Chapter 3 L p -space. Preliminaries on measure and integration.

Chapter 3

Lp-space

Preliminaries on measure and integration

σ-algebra

AAA c \)2(

,)1(

11)3(

nnnn AA

Ω ≠ψ is a set

Σis a family of subsets of Ω with

Σis called σ-algebra of subsets of Ω

measure space

0)()1(

)(

int)2(

11

1

additivityAA

disjoisA

nn

nn

nn

Ω ≠ψ is a set

μ:Σ→[0, ∞] satisfies

Σis aσ-algebra of subsets of Ω

(Ω , Σ, μ) is called a measure space

measurable function p.1

Rwfwf )(;

f:Ω→R is measurable if

(Ω , Σ, μ) is a measure space

The family of measurable functions is a

real vector space.

measurable function p.2

1nnf if

The family is closed under limit, i.e.

is a sequence of measurable functions ,

which converges pointwise to a

finite-valued function f , then f is

measurable.(see Exercise 1.1 and 1.3)

Exercise 1.1

R

If f, g are measurable , then


f+g is also measurable.

Hint: for all

gfgfQ

gfx

xgf

xgxf

xgandxfthen

Qsomeforgfx

thengfxIf

gfgfClaim

Q

Q

))((

)()(

)()(

,""

:

.

,

lg

,,

)()(

)()(

)()(

)()(

,""

measurableisgfhence

gfgf

ClaimBy

gf

countableisQand

ofsubsetsofebraaisSince

QRgfthen

functionsmeasurablearegandfthatAssume

gfx

xgandxf

Qsomeforxfxg

xfxg

xgxf

thengfxIf

Q

Q

Q

Exercise 1.3

1 1

1

m k knn mff

ffnn

limLet f1,f2,… be measurable and


and f(x) is finite for each

Hint: for all

Show that f is measurable

x

1 1

1 1

1

1

1)(..1

),()(lim

1)(..1

)(

""

:

1:

m k knn

knn

n

nn

m k knn

mfw

mfw

knm

wftsk

wfwfSincem

wftsm

wf

fwanyFor

pf

mffClaim

RanyFor

.

1

,1

,1

)(

1)(

,

1)(

1,1

,1

""

1 1

1 1

measurableisfHence

mffthen

Nnmm

f

nmeasurableisfSince

fw

wfm

wf

havewenlettingBy

knm

wf

kmsomeform

fw

thenm

fwIf

m k knn

n

n

n

knn

m k knn

indicator function

Afor


χA is the indicator function of A

χA is measurable

<W>

<W> denotes the smallest vector

subspace containing W in a vector space.

Simple function p.1

AA :

are called simple functions

Elements of

Simple function p.2

iA

k

iif

1 k ,,1

if the right hand side has a meaning

, where

A simple function can be expressed as

are different values and Ai = {f =αi} ,

we define then )(1

i

k

ii Adf

Simple function p.3

fd

In particular

is meaningful if f is simple and

nonnegative, although it is possible that

df

Integration for f 0,measurable≧

For f 0, measurable ,define ≧

gddf

fgsimpleg

0:sup

f+ ,f-

0)(0

0)()()(

wfif

wfifwfwf

fff

If f is measurable, then

f+ and f- are measurable

0)(0

0)()()(

wfif

wfifwfwf

Integration for measurable function p.1

For any measurable function f ,define

dfdfdf

if R.H.S has a meaning


df

df

and

is finite if and only if both

df are finite

f is called integrable


dfdfdf

is integrablef

f is integrable if and only if

limsupAn , liminf An

k kn

nnn

AA1

suplim

k kn

nnn

AA1

inflim

Ω is a set and {An} is a sequence of

subsets of Ω. Define

limAn

nn

nn

AA

inflimsuplim

nn

A

lim

then we say that the limit of the sequence

{An} exists and has the common set as the

limit which is denoted by

If

An: monotone increasing

121 nn AAAA

n

nnn

AA1

lim

then

If

An: monotone decreasing

121 nn AAAA

n

nnn

AA1

limthen

If

Lemma 2.1

1nnA

)(lim1

nnn

n AA

be monotone increasing, then

If


nn

k

n

kn

kkk

kk

k

k

kk

kk

n

kkn

kkk

AB

BBA

disjoisBSince

BAandBAthen

kAABandALet

limlim

int,

,2,1\

1

111

111

10

Lemma 2.2

1nnA

)(lim1

nnn

n AA

be monotone decreasing, then

If


)(lim

)(lim)(

)\(lim)(lim\

1.2

sin

\

1

11

1

111

1

1

nnn

n

nnn

n

nn

nnk

kn

n

n

nn

AA

AAAA

AABBAA

LemmaBy

gincreamonotoneisBthen

AABLet

NnFor

Egoroff Theorem

A BAB ,

Let {fn} be a sequence of measurable function

and fn→f with finite limit on


then for any ε>0 , there is

such that μ(A\B)<ε and fn→f uniformly on B

N

mEx

N

nn

n

nmmn

nCx

ECTake

Nmxfxf

andEAtsNthen

AElemmaBy

AEthen

nxfxfAxElet

pf

xfxfandCAAC

tsCandNegerForClaim

N

)()(sup

)\(..

)()(lim1.2

,2,1)()(;

:

)()(sup,)\(,

.int,0,0:

.

1)()(sup

2

)\()\()\(

,

1)()(sup

2)\(

..

,0

1

11

1

BonuniformlyffHence

Nnm

xfxfAnd

EAEABA

thenEBtakeThen

Nnm

xfxfandEA

tsEwithAEandZNClaimBy

ZmFor

n

mnBx

mm

mm

mm

mm

mnEx

mm

mmm

m

Monotony Convergence Theorem

ffnn

lim

dffd n

nlim

Let {fn} be a nondecreasing sequence of

nonnegative measurable functions

Suppose


is a finite valued,then

Theorem(Beppo-Levi)

,sup dfnn

nasdffn

n0lim

Let {fn} be a increasing sequence of integrable

functions such that

fn f . Then f is integrable and ↗


00)()(

)(

int

int

sup

)(lim)(

)(lim0

11

1

1

11

111

nasdffdff

dff

andegrableisf

egrableisff

dfdf

dffdff

TheoremConvergentMonotonyBy

ffffandff

n

n

nn

nn

nn

n

Fatous Lemma

dfdf n

nn

ninfliminflim

Let {fn} be a sequence of extended real-value

d

measurable functions which is bounded from

below by an integrable function. Then


dfdg

dgdfdf

TheoremeConvergencMonotoneBy

functionegrableanbybelowfromboundedis

andgnondecreaisgthenfgLet

nn

nn

nnnk

kn

nn

nknk

n

inflimlim

liminfliminflim

.int

sin,inf

Remark

dfdf nn

nn

suplimsuplim

Let {fn} be a sequence of extended real-

valued

measurable functions and fn 0. Then ≦


dfdf

dfdf

dfdf

LemmaFatouBy

f

nn

nn

nn

nn

nn

nn

n

suplimsuplim

suplimsuplim

)(inflim)(inflim

0

Lebesque Dominated Convergence Theorem

gfn

dffd n

nlim

If fn ,n=1,2,…, and f are measurable functions

and fn →f a.e. Suppose that

a.e. with g being an integrable function.Then


dffd

dfdfdf

LemmaFatouByfunctionegrableby

abovefromandbelowfromboundedisfSince

nn

nn

nn

nn

n

lim

inflimlimsuplim

.int

Corollary

gfn

0lim dffn

n





0limlim

,

..0

dffdff

LDCTBy

fgffff

eaff

nn

nn

nn

n

5

The space Lp(Ω,Σ,μ)

For measurable function f, let

pp

pdff

1

pif 1

..:0inf eaMfMf

f is called the essential sup-norm of f.

Exercise 5.1

Show that

..eaff

..)(

\)(

,

\1

)(

0)()(,

\1

)(

0)(\)(

1..0

11

eafxfHence

Axfxf

havewenlettingBy

Axn

fxf

andAAthenAALet

Axn

fxf

AandAwhereAxMxf

andMn

ftsM

NnFor

nnn

n

n

nnnn

nn

Conjugate exponents

1, qpIf are such that

111

qp

then they are called conjugate exponents

Theorem 5.1(Hölder’s Inequality) p.1

1, qpIf are conjugate exponents,

qp

gfdfg then

qp

qp

thenLet

qandperchangesimplyotherwise

thatassumemayWe

oreitherNow

forqp

haveweforp

Since

qpthenIfClaim

qp

p

p

p

qp

11

111

1

1

11

1

1

11

1

,

.int

,1

.11

11

,01)1(

,,0:

qp

q

q

q

p

p

p

qp

q

q

q

p

p

p

qp

q

q

p

p

qp

gfdfg

dg

g

qd

f

f

pd

gf

gf

g

g

qf

f

pgf

gfhavewe

Claiming

gand

f

fthatNow

eagfhence

gfthatassumemayWe

111

11

..,

,,0


kiLf ipi 1)(

More generally, if f1,f2,…,fk are functions s.t

with

11111

21

kpppp

then


Pk Lffff 21

and

kpkpppffff

2121

21

21

21

21

221

21

1

21

1

221

2

1

21

1

1

2121

2

2

21

21

2121

21

21

21

21

2121

21

21

2121

)(

,111

,,2

ppp

p

p

p

p

pppp

ppppp

p

ppp

pppp

p

ppp

ppp

ppp

p

ppp

p

ppp

pp

ffff

ff

ff

ff

InequalityHolderby

ff

ffff

andpp

ppp

thenppp

andLfLfkIf

Theorem 5.2(Minkowske Inequality)

p1If f, g be measurable ,

whenever f+g is meaningful a.e. on Ω

pppgfgf

then

)11

)1

1((

)111

(

,1

.1

1

1)1(

11

1

pp

qp

q

pp

gfgf

gfgf

gfgf

qp

gfdgf

gfdgf

dggfdfgf

dgfgfdgfgf

thenpcasetheconsidernowWe

porpwhenobviousisIt

ppp

ppq

pp

p

ppq

p

p

pp

qp

pp

qqp

pp

ppp

p

is the family of all measurable function f with

pf

From Minkowski Inequjality, it is readily seen that

Lp(Ω, Σ, μ)

Lp(Ω, Σ, μ)

Lp(Ω, Σ, μ) is a vector space.

Theorem

pLp is a normed vector space

with

p1for

andLgf

gfgfgf

LgfgfgfthatshowtoNow

LfandRff

fhold

andLffthen

pSuppose

porpifobviousisIt

p

ppppp

pppp

p

pp

pp

)(2

,

)2(

0

,0)1(

1

.0

ppp

pppp

ppp

pp

pp

p

p

p

pp

pp

pp

gfgf

gfgf

gfgf

InequalityHolderbyand

pppp

p

p

gfgf

ggffgf

gfgfgf

1

1

1

11

11

1

1

11111

then

0:),,( p

p fLfN

if and only if f=0 a.e. on Ω

Let

Nf

then Lp(Ω, Σ, μ) is a vector space which

consists of equivalent classes of Lp(Ω, Σ, μ)

Lp(Ω, Σ, μ)

Lp(Ω, Σ, μ) =Lp(Ω, Σ, μ)/N

w.r.t the equivalent relation ~definded by

f~g if and only if f=g a.e. on Ω

is a Banach space.

p

Theorem 5.3

(Fisher)

Lp(Ω, Σ, μ) with norm

negligibleisEei

EthenEELet

NnmExk

xfxf

tsEsetnegligibleaistherethen

Nnmforff

thatsuchNistherekegerpositiveGiven

LinsequenceCauchyabefLet

pCase

k

kkmn

k

kkmn

k

pn

.

0,

,,\1

)()(

.

,

,int

.

:1

1

nasff

Nnk

ff

andLf

NnExk

xfxf

NnExk

xfxf

thenExxfxfLet

RinsequenceCauchyaisxf

ExFor

n

kn

kn

kn

nn

n

0

1

,\1

)()(

,\1

)()(

,\)(lim)(

)(

\

p

pk

p

kk

p

pnn

n

k

p

nnn

n

k

p

nnn

p

p

knn

kpnn

nn

pn

Lg

ff

ffffg

InqMinkowskiandeConvergencMonotoneBy

xffxgLet

kff

tsfoffesubsequencaisThere


pCase

kk

kkkk

kk

kk

k

12

1

2

1

limlim

.

)(

,2,12

1

.

.

1:2

11

11

1

1

11

1

1

),,(

,1

,

,2,1,2

int

sin

).,,(

1

1

1

1

1

1

1

p

k pnnp

knn

k

pnn

kk

p

n

Lghence

ffg

thatimpliesInequalityMinkowski

andTheoremeConvergencMonotone

ffgPut

kff

thatsuchegerspositive

ofnsequencegincreaanisThere

LinsequenceCauchya

befletandpthatAssume

pwhenobviousisThis

kk

kk

kk

kasffknowweagainLDCTby

thusoneagffffNow

LDCTbyLf

thatimplieskgffBut

eafinitefwith

fflyconsequentandeafinite

andconvergesffff

thatfollowsit

oneagSince

pn

p

n

p

n

p

nn

nk

nnkk

nn

k

k

k

k

kkk

0

,..

.),,(

,2,1,

..

lim..

lim

,..

1

1

11

1

.),,(

,2

arg

;,2

..int

,0

0

0

0

0

completeisLHence

ffffff

havewennifeconsequencaas

ffandnnthatso

ellysufficientkthenchoose

nmnwheneverff

tsnegerpositiveaisthere

Given

p

pnpnnpn

pnk

pmn

kk

k

Theorem

nasffandLfLfLetpn

ppn 0,,

kn

fthen there is a subsequence

such that

pn

n

Lhwithkeaxhxf

eaxfxf

k

k

.)()()2(

..)()()1(

eaff

nasff

andkaseaxfxf

tsfesubsequencaisthere

ThmpreviousofprooftheAs

LinsequenceCauchyaisf

nasff

pthatassumeMay

n

n

n

pn

pn

k

k

.

0

.)()(

.

0

1

*

*

*

NleaxhxfandLh

thenxxgxfxhLet

Nleaxgxfxf

Nleaxgxfxf

haveweklettingBy

Nlkxgxfxfxfxf

l

l

l

kklk

np

n

n

knnnn

.)()(

,)()()(

.)()()(

.)()()(

,

,)()()()()(

*

*

*

11

f is called an essential bounded function if

f

Exercise 5.4

L∞ (Ω, Σ, μ) is a Banach space.

fthen

eaxfxfthen

RinsequenceCauchyabexfthen

nmnwhenevereaffthen

nmnwheneverff

tsNnthen


nn

nn

mn

mn

nn

..)()(lim

)(

,..

,

..,0

),,(

1

0

0

0

1

Outer Measure

Outer Measure

0)()1(

R2:

BAifBA )()()2(

Ω ≠ψ: a set

μis called outer measure on Ω if

)()()()3(11

additivitysubAAn

nn

n

μ-Measurable p.1

AACandAB c \

)()()( BABAB C B

A subset A of Ω is called μ-measurable if

for any

i.e. for any

)()()( CBCB

Exercise 1.1

otherwise

setfiniteisAifAofycardinalitA)(

Let μ:2Ω→[0,+∞] be defined by

(μis called the counting measure of Ω and

every subset of Ω is μ-measurable.

Show that μis an outer measure

)()(

,inf:)(

)()(

,:)(

)2(

0)()1(

BA

thensetiniteanisBIfiiCase

BA

BofycardinalittheAofycardinalitthe

andsetfiniteaisA

thensetfiniteaisBIfiCase

BALet

thatobviousisIt

reoutermeasuanisBy

AA

NnsomeforsetiniteanisA

thensetiniteanisAIfiiCase

AAei

Aofycardinalitthe

Aofycardinalittheand

nsetfiniteaisA

thensetfiniteaisAIfiCase

thenofsequenceabeALet

nn

nn

n

nn

nn

nn

nn

nn

n

nn

nn

)3(~)1(

)(

inf

,inf:)(

)()(.

,2,1

,:)(

,)3(

11

1

11

1

1

1

1

.

)()()(

inf

,inf)(

)()()(

,)(

measurableisAHence

BABAB

andsetsiniteareBAorBA

thensetiniteanisBIfiiCase

BABAB

andsetsfiniteareBAandBA

thensetfiniteaisBIfiCase

Banyfor

ALet

measurableare

ofsubseteverythatshowtoNow

c

c

c

c

Exercise 1.2 p.1

SAthenSAIfiii

SBthenABandSAIfii

Si

nnnn

11 ,)(

,)(

)(

Let S be a subset of 2Ω having the following

properties:

Exercise 1.2 p.2

Define μ : 2Ω→[ 0, +∞] by

What are theμ-measurable subsets of Ω

Show that μis an outer measure

otherwise

SAifA

0)(

Exercise 1.2 p.2

Define υ : 2Ω→[ 0, +∞] by

What are theυ-measurable subsets of Ω

Show that υis an outer measure

otherwise

SAifA

1

0)(

11

1

1

0)(

,

,:1

2)(

)()(

,:2

0)()(

,:1

)(

0)(,)(

)1(

nn

nn

n

nn

nn

AA

andSAnallfor

thenSAIfCase

AanyForiii

BA

thenSBIfCase

BA

andSAthenSBIfCase

BAanyForii

SSincei

measureouteranisthatshowTo

SAorSAeither

ofsubsetsmeasurableaisAClaim

measureouteranisiiiiBy

AAthen

NnsomeforSA

thenSAIfCase

c

nn

nn

n

nn

:

)2(

.)(~)(

)(

,:2

11

1

)()()(

)(

,:2

0)()()(

,,:1

""

)()(

)(,

)()()(

":"

BABABthen

SBABABotherwise

SBAorSBAthenSBIfCase

BABABthen

SBABAthenSBIfCase

BsubsetanyFor

SAthatassumeMay

generalityofloseWithout

SAorSAeither

BAorBAeitherthen

BthenSBIf

BABAB

ofBsubsetanyforthen

ofsubsetsmeasurableaisAIfpf

c

c

c

c

c

c

c

c

11

1

1

)(1

,

,:1

2)(

1)()(

,:2

0)()(

,:1

)(

0)(,)(

)1(

nn

nn

n

nn

nn

AA

andSAnallfor

thenSAIfCase

AanyForiii

BA

thenSBIfCase

BA

andSAthenSBIfCase

BAanyForii

SSincei

measureouteranisthatshowTo

SAandSAthator

SAandSAthateither

ofsubsetsmeasurableaisAClaim

measureouteranisiiiiBy

AAthen

NnsomeforSA

thenSAIfCase

c

c

nn

nn

n

nn

:

)2(

.)(~)(

)(1

,:2

11

1

1)()()(

)(

,:2

0)()()(

,,:1

""

1)(0)(

0)(1)(

1)(,

)()()(

":"

BABABthen

SBABABotherwise

SBAandSBAthenSBIfCase

BABABthen

SBABAthenSBIfCase

BsubsetanyFor

SAandSAthatassumeMay

generalityofloseWithout

SAandSAthator

SAandSAthateither

BAandBAthator

BAandBAthateitherthen

BthenSBIf

BABAB

ofBsubsetanyforthen

ofsubsetsmeasurableaisAIfpf

c

c

c

c

c

c

c

c

c

c

c

Exercise 1.3

A

Suppose μis an outer measure onΩ and

then the restriction of μto A

denoted byμ A(B)=μ(A∩B) for ∣ B

Show that μ A is an outer measure on∣Ω and every μ-measurable set is also

μ A –measurable.∣

.

)()(

)()(

)(

)()()()(

,)(

0)()()()(

)1(

11

111

1

measureouteranisAHence

BABA

BABABA

BanyForiii

DADACACA

DCwithofDCsubsetsanyForii

AAi

measureouterisAthatshowTo

nn

nn

nn

nn

nn

nn

.

)()(

)()()()(

)()()(

.

)2(

measureAalsoisBHence

CBACBA

CABCABCACA

CBCBC

Canyforthen

setmeasurablebeBLet

measureAalsois

setmeasurableeverythatshowTo

c

c

c

Properties of Measurable sets p.1

Suppose μmeasures Ω.

(1) If A is μ-measurable, then so is Ω\A=Ac

(2) If A1, A2 are μ-measurable, then so is

A1 A∪ 2

))(())((

)()()(

,)()()(

,,)2(

)())((

)()()(

,)1(

2121

21211

11

21

BAABAA

ABAABABA

andBABAB

Banyfor

thenmeasurableareAAIf

measurableisAHence

BABA

BABAB

Banyfor

thenmeasurableisAIf

c

ccc

c

c

ccc

c


Remark :

By induction the union of finitely

many μ-measurable sets is μ-measurable.

This fact together with (1) implies that

the intersection of finitely many

μ-measurable set is μ-measurable.


B

1jjA is a disjointed sequence of(3) If

then

μ-measurable sets in Ω and

11 jj

jj ABAB

11

111

1

1

1

1

1

1

11

,

,int

jj

jj

n

jj

n

jj

jj

n

jj

n

n

jj

n

n

jj

n

jj

ABAB

hencenallfor

ABABABthen

AB

ABAB

ABABAB

thenegerpositiveabenLet


1nnA

1jjA is a disjointed sequence of(4) If

is μ-measurable .

μ-measurable sets in Ω, then

1

111

11

)(

\

\

,

njj

njj

n

jj

n

jj

jj

jj

ABB

ABABAB

ABAB

thenBLet

.

)(\

)(

,:2

)(\

,

,:1

1

11

11

1

11

1

measurableisAHence

BABABthen

ABABB

thenABIfCase

BABAB

haveweinequalityabovethein

nlettingbyABIfCase

jj

jj

jj

jj

jj

njj

jj

jj

njj


11 jj

jj AandA

1jjA is a sequence of(5) If

μ-measurable sets in Ω, then so are

.)1(

,

.)4(

,

1

11

1

1

1

11

measurableisAby

AASince

measurableisAby

AAASince

jj

c

jj

jj

jj

j

cj

iij

jj


sets in Ω, then (Ω, Σ, μ) is a measure

(6) Let Σ be the family of all μ-measurable

space.

Exercise 1.4 p.1

nnn

n AA

lim1

121 nn AAAA(i) If

is an increasing μ-measurable sets in

Ω, then

mm

m

nnn

mnn

m

nm

n

m

nmn

nnn

nn

nn

nn

nn

nnn

A

AAAA

BBBA

thenBA

andmeasurableofsequencedisjoaisB

thenAABandALet

lim

limlim

lim

,

int

,\

111

1

1111

11

1

10

Exercise 1.4 p.2

nnn

n AA

lim1

121 nn AAAA(ii) If

is a decreasing μ-measurable sets in

Ω with μ(A1)<+∞, then

nnn

nn

nnn

nn

nnn

n

nn

nnn

n

nnn

n

nn

nn

AA

AAAA

AAAA

AAAAAA

BBand

measurableofsequencegincreaanisB

thennforAABLet

limlim

limlim

lim\

lim\lim\

lim

sin

,,2,1\

1

11

1

11

1

111

1

1

1

1

Regular

BA

B

A measure μ on Ωis called regular if for each

there is a μ-measurable set

such that μ(A)=μ(B)

Exercise 1.5

nnn

n AA

lim1

121 nn AAAAIf

is a sequence of sets in Ω and μis a

regular measure on Ω, then

nnj

n

nnj

n

n

n

jj

jn

nn

nn

nnnn

nnj

nj

nj

j

nnj

n

nn

nnnn

nn

nn

AAhence

AAthen

nallforAAABut

A

BBBB

CCBA

CC

iExercisebytheninsetsmeasurable

ofsequencegincreaaisC

thenBBBCLet

BAthatsuch

ABsetmeasurableaisthere

NnanyforregularisSince

lim

lim

lim

lim)\(lim

lim

lim

)(4.1,

sin

),\(

)()(

,

1

1

11

11

111

1

1

11

Theorem

The family Σμ of μ-measurable subsets

of Ω is σ- algebra and μ=μ ︳ Σμ is

σ- additive . i.e.

(Ω, Σμ, μ) is a measure space.

Premeasure

premeasure

If Ω is a nonempty set,

G a class of subsets of Ω containing ψ, and

τ: G→[0,+∞] satisfy τ(ψ)=0.

τis called a premeasure.

Outer Measure constructed from τ

1

1

1

inf)(i

i

AC

GCCA

ii

ii

For a premeasure τ,

Define μ:2Ω→[0,+∞] by

Then μ measures Ω and is called the

outer measure constructed fromτby

Method I.

Example 2.1 The Lebesgue measure on Rn

Let G be the class of all oriented rectangles

in Rn with ψ adjoined and let

τ(I)=the volumn of I if I is an oriented

rectangle

τ(ψ)=0

the measure on Rn constructed fromτ by

Method I is called the Lebesgue measure

on Rn.

Exercise 2.1

GI

For ε>0, Let Gε be the class of all open oriented rectangles in Rn with diameter <εand

τε (I)=volume of I for

Show that the measure on Rn constructed

from τε by Method I is the Lebesgue

measure.

Exercise 2.2 p.1

Let μ be the Lebesgue measure on Rn

(i) Show that μ(I)=volume I if I is an open

oriented rectangle.

)(

)(,

)(

)()(

.tan

1

11

IIHence

IIIIBut

IIthen

II

IIwithIsequenceanyforthen

glerecorientedanbeILet

nn

nnnn

Exercise 2.2 p.2

(ii)Show that every open oriented rectangle

is μ-measurable and hence so is every

open set in Rn

( μ-measurable set is called Lebesgue

measurable set in this case.)

)(inf)(inf

)(inf)(inf

)()(inf)(inf)(

.tan

cn

IBIGI

n

IBIGI

cn

BIGI

n

BIGI

cnn

BIGI

n

BIGI

n

n

III

IIII

IIIIIB

RofBsubsetanyFor

RinglerecorientedanbeILet

c

n

n

n

n

n

n

n

n

n

n

n

n

Exercise 2.2 p.3

(iii) If A and B are subsets of Rn and

dist(A,B)>0, then

μ(A B)=μ(A)+μ(B)∪

Metric spaces

Metric Space

Mzyxzyyxzxii

yxyx

andMyxxyyxi

,,),(),(),()(

0),(

,0),(),()(

Let M be a nonempty set and

ρ:MXM→[0, ∞) satisfies

ρis called a metric on M

(M,ρ)is called a metric space.

Example 1 for Metric Space

nn

n

ii

n

Raaaif

aawhere

Ryxyxyx

,,

,),(

1

21

1

2

Let M=Rn and let


nii

niRyxyxyx

,max),(

1

Let M=Rn and let


],[,)()(max),( baCgftgtfgfbta

Let M=C[a,b] (-∞<a<b<∞) and let


)(,)()(max),( KCgftgtfgfKx

Let M=C(K), where K is a compace set in Rn

and let

Unless statement otherwise, C(K) will denote

the metric space with the metric so defined.


pifgf

pifdgfgf

LgfFor

pp

p

1),(

),,(,1

Let M= Lp (Ω, Σ, μ)and let

Converge

Mx 0

nxNn 0

Let (M, ρ) be a metric space.

A sequence

0nn

is said to converge to

if for any ε>0, there is

such that ρ(xn ,x0)<ε whenever

Since x0 is uniquely determined, x0 is denoted by limn→∞ xn

If limn→∞ xn exists, then we say that{ xn } converges in M.

Example 3.1 for converge

)(xfn

1,0Cfn converges if and only if

converges uniformly for 1,0x

Cauchy sequence

Nn 0

Mxn A sequence

is called a Cauchy sequence if for any

ε>0 there is

0,),( nnmwheneverxx nm

such that

Exercise 3.1

nx

Mxn Show that if converges, then

is a Cauchy sequence.

22

),(),(),(

,2

),(

0,lim

00

0

00

0

0

mnmn

n

nn

xxxxxx

nnmanyforthen

nnwheneverxx

thatsuchNnisthere

anyforthenxxIf

Complete Metric space

A metric space M is called complete

if every Cauchy sequence in M converges

in M

Examples for Complete

)(KC

nRK (1) Let be compact, then

is complete.

(2) Lp(Ω, Σ, μ) is complete.

Normed vector space p.1

Ex

x

Let K=R or C and let E be a vector space

over K. Suppose that for each

there is a nonnegative number

associated with it so that


Eyxyxyxiii

ExKxxii

xxi

,)(

;,)(

00)(

Then E is called a normed vector space

(n.v.s) with norm


Eyxyxyx ,),(

Then ρ is a metric on E and is called

the metric associated with norm

Let E be a n.v.s and

Unless stated otherwise, for a n.v.s., we

always consider this metric.

Banach space

Both C(K) with K a compact subset of Rn

and Lp(Ω, Σ, μ) are Banach spaces.

A normed vector space is called a

Banach space if it is a complete metric space

Continuous mapping

10 Mx

),())(),(( 0102 xxwheneverxTxT

Let M1 and M2 be metric spaces with

metrics

ρ1and ρ2 respectively.

A mapping T: M1→M2 is continuous at if for any ε >0 , there is δ>0 such that

Open and Closed set in a metric space

Gx

),( yxwheneverGy

A set G in a metric space is called an open

set if there is δ>0 such that

The complete of an open set is called

a closed set.

Exercise 3.2 p.1

10 Mx

2121

2022

)(:)(

,)(

GxTMxGT

settheGxTwithMG

Let M1 and M2 be metric spaces with metrics

ρ1and ρ2 respectively and let T: M1→M2

(1) Show that T is continuous at

if and only if for any open set

contains an open subset which contains x0 .

.

),())(),((

),()),(()(

),())),(((

))),(((),(

..0

))),(((

,)),(()(

)),((,0""

)())),(((),(

),()),(()(

..0,

)),((..0

,)(

""

0

0102

010

0101

01

10

10

01

1

00

20

21

01

0

00

0

20

2022

xatcontinuousisTHence

xxwheneverxTxT

xxwheneverxTBxT

xxwheneverxTBTx

xTBTGxxB

tsthenGxwith

xTBTGsetopenanisthere

thenxTBxTwith

MinsetopenisxTBFor

GTxTBTxB

xBxwheneverxTBxT

tsxatcontinuousisTSince

GxTBts

thenGxTwithMinsetopenanbeGLet

Exercise 3.2 p.2

22 MG

)( 21 GT

(2) Show that T is continuous on M

if and only if for any open set

is an open subset of M1 .

.

),())(),((

),(),()(

)),((),(..0

)),((

),(

0,""

)(

)(),(

),()()),((

),()),(()(

..0,

)),((..0

)(),(

""

1

0

0102

00

01

0

101

0

20

10

121

21

0

021

01

00

0

20

2021

0

22

MoncontinuousisTTherefore

xatcontinuousisTHence

xxwheneverxTxT

xBxxTBxT

xTBTxBts

MinopenisxTBTx

MofsubsetopenanisxTB

anyforthenMxLet

MofsubsetopenanisGTHence

GTxB

xBxGTxTBTx

xBxxTBxT

tsxatcontinuousisTSince

GxTBts

GxTthenGTxLet

MinsetopenanbeGLet

Exercise 3.3

.

,)(

;,)(

setindexanyisIwhere

OAthenOAIfii

OBAOBAi

IiiIii

Let O be the family of all open

subsets of a metric

space. Show that

OBAHence

openisBA

BAxB

BxBandAxB

thenLet

BxBandAxB

ts

BxandAx

BAxanyFori

),(

),(),(

,,min

),(),(

..0,

)(

21

21

21

BABA

OAHence

openisA

AxB

AxBtsthen

IisomeforAx

thenAxLetii

Iii

Iii

Iii

i

i

Iii

),(

),(..0

,)(

4.Carathéodory measure

Carathédory Measure

)()()( BABA

0),(inf),(

yxBAdistByAx

If Ω is a metric space, then a measureμ is c

alled Carathédory measure if

whenever

Example 4.1

The Lebesgue measure on Rn is a Carathéd

ory measure.

Lemma 4.1

121 nn AAAALet

nnn

n

cnn

AA

ThenAA

sup

.0),(

1

1

be an increasing sequence of subsets of Ω

and for each n

thenDDdisthavewe

nmandnanyforassumptionBy

AADAADADLet

Athatassumemaywe

AAthatshowTo

AAObviously

mn

nnn

nn

nnn

n

nnn

n

,0),(

2,

,\,,\,

sup

,sup

sup,

112211

1

1

nnj

j

jnj

n

njjn

njjn

njjn

jj

ii

ii

nn

k

k

iik

AA

havewenlettingby

DA

DA

DAAAA

NowDSimilarly

DThenkeachfor

AA

DDDD

sup

,

.

.

sup

1

1

1

111

12

112

12

1121231

Theorem 4.1

If μ is Carathédory measure onΩ, then

every closed subset of Ωis μ- measurable.

BABABA

havewenlettingbythusneachfor

BABABAand

BBB

BtoappliedLemmabyhence

nnBBBdistNow

BBBB

ObviouslyBB

haveweclosedisFceThen

nFxdistBxB

letNneachFor

FBandFA

letandsetclosedabeFLet

nn

nn

nnn

n

nn

nn

nn

n

c

)()(sup)(

,,;

)()(

sup

),,(1.4

0)1(

1)\,(

,.

,sin,

1),(;

,

,

1

1

121

1

Borel sets

B(Ω) is the smallest σ-algebra of subsets

Elements of B(Ω) are called Borel sets of

Ω.

of Ω that contains all closed subsets of Ω

Corollary 4.1

If μ is Carathédory measure onΩ, then

all Borel subsets of Ωareμ- measurable.

Lebesgue Measure p.1

RA

n

n

AII

I

n

n

inf

Ω =R , I: open finite interval of R

Define L(A)

then L is a Carathédory measure.

L is called the Lebesgue measure.

Lebesgue Measure p.2

RA

nIII 21

(R, ΣL ,L)

Similar construction on Rn with

I replaced by n-dimensional intervals

Ln is a Carathédory measure.

Ln is called the Lebesgue measure on Rn .

Regularity of Measure

Regular measure

BA

B

A measure μ on Ωis called regular if for each

there is a μ-measurable set


Borel regular measure

BAB

A measure μ on Ωis called Borel

if every Borel set is μ-measurable.

It is called Borel regular if it is Borel an

d for every


there is a Borel set

Radon measure

A measure μ on Ωis called Radon

measure if it is Borel regular and

μ(K)<∞ for each compact set K.

We already known that Carathéodory m

easure is Borel .

Theorem 6.1

ALet μ be a Borel regular on a metric

space Ω and suppose

is μ-measurable andμ(A)<∞

Then μ ︱ A is a Radon measure.

CDthen

CCAEAEADD

AEADSince

DACAC

andsetBorelaisDthenAEDLet

CAEthatsuch

CAEsetBorelaisthereCnowLet

BorelisAthatassumemayWe

CBCHence

CACABAC

ACBACBCBCB

haveweCFor

BAAB

BAtsABsetBorelaisThere

regularBorelisthatshowtoremainsIt

measureBorelais

setmeasurableissetmeasurableeverySince

setcompacteveryforKClearly

ALet

c

c

)(

)(

,

,

.

,

.

\

\

,

0)()(\

)()(..

.

.

,

.

.

Measure Theoretical Approximation of Sets in Rn

Lemma 7.1 p.1

CB \

BC

Let μ be a Borel measure on Rn and

B is a Borel set

(i)If μ(B)<∞ , then for each ε >0 there is a

closed set such that

Lemma 7.1 p.2

BU \

BU (ii) If μ is Radon measure , then for each

ε >0 there is an open set

such that

1

111

1

11

\

\\\

,

2\

,0

,:1

\

0

.,

iii

iii

ii

ii

ii

iii

ii

iiii

n

CA

CACACA

andclosedisCthenCCLet

CAthatsuch

ACclosedaisthereieachforFix

AAthenAIfClaim

CAthatsuchACset

closedaisthereeachforthatsuchRA

measurablethosealloffamilythebeLet

setBorelfiniteaisvALet

CAandAC

thenCCletandCA

thatsoellysufficientmChoose

CA

CACACA

ThenClaimof

prooftheinasiCChooseFor

AAthenAIfClaim

m

ii

m

ii

iii

iii

ii

ii

m

ii

m

i

iiii

\

,\

arg

\

\\\lim

.1

,2,1,0

,:2

00

11

0

1

1111

11

GAhence

AAClaim

ASince

AClaim

GAAthenGAIfClaim

setsopenallcontainsGandGAGAThen

AAGnowLetsets

openallcontainsthatClaimbyfollowsitsets

closedofunioncountableaswrittenbecanset

openeveryandsetsclosedallcontainsSince

i

ci

c

ici

iiii

c

c

1

1

11

1

,

.2

,:3

:.

2

.0

),0(int

)(

)(

.\\

0,,

.

mradiusandcenterwithballopenthe

BUletmegerpositiveeachfor

iiproveTo

iprovesThis

CBCB

thatsuchBCsetclosedaisthere

forhenceGBparticularIn

setsBorelallcontainsGThen

mm

1

1

11

1

\\

\\\,

\

,\

2\\\\

..\)(

0\

\

mmm

mmm

mmm

mm

mmm

mmmmm

mm

mm

m

BCU

BCUBUNow

UCUBUB

andopenisUthenCUULet

CBUBCU

tsBUCsetclosedaisthereiby

forsoandUBUwith

setBorelaisBUThen

Theorem 7.1Approximation by open and compact sets

nRA

openisGAGGA ,);(inf)(

Let μbe a Radon measure on Rn, then

(1) For

(2) If A is μ-measurable on Rn, then

compactisKKAKA ,);(sup)(

.

),(

,,:inf

,:inf)(

)(

.

.)(

,\

,\..

01.7

.

.)(

obviousisinequalityreversethebecause

iestablishwhich

openisUAUU

openisUBUUBAThen

BAwithABsetBorelaisThere

arbitaryisAnowLet

holdsithatshowwhich

AAUAUhence

AUtsAUsetopenanisthere

forLemmaBy

setBorelaisAthatfirstSuppose

AthatassumemayWei

closedisCACCAhenceand

CAwhichfrom

UCCRCA

andACclosedisCUCLet

UwithAUsetopenanisthere

givenforiBy

measureRadonaisTheoremBy

APut

AwithmeasurablebeALetii

cn

c

c

,:)(sup)(

)(0

,\\

,,,

,0)(

.1.6

,

)(

kkkkk

k

kk

k

k

ACwithACsetclosedaisthere

aboveprovediswhatBy

AmeasureRadonaisSince

AA

andmeasurableisAeachThen

kkxkAxAletAIf

2

1

,

)(,

,2,1,1:,

1

,2,1:sup

,:sup

.

,,

2

1lim

1

1

1111

1

nC

compactisKAKKThus

neveryforCisso

compactisCeachboundedisCeachSince

ACCC

andACNow

n

kk

n

kk

kk

kkk

kk

kk

n

kk

n

kk

Exercise 7.1 p.1

nRA

AH

(i)Show that the Lebesgue measure on Rn

is a Radon measure.

(ii) Let show that there is a Gδset

such that Ln(H)=Ln(A),where

Ln denotes the Lebesgue measure on Rn.

.1.4

,

.)(

BorelisLCorollaryby

measureCaraeodoryaisLSince

ROnmeasureLebesguethebeLLeti

n

n

nn

Exercise 7.1 p.2

nRA

AM

(iii) Let be Lebesgue measure

show that there is a Fσset

with Ln(M)=Ln(A).

Exercise 7.1 p.3

RRf n :(iv) Let be Lebesgue measurable

show that f is equivalent to a Borel

measurable function.

Theorem

nRA

openisGAGGA ,);(inf)(

Let μ=Ln be the Lebesgue measure on Rn, then(1) For

(2) If A is Lebesgue measurable on Rn, then

compactisKKAKA ,);(sup)(

(A, Σ︳ A, μ)

A

dffd AA

(Ω, Σ, μ): measure space

(A, Σ︳ A, μ ) is a measure space

Lp(Ω)nR

Lp(Ω, Σ,μ)= Lp(Ω)

Σ: the family of Lebesgue measurable

subsets of Ω

μ: the Lebesgue measure

Cc(Ω)

)( cCf then f is continuous and

Cc(Ω) is the space of all continuous

functions with compact surport in Ω i.e. if

closurexfx 0)(;

is a compact set in Ω

Lemma

)( cCg

nR

B

such that

Let B be a measurable subset of Ωwith

=Lp(B)<∞, then for any ε>0, there is

pBg

ppp

c

BG BFG B

FG BFG B

F BB

c

p

p

FBFG

BGFG

dxgdxg

dxgdxg

dxgdxgthen

Gxxg

Fxxg

g

satisfyingfunctioncontinuousabegLet

FGthatsuchGGF

withGsubsetopenanisthere

thenandFB

tsofFsubsetcompactaisthere

anyFor

c

c

22\\

\

,0)(

,1)(

;10

2

2

..

,0

\

\\

\

Corollary

)( cCg

ii

k

iBi Bandf

i0,

1

such that

Let

be a simple function on Ω, then for any ε>0

p

fg

i

k

ii

k

i pBii

k

i pBiip

k

iBiip

c

k

iii

ipBici

kg

ggfg

andCgthenggLet

kgtsCg

kiFor

i

ii

i

11

11

1

)(

)()(

)(,

..)(

,,2,1

Theorem

Cc(Ω) is dence in Lp(Ω), 1 p<∞≦

nR

0)()(lim)()(lim

min

)(2)()(

)()()()(lim

..

:Pr

2

..)(2

,0:

)(

1

pk

k

pk

k

pppk

kkk

kk

ppp

p

c

p

p

xuxfxuxf

TheorematedDoLebesgue

xforxuxuxfthen

xforxuxfandxuxf

ts

functionssimpleoffsequenceaisThere

Claimofoof

gffugu

gf

tsCgisthere

futhatsuchonffunction

simpleaisthereanyForClaim

LuLet

Chapter IV

Lp space

IV 1

Some result for integration

which one must know

Theorem(Beppo-Levi)

,sup dfnn

nasdffn

n0lim

Let {fn} be a increasing sequence of integrable

functions such that

fn f . Then f is integrable and ↗


Lebesque Dominated Convergence Theorem

gfn

dffd n

nlim





Fatous Lemma

dfdf n

nn

ninfliminflim

Let {fn} be a sequence of extended real-value

d

measurable functions which is bounded from

below by an integrable function. Then


Theorem IV.3 (Desity Theorem)

Cc(Ω) is dense in Lp(Ω), 1 p<∞≦

nR

Theorem IV.4(Tonelli)

openRN :1

1 openRN :2

2

measurableRF :: 21 1..),(

2

oneadyyxF

dxyxFdydyyxFdx

1221

),(),(

)( 211 LF

Suppose that

and that

Then

Theorem IV.5(Fubini) p.1

)(),( 21 yLyxF

1x

)(),( 11

2

xLdyyxF

)( 211 LFSuppose that

and that

then for a.e.


)(),( 11 xLyxF

2y

)(),( 21

1

yLdxyxFand that

Similarly, for a.e.


21

1221

),(

),(),(

dyxF

dxyxFdydyyxFdx

Furthermore, we have

IV 2

Definition and elementaryproperties of the space Lp

Exercise 5.1

Show that

..eaff

..)(

\)(

,

\1

)(

0)()(,

\1

)(

0)(\)(

1..0

11

eafxfHence

Axfxf

havewenlettingBy

Axn

fxf

andAAthenAALet

Axn

fxf

AandAwhereAxMxf

andMn

ftsM

NnFor

nnn

n

n

nnnn

nn

Chapter 3 L p -space. Preliminaries on measure and integration.

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Transcript of Chapter 3 L p -space. Preliminaries on measure and integration.