ma711LecturesN16Arial - Boston Universitymath.bu.edu/people/mkon/MA711/L9RA.pdf · 2020. 10. 1. ·...
Transcript of ma711LecturesN16Arial - Boston Universitymath.bu.edu/people/mkon/MA711/L9RA.pdf · 2020. 10. 1. ·...
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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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398
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 .