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Lecture 9

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Lecture 9L13

1 Review of signed measuresÞ

Recall: is a measurable space; Ð\ß ÑU /is a signed measure on if is aÐ\ß ÑU /‘ - valued set function such that

is countably additivea) / b) / 9Ð Ñ œ ! there are no sets such that c) EßF , ./ /ÐEÑ œ ∞ ÐFÑ œ ∞

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Definition 1: E is a set if allpositivesubsets of have non-negative measure.E

Definition 2: E is a if all subsets null setF E ÐFÑ œ ! of have /

[We will restate but not prove lemmas.]

Lemma A) If are positive, thenE8

is positive.-8œ"

∞

8E

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Lemma B) If , there's a/ÐIÑ !positive set , such thatE § I

./ÐEÑ !

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2. Hahn Decomposition Theorem:

If is a signed measure on / `Ð\ß Ñß bpositive set , negative set E F

such that , These\ œ E ∪F E ∩ F œ Þ9sets are unique up to possible changes bynull sets.

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Proof: Assume (without loss of generality)that is omitted by .∞ /

Let

- /œ ÐEÑsupE positive

Ê Ethere exist sets that get measure8

arbitrarily close to i.e., such that-ß

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/ -ÐE Ñ Å88Ä∞

.

Let E œ E Þ-8œ"

∞

8

The set is positive by previous lemma, soE

/ -ÐEÑ Ÿ Þ

But: / /ÐEÑ   ÐE Ñ3

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Ê ÐEÑ œ/ -

Let . Then if F œ µ E I § F is positiveand has positive measure we have

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/ / / /ÐI ∪ EÑ œ ÐIÑ ÐEÑ ÐEÑ

which is a contradiction since there are noßpositive sets of measure greater than .-

So: there are no positive sets in thatI Fhave positive measure.

is negative. Ê F

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Definition 3: The above decomposition isknown as the Hahn decomposition

Now: take : for any measurable set we/ Ihave

/ / /

/ /

ÐIÑ œ ÐI ∩ EÑ ÐI ∩ FÑ

´ ÐIÑ ÐIÑ

Can check that as defined above are/ / ßstandard measures, denoted as the positiveand parts of negative /

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Definition 1. This decomposition / / /œ

is called the of .Jordan Decomposition /

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3 Radon-Nykodym TheoremÞ

Definition 2: If and are/ .measures on the -algebra in the set ,5 ` \then is with/ absolutely continuous regard to (written << ) if, whenever. / .. /ÐEÑ œ !ß ÐEÑ œ ! it follows that .

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Example 1: Let on0ÐBÑ   !measure space .Ð\ß ß Ñ` .

Define the measure by:/

/ .ÐIÑ œ 0 .(I

Can show that is a measure (exercise)/and <</ .Þ

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To show that is a measure, can use:/

/Œ '∪ I œ 03œ"

∞

3 ∪I

ßdisjoint

3

œ 0 . œ ÐI Ñ(3œ" 3œ"

∞ ∞

I3

3

. /

where third equality can be proved usingmonotone convergence theorem.

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Example 2: Let Lebesgue. œmeasure on ‘

( , ) Lebesgue measure space‘ ` .ß œ

Let be the measure on / ‘ `Ð ß Ñdefined by

/ ÐIÑ œ / .B(I

B#

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Then is a measure, / and

<< . / / .ÐIÑ œ ! Ê ÐIÑ œ ! Ê

The measure is denoted as / Gaussianmeasure

Definition 3: If and are 2/ /" #

measures, then and are / /" # mutuallysingular

if , , and\ œ E ∪ F E ∩ F œ 9

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/ /" #ÐFÑ œ ! ÐEÑ œ !,

Write: / /" #¼ Þ

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Example 4. Consider Lebesgue. œ 7 œmeasure and counting measure on the/ œintegers, i.e.

/ $ÐIÑ œ I œ ÐIÑÞ# integers in 5œ∞

∞

5

where for fixed integer the measure is the5 $5point mass measure at , i.e. is the5 $5measure defined by

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$5ÐIÑ œ" 5 − I!œ if

otherwise.

Then can show that .. /¼

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Back to Example 1:

Example 1: on 0ÐBÑ   ! \

/ .ÐIÑ œ 0 .(I

then << / .Þ

Now we now show that in fact absoluteeverycontinuous measure arises in this way.

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Radon-Nykodym Theorem: Let , Ð\ ß Ñ` .be a - finite measure space. If << 5 / .and on then function . Á ! \ß b 0such that for all (meas.) sets ,I

/ .ÐIÑ œ 0ÐBÑ .(I

. (1)

The function is unique; i.e., if also0 1satisfies equation (replacing by ),(1) 0 1then a.e.1 œ 0

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Proof: Assume is finite without loss\of generality (easy extension to -finite).5

Can also assume without loss that does not/vanish everywhere.

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Step 1: Show that there is a function0ÐBÑ   ! such that

( (\ I

0ÐBÑ. ! 0. Ÿ ÐIÑ aI − Þ. . / ` and

(2)

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Proof of Step 1:Consider the measure for Let/ -. - !ÞÐT ßR Ñ- - be the Hahn decomposition of/ -. T R, with and positive and- -

negative disjoint (and exhaustive) sets for\.

Claim: there is a for which .- .ÐT Ñ !-

Pf. of claim: Assume otherwise. Then. / -ÐT Ñ œ ! ÐT Ñ œ ! !- - and thus for all .

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Thus for all . Thus forÐ ÑÐT Ñ œ ! !/ -. --

any measurable , we haveIÐ ÑÐIÑ Ÿ ! !/ -. - for all .

Letting , we get in the limit that- Ä !/ /ÐIÑ Ÿ ! ÐIÑ œ ! I, i.e., for all , which is acontradiction of initial assumption. Thusthere is a such that , and- .! ! ÐT Ñ !-!

above claim is provedÞ

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Now define Easy to see0ÐBÑ œ ÐBÑÞ- ;! T-!

that . Also, since is a'\0. ! T. -!

positive set for it follows every/ - . !

subset of is non-negative, so forT-!

I − ß` we have

Ð ÑÐT ∩ IÑ   !Þ/ - .! -!(3)

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Thus

( (I I

! T !0. œ ÐBÑ œ ÐT ∩ IÑ. - ; - .-! !-

Ÿ ÐT ∩ IÑ Ÿ ÐIÑÞby (3) above

/ /-!

Thus it follows that above holds for this ,(2) 0as desired, completing Step 1.

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Step 2: Define Y . / `œ Ö0 À 0. Ÿ ÐIÑ aI − ×Þ'

I

Let Claim that is achievedQ œ 0. Þ Qsup0−

\Y

' .

by some and that this is the satisfying0 0(1) (i.e. R-N Theorem).

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Proof of Step 2: Easy to show that if are0 ß 1in , so is . Select sequenceY 2 œ Ö0ß 1×maxÖ0 × § 0 . œ Q8 8

8Ä∞ \ 8Y . such that . Canlim 'assume they are pointwise increasing, forotherwise replace 0 Ä Ö0 ßá ß 0 ×Þ8 " 8max

Define . By Monotone0ÐBÑ œ 0 ÐBÑlim8Ä∞

8

convergence . Also can show'\0. œ Q.

0 − Y using Monotone convergence.

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Define for I − `

( / .ÐIÑ œ ÐIÑ 0.   !ß(I

by definition of . Easy to show is a (non-0 (negative) measure.

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Claim: on all measurable sets (proving ( œ !(1) and the theorem).

To see this, assume it is false. Then by thesame proof as that of (note << )Step 1 ( .

there exists function such that for0ÐBÑ   !s

I − ß`

( ( (\ I I

0. ! 0. Ÿ ÐIÑ œ ÐIÑ 0.s s. . ( / . and

(4)

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This gives an easy contradiction

[it shows that (since we can just0 0 −s Yrearrange ) and we know that(4) '\0. œ QÞ. ]

[This gives a contradiction of the choice of ,0since , but we now know that0 0 −s Y' Š ‹

\0 0 . Qs . . However it was given

earlier that gives the largest value of0 − Y'\0 . œ Q Þ. ]

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Uniqueness of also follows easily. 0

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L154. Recall Radon-Nikodym theorem:If << then there exists function/ . 0 ´ .

./.

such that for all sets Iß

/ .ÐIÑ œ 0ÐBÑ .(I

Recall definition: if . /¼ \ œ E ∪Fß E ∩ F œ ß9and . /ÐFÑ œ ! à ÐEÑ œ !

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5. Lebesgue Decomposition Theorem:Let -finite measure space,Ð\ß ß Ñ œ` . 5

and be another measure besides on/ .`. Then

/ / / œ ! " ß ß

non-negative non-negative

where/ .! ¼

and/ ." << .

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are unique/ /! "ß

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Proof: Let - . /œ

Ê . -<< / -<<

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Thus there exist , such that0ÐBÑ   ! 1ÐBÑ   !for all ,I

. -ÐIÑ œ 0 .(I

/ -ÐIÑ œ 1 .(I

.

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Let E œ ÖB À 0ÐBÑ !×F œ ÖB À 0ÐBÑ œ !×

Note , are disjoint, andE F

. -ÐFÑ ´ 0 . œ !(F

since on .0 œ ! F

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Define: / /!ÐIÑ œ ÐI ∩ FÑ

/ /"ÐIÑ œ ÐI ∩ EÑ

Then and we know / .!ÐEÑ œ ! ÐFÑ œ !

Ê ¼ Þ/ .!

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Also for any set if , then claimI ÐIÑ œ !./"ÐIÑ œ !à

This is because /"ÐI ∩ FÑ œ !automatically,

and / / -"ÐI ∩ EÑ œ ÐI ∩ EÑ Ÿ ÐI ∩ EÑÞ

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But by above

! œ ÐI ∩ EÑ œ 0ÐBÑ. ÐBÑ. -(I∩E

. (1)

Note on , so by 0ÐBÑ ! I ∩ E (1) we have

-ÐI ∩ EÑ œ !

Ê ÐI ∩ EÑ œ ! Ê ÐIÑ œ !ß/ /"

proving claim that when ,/ ."ÐIÑ œ ! ÐIÑ œ !so that << ./ ."

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Example 2: Let \ œ Ð!ß "Ñ Þ

Let the set .G œ Ö ß ×" #$ $

Let be Lebesgue measure on .7 \

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Define measure

.ÐIÑ œ B.B(E

.

Define

(ÐIÑ œ ÖI ∩ G×# points in

(i.e., or ).!ß "ß #

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Define measure

/ ÐIÑ œ 7ÐIÑ Ö I ∩ G×# points in

´ 7ÐIÑ ÐIÑÞ(

Now want to find the Lebesgue decomposition/ / / / .œ ! " of with respect to .

First define measure

- . /ÐIÑ œ Ð ÑÐIÑ

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œ Ð" BÑ.B Ö I×(E

# points in

Since << , by R-N theorem there is a. -function such that for all sets ,0ÐBÑ   ! I

. - . /ÐIÑ œ 0ÐBÑ . œ 0ÐBÑ . 0ÐBÑ .( ( (I I I

œ B 0ÐBÑ .B 0ÐBÑ .B 0ÐBÑ .( ( (I I I

(

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What is ?0ÐBÑ œ ...-

First: look at last term:

(ç

Š ‹ Š ‹çI

0ÐBÑ . œ 0 0 œ !Þ" #

$ $(

if is in if is in "$ #

$I

I

[otherwise changing by one point canIchange ( - impossible!]. IÑ

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Thus we want 0 œ 0 œ !Š ‹ Š ‹" #$ $

So want .ÐIÑ œ 0ÐBÑ ÐB "Ñ .B'I

But:.ÐIÑ œ " .B'

I

Ê 0ÐBÑÐB "Ñ œ "

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Ê 0ÐBÑ œB Á ß

! B œ ß

" " #B" $ $

" #$ $

ÊE œ ÖB À 0ÐBÑ !× œ B À B Á ß" #

$ $

F œ ÖB À 0ÐBÑ œ ! œ ß" #

$ $

œœ

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We can now get Lebesgue decomposition of /with respect to ..

This has parts and , which are measures/ /" !

defined by/ /" ÐIÑ œ ÐI ∩ EÑ œ 7ÐIÑ

[note excludes ]E Ö ß ×" #$ $

/ /! ÐIÑ œ ÐI ∩ FÑ œ

(ÐIÑ œ Ð Ö ß × I×Þ# points in set contained in " #$ $

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So:/ (! œ/" œ 7 Ð Ñ i.e., Lebesgue measure .