

Zq

|sin y|

y2dy

6= 0.

T he claim i s proved.

Next we est imat e

li mh0+

1

h

Zh0

sin1

d = lim

q

q

Zq

sin y

y2dy

where Zq

sin y

y2dy =

Zq

1

y2d ( cosy) = cosq

q2 2

Zq

cosy

y3dy

and so

li mq

q

Zq

sin y

y2dy

= lim

q

cosq

q 2q

Zq

cosy

y3dy

= 0.

Simil arl y we have

li mh0

1

h

Zh0

sin1

d = 0

and so G0 (0) = g (0).


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Real A nalysis Homework 14, due 2008-1-2 in class

1. (10 points) Let E Rn be a measurable set (|E| < or not). If for any 0 < p < we havef Lp (E) and kfk

p

K, where K is a const ant independent of p. Show that f L (E)

and kfk K also.

Solut ion:

We first assume |E| < . In t his case, we have (see T heorem 8.1) li mp kfkp = kfk . Hencef L (E) and kfk

K.

If |E| = , decompose E =S

k= 1 Ek, where each |Ek| < . On each Ek, we have f L (Ek)

and kfk,Ek

K. Since the constant K is independent of t he set Ek, we have f L (E) and

kfk,E K.

2. (10 point s) Do Exercise 5 in P. 143.

Solut ion:

We have 1 p < , 0 < |E| < , and

Np [f] =

1

|E|

Z

E|f|p

1/p=

1

|E|1/pkfkp .

If p1 < p2, 1 p1, p2 < , t hen set s =p2p1 (1,) . It s conjugat e exponent is t = ss1 =

p2p2p1

(1,) . By Hlder inequali ty we get

kfkp1

p1

=

Z

E

|f(x)|p1 1dx

Z

E|f(x)|p1s dx

1/s Z

E1tdx

1/t=

Z

E|f(x)|p2 dx

p1/p2|E|(p2p1)/p2 .

Taking 1p1 power on bot h sides gives t he conclusion.

Also by M inkowski inequali ty we have

Np [f + g] =1

|E|1/pkf + gkp

1

|E|1/p

hkfkp + kgkp

i= Np [f] + Np [g] .

Furthermore, Hlder inequality implies

1

|E|ZE|f g|

1

|E|1/pkfkp

1

|E|1/p0kgkp0 = Np [f] Np0 [g] ,

1

p+

1

p0= 1.

Finally since 0 < |E| < , we have

li mp

Np [f] = l imp

1

|E|1/pkfkp

!

= limp

kfkp = kfk .

3. (10 points) Prove the converse of Hlder inequalit y (T heorem 8.8) for the case p = 1 and

p = .


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Solut ion:

Assumep = 1. Clearl y we have

kfk1 sup

Z

Ef g (1)

for all g L (E) with kgk 1.

Conversely, t ake g = sign f L (E) . Then kgk 1 and

kfk1 =

Z

E|f| =

Z

Ef g

which i mplies RHS of (1) LHS of (1).

For p = , again we have

kfk sup

Z

Ef g

for all g L1 (E) with kgk1 1.

Conversely, if kfk = 0, then it is clear. For 0 < kfk < , we may assume kfk = 1. Let

En =

x E : |f (x)| > 1

1

n, n N

.

Then |En| > 0 for all n. On each En one can choose gn (x) satisfying gn 0,REn

gn = 1, and let

gn = 0 outside En. Now

Z

E|f| gn =

Z

En

|f| gn

1

1

n

Z

En

gn = 11

n,

Z

Egn = 1

and so

kfk

= sup

Z

E|f| g

for all g L1 (E) with kgk1 1. Finally it is easy t o see t hat supRE |f| g = sup

REf g.

For t he case kfk

= , just repeat t he above process wit h

En = {x E : |f(x)| > n, n N} .

4. (10 point s) Assume 1 p < . Do Exercise 12 in P. 144.

Solut ion:

In t his problem, we assume 1 p < (in fact, as long as 0 < p < , we have the sameconclusion).

(=) By Mi nkowski inequality we havekfkp kfkkp

kf fkkp 0 as k .

Hence we have kfkkp kfkp as k .

(= ) We assume t hat fk f a.e. and kfkkp kfkp as k . By t he inequalit y

2p |f|p + 2p |fk|p |f fk|

p 0

we have (by Fatous Lem m a)Z Z


14/46

where Z

Eliminf

k[2p |f|p + 2p |fk|

p |f fk|

p] =

Z

E[2p |f|p + 2p |f|p]

and

liminfk

Z

E[2p |f|p + 2p |fk|

p |f fk|p] =

Z

E[2p |f|p + 2p |f|p] limsup

k

Z

E|f fk|

p .

Since we assume f Lp, the integralRE [2

p |f|p + 2p |f|p] is finite (this is essential). Hence weconclude

lim supk

Z

E|f fk|

p 0.

T he proof is done.

R e m a r k 1 When p = , the conclusi on in (= ) fails. Just take f = 1 on R and fk = (k,k) .


15/46

Solut ions t o Homework 1

1. (10 points) Let Q be the set of all rationals in the interval [0, 1] . Let S = {I1, I2,...,Im}be a fi n i t e collection of closed intervals covering Q. Show that

mX

k= 1

v (Ik) 1. (0.1)

On the other hand, for any > 0, one can find S = {I1, I2,...,Im, } , which is a count-able collection of closed intervals covering Q, such that

X

k= 1

v (Ik) < . (0.2)

In particular, (0.2) implies that |Q|e

= 0. (Now you see the difference between the use of finit e cover and countable cover .)

Solution: For (0.1), we first assume that the intervals I1, I2, ... , Im are nonoverlapping.In such case we clearly have (0.1).

For arbitrary intervals I1, I2, ... , I m with overlapping, one can throw away the over-lapping part and the remaining nonoverlapping part, which we denote it as J1, J2, . . . , J n,satisfies

Pn

k= 1v (Jk) 1. Therefore we have (0.1).

For (0.2), it has been done in class.

2. (10 points) Find a set E R with outer measure zero and a function f : E R such thatf is continuous on E and f(E) = [0, 1] . This exercise says that a continuous function canmap a set with outer measure zero onto a set with outer measure one.

Solution: Choose E = C to be the Cant or set contained in [0, 1] and let f(x) be thecontinuous Cantor Lebesgue function defined on [0, 1] (see book p. 35). We know that whenrestricted to C, f(x) is still a continuous function on C. Moreover, one can easily see thatf(C) = f([0, 1]) = [0, 1] (for example, we have f

1

3, 2

3

= f

1

3, 2

3

, etc.).

3. (10 points) Let E1 and E2 be two subsets ofRn

such that E1

E2 and E2

E1 iscountable. Show that|E1|e = |E2|e .

Solution: Clearly we have |E1|e |E2|e . Also

|E2|e |E1|e + |E2 E1|e = |E1|e

which implies |E1|e = |E2|e .

4. (10 points) Find a continuous function f(x) defined on [0, 1] such that f(x) is differentiableon a subset E [0, 1] with |E|

e= 1 and f0 (x) = 0 for all x E, but f(x) is not a constant

function.


16/46

Solution: Let f(x) be the Cantor Lebesgue function defi

ned on [0, 1] as given in p. 35 ofthe book. We know f(x) is differentiable on the open set O = O1 O2 O3 , where

O1 =

1

3,

2

3

, O2 =

1

9,

2

9

7

9,

8

9

, O3 =

1

27,

2

27

, O4 =

The total length of these open intervals is given by

1

3

"

1 +2

3+

2

3

2+

2

3

3+

#

= 1.

The example below shows you how to obtain a set which is measurable, but not Borelmeasurable.

Let f(x) : [0, 1] [0, 1] be the Cantor-Lebesgue function and let g (x) = x + f(x) . It iseasy to see that g (x) : [0, 1] [0, 2] is a strictly increasing continuous function. Hence g (x) is ahomeomorphism of [0, 1] onto [0, 2] . On each interval I1, I2, I3, ..., removed in the constructionof the Cantor set, say the interval I1 =

1

3, 2

3

, the function g (x) becomes g (x) = x + 1

2. Hence

g (x) sends I1 onto an open interval wit h t he same length . Using this observation one can seethat

|g (k= 1 Ik)| = |

k= 1 g (Ik)| =X

k= 1

|g (Ik)| =X

k= 1

|Ik| = 1

which implies |g (C)| = 2 1 = 1, where C is the Cantor set.Since g (C) has positive measure, by Corollary 3.39 in the book, there exists a non-measurable

set B g (C) . Now consider the set A = g1 (B) C. It has measure zero, hence it ismeasurable. However it can not be Borel measurable. If A were Borel measurable, then sinceg (x) is a homeomorphism, it would imply that B = g (A) is also Borel measurable. But this isimpossible since B is a non-measurable set.


17/46

Real Analysis Homework 2, due 2007-9-25 in classShow Your Work to Each Problem

1. (10 points) Use Lemma 3.16 to prove Lemma 3.15. Note that in proving Lemma 3.16 wedo not have to use Lemma 3.15.

Solution : Let {Ik}N

k= 1be a finite family of nonoverlapping intervals. It suffices to show that

N[k= 1

Ik

NXk= 1

|Ik| .

For each k, consider smaller interval Ik Ik with |I

k| |Ik|/2

k. Since {Ik}N

k= 1are nonover-

lapping, the set I1 and N

k= 2Ik

has positive distance. We can apply Lemma 3.15 to getN[k= 1

Ik

N[k= 1

Ik

=

NXk= 1

|Ik | NXk= 1

|Ik| .

The proof is done.

2. (10 points)

(a) (5 points) Assuming the validity of Theorem 3.30 and the existence of non-measurablesets in Rn at this moment. Show that there exist two nonempty sets E1 and E2 inRn such that E1 E2 = , but

|E1 E2|e < |E1|e + |E2|e .

Hence the condition d (E1, E2) > 0 in Lemma 3.16 can not be replaced by just E1 E2 = .

(b) (5 points) Construct a sequence of nonempty sets Ek [0, 1] , k = 1, 2, 3..., so that

limsup Ek = [0, 1] , lim infEk = . (0.1)

Solution : (a). This is easy by Theorem 3.30. Choose E Rn to be a nonmeasurable set.Then there exists some nonempty set A Rn such that

|A|e

< |A E|e

+ |AE|e

(0.2)

where A E and A E are disjoint. One can easily see that if (0.2) holds, then both A E and AE are nonempty. (b). Choose for example the sequence

E1 = [0, 1]

E2 =

0,

1

2

, E3 =

1

2, 1

E4 =

0,

1

3

, E5 =

1

3,

2

3

, E6 =

2

3, 1

E7 =

0,

1

4

, E8 =

1

4,

2

4

, E9 =

2

4,

3

4

, E10 =

3

4, 1

.Then we have (0.1).


18/46

3. (10 points) Assuming there exists a non-measurable set contained in [0, 1] , do Exercise 17in p. 48.

Solution : Let E = f1 (A) , where A is a n o n me a su r a b l e set contained in [0, 1] . Heref : C (Cantor set) [0, 1] is the Cantor-Lebesgue function in p. 35. It is continuous. Then since|E| = 0, E is measurable. But f(E) = A is nonmeasurable.


Solution : We first asume that 0 |E|e

< .

It is clear that for any interval I, |Ih| = |I| . For any > 0, choose a sequence of intervals Ik,

covering E, such that Xk

|Ik| < |E|e + .

Now the translation Eh S

k(Ik)h and so

|Eh|e Xk

|(Ik)h| =Xk

|Ik| < |E|e +

which gives |Eh|e |E|e . Conversely |E|e =(Eh)h

e |Eh|e . We are done.

If |E|e

= , then we must have |Eh|e = also (note that from the above observation, if

|Eh|e < , it will force |E|e < ).

Assume E is measurable. For any > 0, there exists open set G E such that |GE|e

< .This implies

|Gh Eh|e = |GE|e <

where Gh is also an open set. Hence Eh is also measurable.


19/46

Real Analysis Homework 3, due 2007-10-3 in classSh ow Y o u r W o r k t o E ach P r ob l em

1. (20 points) Let f : Rn R be a continuous function. Define the collection of setsP

as

X=

B R : f1 (B) is measurable

.

DoesP

form a -algebra? If B R is a Borel set, does it follow that f1 (B) is a Borelset? Give your reasons.

Solut ion: Clearly , R P

. If EP

, by

f1 (Ec) = f1 (E)c

(0.1)

we know that Ec P

also. Similarly if {Ek}

k= 1 is a collection of subsets ofP

, then by

f1

[

k

Ek

!

=[

k

f1 (Ek) and f1

\

k

E

!

=\

k

f1 (E) (0.2)

we know that kEk P

and kE P

. HenceP

is a -algebra.Similarly we can show that the set

=

B R : f1 (B) is Borel measurable

forms a -algebra. For any open set O R, the set f1 (O) is also open. Hence f1 (O) . Thesame for closed set. Hence all open sets and closed sets are contained in . Since is a -algebra,if B R is a Borel set, then f1 (B) must also be a Borel set.

2. (20 points) Do Exercise 12 in p. 48. Hint: You can use Theorem 3.29.

Solut ion: We first show that E1 E2 R2 is measurable.

Write E1 = H1 Z1, H1 is G set and |Z1| = 0. The same for E2 = H2 Z2. Then

E1 E2 = H1 H2 Z1 H2 H1 Z2

where

H1 H2 =

\

n= 1

On

!

\

n= 1

On

!

=\

n= 1

On On

is a G set in R2 (On and On are open set in R

1). We see that H1 H2 R2 is measurable.Assume first that |H2| < . One can find open sets G1 Z1, G2 H2 such that |G1| 0 one can choose open set G E and closedset F E so that

|E|e < |F| |G| < |E|e + . (0.9)

Since |F| |E|e < and then (0.9) gives

|G F| = |G| |F| < 2.

In particular we have |GE| |G F| < 2 and so E is measurable.Let E = (, 0) A, where A (1, 2) is some nonmeasurable set. Clearly E satisfies

|E|e = , infG

E, G open inRn

|G| = supFE, F closed in R

n

|F| =

but it is nonmeasurable.


22/46

Real Analysis Homework 4, due 2007-10-9 in classShow Your Work to Each Problem

1. (20 points)

(a) (10 points) Use definition (do not use Theorem 3.33) to show that the Cantor-Lebesgue function f(x) : [0, 1] [0, 1] is not a Lipschitz continuous function.

(b) (10 points) Show that the Cantor-Lebesgue function f(x) : [0, 1] [0, 1] satisfies thefollowing

|f(x) f(y)| 2 |x y| , x, y [0, 1]

where (0, 1) is a constant given by = log2/ log3. (Hint: Use the fact thatifx, y [0, 1] with |x y| 3k for some k N, then the difference |f(x) f(y)| isat most 2k. For arbitrary x, y [0, 1] one can choose an unique k N suchthat 3k1 < |x y| 3k, which implies |f(x) f(y)| 2k. Rewrite the es-timate without involving k.)

Solution: (a) . For example, look at the function near the point x = 29

= 627

= 1881

. We havef2

9

= 1

4. We also have f

7

27

f

2

9

= 1

8, which gives

f7

27

f

2

9

7

27 2

9

=

3

2

3.

Keep going to get (note that 627

= 1881

, 727

= 2181

)

f19

81

f

2

9

19

81 2

9

=

3

2

4

, etc. Thus it is impossible to find a finite constant M > 0 such that

f(y) f(x)

y x

M for all x, y [0, 1] .

(b) . First note that if x, y [0, 1] with |x y| 3k for some k N, then the

difference |f(x) f(y)| is at most 2k

. For arbitrary x, y [0, 1] one can choose l ar g estk N such that |x y| 3k; therefore

3k1 < |x y| 3k (0.1)

which will gives t he best est imat e

|f(x) f(y)| 2k. (0.2)

(0.1) is equivalent to (k + 1)log 3 < log |x y| k log3

which gives

k < 1 + log |x y|

log3.


23/46

By (0.2) we get

log |f(x) f(y)| k log2 < log 2 + log |x y| , =log2

log3(0.3)

and so|f(x) f(y)| 2 |x y| , x, y [0, 1] .


Solution: Let E be a non-measurable subset of [0, 1] , as established in Corollary 3.39 of

the book. We know |E|e > 0. Let r0 = 0, r1, r2, ..., be the set of all rationals in [0, 1] and letEk = E+rk for k = 0, 1, 2, 3, .... Each Ek [0, 2] for all k with |Ek|e = |E| , and EiEj = fordifferent i and j. We clearly have

[

k

Ek

e

2 0 for all k. Hencewe have limk |Ak|e > 0.


Solution: For each n N let

Zn = Z\

[n, n] .

we have |Zn| = 0 for all n. Also let

Tn (x) =

x2 if x [n, n]n2 otherwise.

Then Tn : R R is Lipschitz continuous on R. Hence |T Zn| = 0 for all n. As T ZS

n TnZn,

we have |T Z| = 0.


24/46



Solution: Assume f t akes dist inct values a1, , aN on disjoint sets E1, , EN. We canexpress t he funct ion f(x) as

f(x) = a1E1 (x) + + aNEN (x) , x E =N[

k= 1

Ek.

If each Ek is measurable, t hen t he charact eri st ic functi on Ek (x) : E R is a measurable

function on E. Hence if E1, , EN are all measurable, so is f(x) .

Conversely, assume f(x) is measurable. T hen by definit ion we know t hat (we may assumea1 < a2 < < aN) t he set {f > aN1} is measurable. Since {f > aN1} = EN, t he set EN ismeasurable. Similarly by

EN1 = {f > aN2}EN

weknow that EN1 is measurable. K eep going t o conclude t hat E1, , EN are all measurable.


Solution: Let F (x) = (f(x) , g (x)) , x Rn. F : Rn R2.

(=) Assume F is measurable. For any open set Gx R, let G = Gx R. Then G R2 is

open and by t he identi ty

F1 (G) = f1 (Gx)

weknow t hat f1 (Gx) is measurable for any open set Gx R. Hence f : Rn R is measurable.

The same for g : Rn R.

(= ) Assume f : Rn R and g : Rn R are both measurable. Let G R2 be an

arbi trary open set . We can express G as a countable union of nonoverlapping closed int ervalsG =

S

k= 1 Ik, where Ik = [ak, bk] [ck, dk] . Not e t hat

F1 (Ik) = f1 [ak, bk]

\g1 [ck, dk] (bot h set s are measurable)

and so F1 (Ik) is measurable. By

F1 (G) =

[

k= 1

F1 (Ik)

t he set F1 (G) is measurable in Rn. Hence F is a measurable functi on.


Solution: For any a R, we have

{x Rn : f(T x) > a} = {x Rn : f(T x) (a,)}[

{x Rn : f(T x) = }

=

x Rn : T1

f1 (a,) [

x Rn : T1

f1 ({})

. (0.1)

As T1 : Rn Rn is li near and Li pschit z continuous, (0.1) is a measurable set . Hence f(T x)is a measurable functi on.


25/46


First solution: We first prove t he following

Lemma 0.1 T he Cantor-Lebesgue function f(x) : [0, 1] [0, 1] satisfies f(x1) = f(x2) , where

x1 and x2 are i n C, if and only if both x1 and x2 are endpoints of some interval removed.

Proof. T he direct ion (= ) is trivial.

(=) Assume at least one of x1, x2 is not endpoint , say x2. On the interval (x1, x2) , x1

1

2

= A

is not measurable.

Second solution (much easier):

Let f(x) : [0, 1] [0, 1] be the Cantor-Lebesgue function and let g (x) = x + f(x) . I t i s

easy t o see that g (x) : [0, 1] [0, 2] is a str ict ly increasing continuous funct ion. Hence g (x) is

a homeomorphism of [0, 1] onto [0, 2] . We denot e it s continuous inverse by h : [0, 2] [0, 1] . On

each int erval I1, I2, I3, ..., removed in t he const ructi on of t he Cantor set , say t he interval


26/46

I1 =

1

3 ,

2

3

, t he funct ion g (x) becomes g (x) = x +

1

2 . Hence g (x) sends I1 onto an open i ntervalwith the same length. Using t his observat ion one can see t hat

g

[

k= 1

Ik

!

=

[

k= 1

g (Ik)

=

X

k= 1

|g (Ik)| =X

k= 1

|Ik| = 1

which implies |g (C)| = 2 1 = 1, where C is t he Cantor set . Since g (C) has positive measure,by Corollary 3.39 in the book, there exists a n o n -me a su r a b l e se t B g (C) . Now consider

t he set A = h (B) C. It has measure zero, hence A is measurable. L et

= A : [0, 1] R, |A| = 0

then is a measurable function on [0, 1] . We now have two measurable functions h : [0, 2]

[0, 1] (continuous) and : [0, 1] R. But t he composit e funct ion h : [0, 2] R is not

measurable since t he set [0, 2] : (h ()) >

1

2

= A

is not measurable.


27/46



Solut ion: For any x0 E (compact) and Mx0 > f (x0) one can choose x0 > 0 such that

f(x) < Mx0 for all x E with |x x0| < x0 .

The collection {x E : B (x0; x0)} forms an open cover of E and so there exist a finite coverB (x1; x1) , , B (xn; xn) of E. Set M = max {Mx1 , , Mxn} . We have f(x) < M for allx E. Hence f is bounded above.

Let M = supxEf(x) . For k = 1, 2, 3,... choose xk E such that

M 1k

< f (xk) < M. (0.1)

As E is compact, by passing to a subsequence if necessary, we may assume that xk Econvergesto some x0 E. Assume f(x0) < M. Choose M with f(x0) < M < M. Since f is usc at x0,there exists some > 0 such that

f(x) < M for all x B (x0; ) E. (0.2)

For k large enough, we have xk B (x0; ) E. (0.1) will contradict to (0.2). Hence we must

have f(x0) =

M, i.e., the maximum is attained.

2. (10 points) Show that the limit of a decr easing sequence of functions (with commondomain E) usc at x0 E is also usc at x0. Give an example of a decreasing sequence offunctions continuous at x0 E but its limit is not continuous at x0 (by the first part ofthe problem we know that the limit is at least usc at x0).

Solut ion: Denote the decreasing sequence of functions by fk (x) and the limit by f(x) . Wehave

f1 (x) f2 (x) fk (x) , x Eand each fk (x) is usc at x0. Since we have f(x) fk (x) on E for each k, we have

lim supxx0 ;xE

f(x) lim supxx0 ;xE

fk (x) fk (x0) for each k N.

As k is arbitrary. Letting k gives the conclusion.Let E = [0, 1] and let {xk : k = 0, 1, 2, 3,...} be the set of all rationals in E. Set f0 (x) 1 and

set for each k N the function

fk (x) =

0 at x = x1, x2, , xk1 otherwise.

Then fk (x) is a decreasing sequence of functions on E; all are continuous at

2/2. But the limitf(x) is not continuous at

2/2. However it is usc at

2/2.



28/46

Solut ion:We only show that h (x) is usc on

Rn

. The proof of the other case is similar.For any x0 Rn, we first assume that h (x0) < (otherwise we are done). It suffices to

show that for any M > h (x0) = inf{f(y) : y B (x0)} there exists > 0 such that h (x) < Mfor all x B (x0; ) .

For any M > h (x0) choose y0 B (x0) such that f(y0) < h (x0)+ < M. Set = |y0 x0| 0. Then for all x with |x x0| < we have

|x y0| |x x0| + |y0 x0| + < r.

Hence y0 B (x) and by definition we get h (x) = inf{f(y) : y B (x)} f(y0) < M. Thus h (x)is usc at x0.

For the case of using closed balls, consider in R2

the function

f(p) =

0 if p = (1, 0)1 if p 6= (1, 0)

, p R2

and take r = 1, B (x) =

p R2 : |p x| 1

. Then h (0, 0) = 0 and for any > 0 we haveh (, 0) = 1. Hence h is not usc at the point (0, 0) .


Solut ion: Assume f(x) : [a, b] R is continuous a.e. on [a, b] . Let k be a sequence ofpartitions of [a, b] with norms tending to zero. We also assume that each k+ 1 is a refinement

ofk. For each k, if x(k)1 < x(k)2 < are the partitioning points ofk, let lk (x) and uk (x) be

defined in each semi-open intervalh

x(k)i

, x(k)i+ 1

as the inf and sup off onh

x(k)i

, x(k)i+ 1

i

. Note that

lk (x) f(x) uk (x) for all x [a, b) and all k. It is easy to see that for each k, lk (x) and uk (x)are measur a bl e functions on [a, b) (even if it is possible that lk (x) = or uk (x) = + onsome intervals).

Let x0 (a, b) at which f(x) is continuous. For any > 0 there exists > 0 such that

x (x0 , x0 + ) implies |f(x) f(x0)| <

and when k is large enough, the interval in k containing x0 must lie inside (x0 , x0 + ) .This implies

|lk (x0) f(x0)| < and |uk (x0) f(x0)| < for all k large enough. Hence limk lk (x) = f(x) a.e. on [a, b] (we also have limkuk (x) =f(x) a.e. on [a, b]). By Theorem 4.12 of the book, we know that f(x) is measurable on [a, b] .

R e m a r k 1 (be careful) I f g (x) is a conti nuous functi on on [a, b] and f(x) = g (x) a.e. on [a, b] ,it does not, in general, imply that f(x) is continuous a.e. on [a, b] . For example, t ake g (x) = 1and let

f(x) =

1, x is i rrational in [0, 1]0, x is rational in [0, 1] .

We see that f(x) = g (x) a.e. on [0, 1] , but f(x) is discontinuous everywhere on [0, 1] .


29/46


1. (10 point s)

(a) (7 points) Do Exercise 15 in p. 62.

(b) (3 points) Use Exercise 15 in p. 62 t o provet hefollowing: Let f : E R {} be a measurablfunction where |E| < and |f| < a.e. on E. Show t hat for any > 0 there exists a const anM > 0 and a closed set F E such that |E F| < and

|f(x)| M for all x F.

This says that a finite funct ion is, up t o a set of small measure, a bounded function.

Solut ion: For (a) . For each n = 1, 2,..., let

En = {x E : |fk (x)| n for all k} .

Then by |fk (x)| Mx < for all k and all x E, we have En % E as n with limn |En| =|E| . Since |E| < , we also have limn |E En| = 0. Choose M such that |E EM| < /2 and choosa closed set F EM such that |EM F| < /2. T hen we have |E F| < and t he following holds

|fk (x)| M for all x F and all k.

For (b) . By (a) , if we choose fk (x) = f(x) for each k N, t hen we are done.


Solut ion:

(=). By definition, for any > 0 we have

li mk

|{x E : |f(x) fk (x)| > }| = 0.

Hence for the same > 0 we have

|{x E : |f(x) fk (x)| > }| <

if k K, for some large K > 0.

(= ). Fixed an arbitrary > 0 first . We want t o show t hat for any > 0 there exist s K > 0 such thaif k K we have

|{x E : |f(x) fk (x)| > }| < . (0.1

Now for any > 0, if , t hen by the assumption we automat ically have t he exist ence of a K > 0 suchthat if k K we have

|{x E : |f(x) fk (x)| > }| < .

Hence we assume < . Again by t he assumpti on we have t he exist ence of a L > 0 such that if k L whave

|{x E : |f(x) fk (x)| > }| < .

But t he set {x E : |f(x) fk (x)| > } {x E : |f(x) fk (x)| > } , and so we have (0.1).

T he Cauchy crit erion is: For any > 0 there exist s K > 0 such that if m, n K we have

|{ E |f ( ) f ( )| > }| < (0 2


30/46


Solut ion: Given f : E R {} measurable and let

f (a) = |{f > a}| , where < a < .

As a function of a, f (a) is decreasing on (, ). If fk % f on E, t hen set

Ek = {fk > a} , k = 1, 2, 3, ...

We have E1 E2 E3 and if f(x) > a, we wil l have fk (x) > a for all k large enough (sincfk % f on E). Thus

{f > a} =[k= 1

Ek

and so fk (a) % f (a) for all a (, ).

If fk f in measure on E, given > 0 let

A1k = {|f fk| > } , A2k = {|f fk| } .

We have E = A1kS

A2k (disjoint union). Hence for each fixed a R we have

Ek =

Ek\

A1k

[Ek\

A2k

, Ek = {fk > a}

where limkEkT

A1k

= 0 (due t o convergence in measure) and

Ek\A2k = {fk > a}\ {|f fk| } {f > a } .

We have

fk (a) = |Ek| =Ek\

A1k

+Ek\

A2k

Ek\

A1k

+ f (a )

and so

lim supk

fk (a) f (a ) for any > 0.

Simil arl y we have

{f > a + }\

{|f fk| } {fk > a}

which gives

liminfk

fk (a) f (a + ) for any > 0.

We conclude

f (a + ) liminfk

fk (a) limsupk

fk (a) f (a ) for any > 0.

T hus if f (x) i s continuous at x = a, we have limkfk (a) = f (a) .


Solut ion: Let S = [0, 1] [0, 1] . The idea is to separate the x variable from the y variable. For eachn = 1, 2, 3, ...., define

fn (x, y) =n1Xk 1

f(rk, y) [ k1n

, kn

) (x) + f(rn, y) [n1n

,1] (x) , (x, y) S


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For each k, t he funct ion hk (x, y) := f(rk, y) [ k1n

, kn

) (x) : S R satisfies

hk (x, y) = 0 on

[0, 1]

k 1

n,

k

n

[0, 1]

and onk1n

, kn

[0, 1] we have

(x, y)

k 1

n,

k

n

[0, 1] : hk (x, y) > a

=

k 1

n,

k

n

{y [0, 1] : f(rk, y) > a}

which is a measurable set (due t o Exercise 12, p. 48). Hence we can conclude t hat hk (x, y) is a measurablfunction on S for each k. As a consequence the functi on fn (x, y) is also measurable on S.

For each (x0, y0) S we have (assume t hat x0 k1n

, kn

)

|fn (x0, y0) f(x0, y0)| = |f(rk, y0) f(x0, y0)| , rk

k 1n

, kn

and so fn (x0, y0) f(x0, y0) as n . Hence f(x, y) is a measurable funct ion on S.


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Solut ion:

In this problem we assume xf(x, y) exist s on I = [0, 1] [0, 1] . we also know t hat it is abounded function on I. Let

Fn (x, y) =f

x + 1n , y f(x, y)

1n

, n = 1, 2, 3, ....

we see that for each fixed x, Fn (x, y) is a sequence of bounded (use mean value theorem to see

t his) measurable funct ions of y. By

x f(x, y) = l imnFn (x, y)

we know t hat for each fixed x, xf(x, y) is a measurable funct ion of y. Now by the Bounded

Convergence Theorem, we obtain

d

dx

Z 1

0

f(x, y) dy = limn

Z 1

0

Fn (x, y) dy =

Z 1

0

xf(x, y) dy.


Solut ion:

For any > 0 we have

|{|fk f|p > }|

1

Z

E

|fk f|p 0 as k

due to t he Tchebyshev inequality. Hence fk converges t o f in measure on E.


Solut ion:

By Exercise 9 we know that fk f in measure. In particular, there exists a subsequence fkjsuch t hat it converges to f a.e. on E. Fat ous lemma implies

Z

E

liminfj

fkj

p=

Z

E

|f|p liminfj

Z

E

fkj

p M.


Solut ion:

C ase1: If f(x) = E1 , E1 E, then LHS of the identit y is |E1| , and t he RHS of t he ident ityis given by Z

1


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Real Analysis Midt erm Exam, November 13, 2007

Show detailed argument to each problem.

1. (10 points) Suppose E is a Lebesgue measurable subset ofR with |E| < . Prove that

|E| = sup {|K| : K E and K is compact} . (0.1

solution:

Let En = ET

[n, n] , n N. We have En % E and so |E En| 0 as n (note that |E| < )For any > 0 one can find m N so that |E En| < /2 for all n m. For Em one can find a closedsubset F Em such that |Em F| < /2. Hence |E F| |E Em| + |Em F| < . In particular F icompact since Em is bounded. We also have

|E| |E F| + |F| < + |F|

where F E and F is compact. (0.1) is proved.

2. (10 points) Assume E(t) is a continuously differentiable increasing function on [0, ) such that 0 E(t) C for all t [0, ), where C is some positive constant. Show that for any > 0, we have

t [0, ) : E0 (t) >

C

.

solution:

By Tchebyshev inequality we have

t (0, ) : E0 (t) >

1

Z

0

E0 (t) dt 1

hlimt

E(t) E(0)i

C

since 0 E(t) C for all t [0, ).

3. (10 points) Let fn : E R be a sequence of measurable functions defined on a measurable seE Rn. Let

A =n

x E : limn

fn (x) existso

.

Is A a measurable set or not? Give your reasons.

solution:

Let F (x) = limsupn fn (x) and F (x) = liminfn fn (x) . We know that both functions are

measurable on E. Hence the sets(S1 := {x E : F

(x) = and F (x) = }

S2 := {x E : F (x) = and F (x) = }

are all measurable. Let S = S1S

S2. Now E S is also measurable, and F F is a measurable function

on E S. By the relationA = {x E S : F (x) F (x) = 0}

we know that A is a measurable set.

4. (10 points) Give an example of a sequence of measurable functions {fk} defined on a measurablset E Rn such that the following strict inequalities hold:

Z

E

lim infk

fkdx < liminfk

Z

E

fkdx < lim supk

Z

E

fkdx 0 be a constant and let f, fk, k = 1, 2, 3,..., be measurable functions onE. If

RE

|fk f|p

0 as k andRE

|fk|p

M (M > 0 is a constant) for all k, show thaRE |f|

p M also.

solution:

For any > 0 we have

|{|fk f|p > }|

1

Z

E

|fk f|p

0 as k

due to the Tchebyshev inequality. Hence fk converges to f in measure on E. In particular, there exists asubsequence fkj such that it converges to f a.e. on E. Fatous lemma implies

Z

E

liminfj

fkj

p=

Z

E

|f|p lim infj

Z

E

fkj

p M.

The proof is done.

Remark 1 (be careful)RE

|fk f|p

0 as k does not , in general, imply that |fk f|p

0 a.e. on

E as k . Also for p > 0 the inequality

|f|p |fk f|p + |fk|

p


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Real Analysis Sample Exam, November 6, 2007

Show detailed argument to each problem.

1. A totally unlucky number is one that contains no sevens in any decimal expansion. Computethe Lebesgue measure of the totally unlucky numbers in [0, 1] .

solution:

Among the numbers 0.1...., 0.2...., , 0.9...., the lucky numbers has measure s = 110

. Amongthe numbers 0.11...., 0.12...., , 0.19...., the lucky numbers has measure 1

10 s. Based on thi

observation, all of the lucky numbers has measure

s + (1 s) s + [1 s (1 s) s] s + [1 {s + (1 s) s + [1 s (1 s) s] s}] s +

= s + (1 s) s + (1 s)2 s + (1 s)3 s +

=

s

1 (1 s) = 1.

Hence the totally unlucky numbers has measure zero. We are lucky!!!

2. Let f : [a, b] R be a finite increasing function. Show that f is a measurable function on[a, b]. For any p (a, b) , evaluate the following limits:

limh0+

1

h

Z

[p,p+ h]

f (Lebesgue integral)

and

limh0+

1

h

Z

[ph,p]

f (Lebesgue integral).

solution:

Iff : [a, b] R is an increasing function, the number ofx [a, b] such that f is discontinuous ax is at most countable (see Rudin, p.96). Hence f is continuous a.e. on [a, b] and so measurable.

We also know that both f(p+) and f(p) exist for any p (a, b) . For fixed p (a, b) , therexists a number A such that for any > 0 there exists > 0 so that if x (p,p + ) then|f(x) A| < . Hence

1

h

Z

[p,p+ h]

(A ) A

1

h

Z

[p,p+ h]

f

A

1

h

Z

[p,p+ h]

(A + ) A

for all 0 < h < . Therefore limh0+1h

R[p,p+ h]

f = A = f(p+) . Similarly we have limh0+1h

R[ph,p]

f

f(p) .

3. Let En R be a sequence of measurable sets. Let

A = {x R : x En for infinitely many n} .

Is the set measurable or not? Give your reasons.

solution:

We actually have

A =\ [

Ek

!.


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4. Give an example of a measurable function h : E R R such that for some measurable setB R the inverse image h1 (B) is N OT measurable.

solution:

Let f(x) : [0, 1] [0, 1] be the Cantor-Lebesgue function and let g (x) = x + f(x) . It ieasy to see that g (x) : [0, 1] [0, 2] is a strictly increasing continuous function. Hence g (x) is ahomeomorphism of [0, 1] onto [0, 2] . On each interval I1, I2, I3, ..., removed in the construction othe Cantor set, say the interval I1 =

13

, 23

, the function g (x) becomes g (x) = x + 1

2. Hence g (x

sends I1 onto an open interval wit h t he same lengt h. Using this observation one can see thatg

[

k= 1

Ik

!

=

[

k= 1

g (Ik)

=X

k= 1

|g (Ik)| =X

k= 1

|Ik| = 1

which implies |g (C)| = 2 1 = 1, where C is the Cantor set. Since g (C) has positive measure, thereexists a non-meas ur abl e s et A g (C) . Now consider the set B = g1 (A) C. It has measurezero, hence it is measurable. Let h = g1. Then it is a measurable function and h1 (B) = A is notmeasurable.

5. Let E be a measurable set in Rn with |E| < and f is a measurable function on E. Let

En = {x E : |f(x)| n} , n = 0, 1, 2, 3....

Show that f L (E) if and only ifP

n= 0 |En| < .

solution:

(=) Assume f L (E) . Then f is finite a.e. in E (without loss of generality, we can assume

f is finite everywhere in E). It is not hard to see that

lim

Z

{|f|}

|f| = 0

which is like the case of an absolutely convergence sequence. This also implies (by Tchebyshevinequality)

limn

n |En| limn

Z

{|f|n}

|f| = 0. (0.1

By the relation E = E0 E1 E2 , |E0| < , we can decompose E as

E = (E0 E1) (E1 E2) (E2 E3) (disjoint union)

and observe that

n |En En+ 1|

Z

EnEn+ 1

|f| (n + 1) |En En+ 1| , n = 0, 1, 2, 3,... (0.2

HenceX

n= 0

n |En En+ 1| X

n= 0

Z

EnEn+ 1

|f| =

Z

E

|f|

where (in below we need to use the fact that |E| < , and so |En En+ 1| = |En| |En+ 1|) by (0.1

we haveX

n |E E + 1| = (|E1| |E2|) + 2 (|E2| |E3|) + 3 (|E3| |E4|) +


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ThereforeP

n= 0 |En| < .

(=) Conversely ifP

n= 0 |En| < , then the set {x E : |f(x)| = } must have measure zeroimplying f is finite a.e. in E . Again we assume f is finite everywhere in E. Since f is bounded onEn En+ 1, it is integrable on it. Note that

X

n= 0

(n + 1) |En En+ 1| = 2 (|E1| |E2|) + 3 (|E2| |E3|) + 4 (|E3| |E4|) +

= 2 |E1| + |E2| + |E3| + |E4| + < .

By (0.2) we know |f| must be integrable on E. Hence f is integrable on E.

6. Assume h (x) is a differentiable function on R. Show that h0 (x) is a measurable function onR.

solution:

Let

fn (x) =h

x + 1n

h (x)

1n

, x R.

For each n = 1, 2, 3,..., fn (x) is a finite measurable function on R with fn (x) h0 (x) for all x R

Hence h0 (x) is a measurable function on R.

7. In the Lebesgue Dominated Convergence Theorem (Theorem 5.36) if we replace the conditionfk f a.e. in E by fk f in measure on E, is the theorem still correct or not? Give

your reasons.

solution:

The theorem is still correct.

First note that by Theorem 4.22 there exists a subsequence fkj f a.e. on E. This impliethat |f| a.e. in E and so f L (E) (since L (E)). By the usual Lebesgue DominatedConvergence Theorem we have

Z

E

fkj

Z

E

f as j . (0.3

We then use contradiction argument to show thatRE

fk RE

f as k . Assume not. Thenthere exists a subsequence of fk, still denote it as fkj , j = 1, 2, 3, ..., so that

Z

E

fkj

Z

E

f

> 0

for all j. But then this subsequence has a further subsequence so that (0.3) holds, a contradiction.


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Excercise 1: Prove that the convergence of {sn} implies convergence of|sn|. Is the converse true?Proof. (Im assuming the sn are real numbers since Rudin doesnt say, andin an arbitrary metric space |x| is meaningless).

Assume sn converges to L. We claim that |sn| converges to |L|. To seethis, let > 0 be an arbitrary real number. We know that there is an N Nsuch that for any n N with n N, |sn L| < , and we claim the same Nworks for |sn|. To see this, let n N. Then ||sn| |L| | |sn L| < (Seehomework 2 problem 4 for the part).

Thus, |sn| converges to L.The converse, however, is FALSE. To see this, consider the sequence

sn = (1)n. Then |sn| = 1 for all n, so of course sn converges to 1. How-ever, sn cant converge. To see this, one can just use the fact quoted inRudin in definition 3.5: It is clear the {pn} converges to p iff every subse-quence of {pn} converges to p. Thus, since a subsequence of sn is simple1, 1, 1,..., its clear this converges to -1. While another subsequence is1, 1,..., which clearly converges to 1. It follows that sn has two subsequenceswhich converge to different numbers, and hence sn cannot converge.

Excercise 2: Ifs1 =

2 and sn+1 =

2 +

sn, prove that {sn} convergesand that sn < 2 for all n.

Proof. Well show the sequence is monotonely increasing and bounded aboveby 2. It will follow from theorem 3.14 that {sn} converges.To see its bounded by 2, notice that s1 clearly is. Now, assume induc-

tively that sn < 2. Then s2n+1 = 2 +

sn < 2 +

2 < 4, so that sn+1 < 2.

Then, by the principle of mathematical induction, it follows that sn < 2 forall n.

To see that its increasing, well again prove it inductively. Notice that

s2 =

2 +

2 >

2 = s1, so we start increasing. Now, assume sn1 < sn.Then

sn1 0 for all n and that limsupsn = althoughlimn = 0?

d) Let an = sn sn1. Show that sn n = 1n+1n

k=1 kak. Assume thatlim(nan) = 0 and that n converges. Prove that {sn} converges.

e) Derive the last conclusion from a weaker hypothesis: Assume M < |nan| M for all n and that lim n = . Prove that limsn = by completingthe following outline....

2


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Proof. a) Let > 0. We wish to find an N N such that for any n N,well have |n s| < . We know there is an N such that for any n N,|sn s| < /4. Now, choose N > N such that

Nn=1 snN+1 < /4 and such that

Ns/(N + 1) < /4. I claim this N works. For if n N, then

|n s| = |N

k=1 sn +N

k=N+1 sn NsN + 1

|

|N

k=1 snN + 1

| + |N

k=N+1 sn sN + 1

| + |Ns/(N + 1)|

/4 + /4 +

Nk=N+1 |sn s|/(N + 1)

/2 +N

k=N+1

/4(N + 1) = /2 + (N N 1)/4(N + 1)

< /2 + /4 = 3/4 < .

Thus N prime works.b) Consider sn = (1)n. Then even = 0 while 2n+1 = 12n+2 . Thus we

see that lim n = 0. However, cleary sn doesnt converge (see problem 2).

c) Yes it can, consider sn

=

k n = k2 for some k

N

2n otherwiseThen if you

look at the subsequence of the form sk2, its clear this has limit . Thus,limsupnsn = . Further, its clear that sn > 0 for each n. What aboutn?

Well, s1+...+snn

1+1/2+...+1/2n+(n+1)n1/4

n= 1+1/2+...+1/2

n

n+ n

3/4+n1/4

n=

1+...+1/2n

n+ n1/4 + n3/4. Now, for the second two terms, the limit as n

is clearly 0, so we must simply argue the first term goes to 0 as well. But1 + 1/2 + ... + 1/2n = 12

(n+1)

11/2 = 2 2(n+1). Thus (1 + ... + 1/2n)/n =2/n 2(n+1)/n, both of which clearly go to 0.

d)Well prove it inductively. If n = 1, then 1 = (s0 + s1)/2 so we see

s1 1 = s1/2 s0/2 while 1/2k ak = 1/2(s1/2 s0/2), so they agree.Now, assume its true for n, well show its true for n + 1.Then

1

n + 2

n+1k=1

kak =1

n + 2

nk=1

kak +n + 1

n + 2an+1

=n + 1

n + 2

1

n + 1

nk=1

kak +n + 1

n + 2an+1

3


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=n + 1

n + 2

(sn

n) +

n + 1

n + 2

an+1

=n + 1

n + 2(sn n + sn+1 sn)

=n + 1

n + 2(sn+1 n)

= sn+1 1n + 2

sn+1 n + 1n + 2

n

= sn+1 1n + 2

(s0 + ... + sn+1) = sn+1 n+1.Ill skip the next part since well prove a stronger result in part e)

e) Assume M < and |nan| M for all n. Let n .Now, if m < n, then

m + 1

n m(n m) +1

n m(sn si)

=m + 1

n m(s0 + .... + sn

n + 1s0 + ... + sm

m + 1)+

1

n m(n(m+1)+1)sn1

n mn

i=m+1

si

= sn +1

n

m

(m + 1

n + 1(s0 + ... + sn) s0 ... sm

n

i=m+1si)

= sn +1

n m(m + 1

n + 1(s0 + ... + sn) s0 ... sn)

= sn +1

n m((m + 1)(s0 + ... + sn)

n + 1 (n + 1)(s0 + ... + sn)

n + 1

= sn +1

n m(m + 1 (n + 1)

n + 1(s0 + ... + sn))

= sn n.Now, for the i in the sum, we have |sn si| = |sn sn+1 + sn+1 sn2 +

....

si

|=

|an + ... + ai+1

| |an

|+

|an1

|+ ... +

|ai+1

| M/n + M/(n

1) +

... + M/(i + 1) (n i)M/(i + 1).I claim that (n i)M/(i + 1) < (n m)M/(m + 1) (this is obtained by

just plugging in i = m into the expression. This is easiest to see by writing(n i)M/(i + 1) = (n + 1 i1)M/(i + 1) = (n + 1)M/(i1)N. But thenits clear that one can make this larger, by taking i smaller - that is i = m.

Thus, we have |sn si| < (nm)Mm+1 .Now, pick > 0 and let m(n) be defined as the integer satisfying m(n)

1 n1+ < m(n).

4


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Then 0 n1+ m = nm(m+1)1+ so that 0 n m (m + 1). Thus, nmm+1 . Then, we see that |sn si| < (nm)Mm+1 < M .

Thus we see that |sn | = |sn n + n | |sn n| + |n |.Now, we can make the second term as small as we want since n , so letsfocus on the first term.

The first term is |sn n| = |m+1nm(n m) + 1nmn

i=m+1(sn si)| |m+1nm(n m)| + 1nm

ni=m+1 |(sn si)|. Now, with the first term of this,

since n it follows that n is bounded, and hence that the first termgoes to 0 for large enough n. Finally, we deal with 1nm

ni=m+1 |(sn si)|.

We just showed |sn si| < M. Then we have 1nmn

i=m+1 |sn si| nm+1nm |sn si| < M. Thus, we can make this term as small as we like.

Finally, since we can make all the terms as small as we like, it followsthat we can make |sn | as small as we like. It follows that sn .

Excercise 5. Suppose {pn} is Cauchy in a metric space X and somesubsequence pnk converges to p X. Prove pn converges to p as well.Proof. Let > 0. We seek an N such that for all n N, |pnp| < . To thatend, we know that there is an N1 such that for any k N1, |pnk p| < /2.Likewise, we know there is an N2 such that for all n, m N2, |pnpm| < /2(since pn is a Cauchy sequence).

Let N = max{

N1, N

2}. Let n

N. Then

|p

pn|

=|p

pnn

+pnn

pn| |p pnn | + |pnn pn| < /2 + /2 = .

Excercise 6. Prov ethe following analogue of Theorem 3.10(b): If En isa sequence of nonempty bounded sets in a complete metric space X, and ifEn+1 En and if limdiam(En) = 0, then E =

n=1 En has exactly one

point.

Proof. For each n, let pn En. I claim the pn form a cauchy sequence. To seethis, let > 0. Then there is an N such that for any n N, diam(En) < .Then for any n, m

N, pn, pm

EN and diam(EN) < so that d(pn, pm) 0. We know there is an N so that for n, m N1, |pnpm|

Combine Solution

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