Analytic sets

Updated April 15, 2020

Goals 2

Plan:Continuous Lebesgue non-measurable functionAnalytic sets: 3 basic characterizationsAnalytic vs Borel, Lusin’s separation theoremUniversal analytic setUniversal measurability

(Non)-measurability of continuous functions 3

Fact: f : R Ñ R continuousñ Borel measurable

Question: Is f Lebesgue measurable?

Answered in:

Lemma

Assuming the Axiom of Choice, there exists a continuous, strictlyincreasing f : R Ñ R that is not LpRq{LpRq-measurable.

Idea: By Banach-Zarecki, there is f continuous, strictlyincreasing s.t.

DE Ď R : λpEq “ 0 ^ λ‹pf´1pEqq ą 0

Now pick a non-measurable subset of E. Need . . .

Dense non-measurable set 4

Lemma

Assuming the Axiom of Choice, there exists A Ď R s.t.

@E P LpRq : λpEq “ λ‹pAX Eq “ λ‹pE r Aq.

In particular, A R LpRq and, in fact,

@E P LpRq : λpEq ą 0 ñ EXA R LpRq.

Construction of A 5

Pick α R Q and set

x „ y :“ Dm, n P Z : x´ y “ m` nα

Use AC to choose representatives using φ : trxs : x P Ru Ñ Robeying φprxsq P rxs for all x. Then set

A :“

2m` nα` φprxsq : x P R ^ m, n P Z(

.

If λ‹pE r Aq ă λpEq then find B P LpRqwith

E r A Ď B ^ λ‹pE r Aq “ λpBq

Now set F :“ E r B and note F Ď A.As λpFq ą 0, F´ F contains open interval around 0. So doesA´A and, since t2m` 1` nα : m, n P Zu is dense in R,

pA´Aq X t2m` 1` nα : m, n P Zu ‰ H

Construction of A continued . . . 6

But x P A´A means x “ 2m1 ` n1α` z´ z1 for somez, z1 P tφprysq : y P Ru and x “ 2m` 1` nα then forces z „ z1and, since each class represented uniquely, so z “ z1. Thus

2m1 ` n1α “ 2m` 1` nα

As α R Q, this forces n “ n1 and 2m1 “ 2m` 1, which is absurd!

So λ‹pE r Aq “ λpEq. The argument for EXA similar:

A1 :“

2m` 1` nα` φprxsq : x P R ^ m, n P Z(

obeys A1 “ R r A and so EXA “ E r A1.

Non-measurable function 7

F :“ Cantor function (a.k.a. Devil’s staricase) and C :“ Cantorternary set. Define

Gpxq :“ x` Fpxq

Then µG “ λ` µF and µGpBq “ λpGpBqq shows λpGpCqq “ 1.For A as above,

AXGpCq R LpRq

yetB :“ G´1pAXGpCqq obeys λpBq “ 0

so B P LpRq. For f :“ G´1 we get

f´1pBq “ AXGpCq R LpRq

Measurability questions 8

The example shows BpRq ‰ LpRq and motivates

F :“č

f : RÑRcontinuous

E Ď R : f´1pEq P LpRq(

Then F “ σ-algebra s.t. all continuous functions areLpRq{F -measurable. We know

BpRq Ď F Ĺ LpRq

but is F ‰ BpRq?

And what if LpRq is replaced by F ; do we get a propersub-σ-algebra of F? And if so, what if we iterate?

And, for what σ-algebras F is every continuous f : R Ñ RF{F -measurable?

Lebesgue’s fallacy 9

Question: Is the continuous image of a BpRq-set in BpRq?

Lebesgue (1905): yes for projections of BpR2q-sets onto R

Suslin (1917): No! In fact, it must be of the formď

tniuiě1PNN

č

kě1In1,...,nk

wheretIn1,...,nk : k ě 1 ^ tniuiě1 P N

Nu

are closed intervals. Note: Uncountable union!

Analytic set 10

Definition

Let X be a Polish space. A set A Ď X is said to be analytic if it isa continuous image of a Polish space.

Not the original (Suslin’s) definition nor connected to Suslin’sform of a projections sets. We thus prove . . .

Universality of the Baire space 11

Theorem

Every Polish space is a continuous image of the Baire space NN

Before giving the proof, note two Corollaries . . .

Equivalent definitions 12

Corollary

Let X be a Polish space and let A Ď X. The following are equivalent:(1) A is analytic,(2) A is a continuous image of NN,(3) A is the projection of a closed set in XˆNN onto X.

Proof:

(1)ñ (2): A “ f pXq and, by Theorem, X “ gpNNq, for f , g cont.

(2)ñ (3): Write A “ f pNNq and define g : NN Ñ XˆNN by

gpn̄q :“ pf pn̄q, n̄q

Then g continuous and gpNNq is closed and projects onto A

(3)ñ (1): If C Ď XˆNN closed, then C is a Polish space. Theprojection is continuous so definition applies.

Suslin’s form in R 13

Corollary

Every subset of R of the formď

tniuiě1PNN

č

kě1In1,...,nk

withtIn1,...,nk : k ě 1 ^ tniuiě1 P N

Nu

non-empty closed intervals is analytic.

Proof: tpx, n̄q P RˆNN : x P In1,...,nku is closed in RˆNN.Hence, so is

č

kě1

px, n̄q P RˆNN : x P In1,...,nk(

.

This projects to the set above.

Proof of Theorem 14

WTS: @A Ď X analytic Df : NN Ñ X continuous : A “ f pNNq

Let txnuně1 be dense in R and B1px, rq :“ ty P X : $px, yq ď ru.

Define non-empty closed sets Cn1,...,nk Ď X indexed by k ě 1 andtniuiě1 P NN recursively by:

For k “ 1, set Cn :“ B1pxn, 1{2q @n P N.If Cn1,...,nk defined for some k ě 1, enumerate

m P N : B1pxm, 2´k´1q XCn1,...,nk ‰ H(

into tm`u`ě1 and set

Cn1,...,nk,` :“ Cn1,...,nk X B1pxm` , 2

´k´1q

Then Cn1,...,nk ‰ H and closed. Moreover we get . . .

Proof of Theorem continued . . . 15

Nesting and partitioning:

Cn1,...,nk “ď

mPNCn1,...,nk,m

Diameter estimate

diam`

Cn1,...,nk˘

ď 2´k`1

It follows (HW problem)č

kě1Cn1,...,nk “ tyn̄u

Claim: n̄ ÞÑ yn̄ continuousUnder dpn̄, n̄1q :“

ř

kě1 2´k1tnk“n1ku, if 2

´k´1 ď dpn̄, n̄1q ă 2´kthen ni “ n1i for i ď k. So yn̄, yn̄1 P Cn1,...,nk and thus

$pyn̄, yn̄1q ď diampCn1,...,nkq ď 2´k`1 ď 4dpn̄, n̄1q

Map is surjective by partitioning property.

Properties of analytic sets 16

For analytic sets, we will now discuss:Behavior under image/preimage by continuous mapsContainment of all closed and open setsCloseness under countable unions and intersectionsInclusion of Borel setsSeparation by Borel sets

Image/preimage under continuous map 17

Lemma

Let X and Y be Polish spaces and f : X Ñ Y a continuous map. Then

@A Ď X analytic : f pAq analytic (in Y)

and@B Ď Y analytic : f´1pBq analytic (in X)

Proof of Lemma 18

Image map:A Ď X analyticñ A “ gpNNq for g : NN Ñ X continuous.Then f pAq “ f ˝ gpNNq. As NN is Polish, f pAq is analytic.

Preimage map:Let B Ď Y analytic. Then DC Ď YˆNN closed that projects on B.For f : X Ñ Y, define h : YˆNN Ñ XˆNN by

hpx, n̄q “ p f pxq, n̄q

Then D :“ h´1pCq is closed and projects on f´1pAq.So: f´1pAq analytic.

Containment of open/closed sets 19

Lemma

All closed and all open subsets of a Polish space are analytic

Proof: Closed subsets of a Polish space are Polish. Same foropen sets (HW). Identity map shows they are analytic.

Generalizes further:

Theorem (Alexandrov’s Theorem)

Every non-empty Gδ-subset of a Polish space is Polish

Converse: A subset of a Polish is Polish iff it is Gδ

Closeness under countable unions/intersections 20

Lemma

The class of analytic subsets of a Polish space is closed undercountable unions and countable intersections.

Let tAnuně1 be analytic in X. Then Dfn : Yn Ñ X : An “ f pXnqCountable unions:Set Y :“

Ť

ně1ptnu ˆYnq and define g : Y Ñ X bygpn, yq :“ fnpyq. Then gpYq “

Ť

ně1 An so union analytic.

Countable intersections:Define f :

Ś

ně1 Yn Ñ XN by f ptynuně1q :“ tfnpynquně1. Then fcontinuous with f p

Ś

ně1 Ynq “Ś

ně1 An. SoŚ

ně1 An analytic.Next define g : X Ñ XN by gpxq “ txuně1. Then

g´1´

ą

ně1An

¯

“č

ně1An

and soŞ

ně1 An analytic.

Inclusion of Borel sets 21

Proposition

All Borel sets in a Polish space are analytic

Could use Borel hierarchy. Instead rely on:

Lemma (Monotone class lemma)

Let X be a set and assume M Ď 2X contains X andH and is closedunder countable unions and countable intersections. Then

F :“ tA PM : Ac PMu.

is a σ-algebra.

Proof of Proposition: M :“ tA Ď X : analyticu is a monotoneclass containing all closed/open sets. F contains closed/opensets. So BpXq Ď F ĎM.

Lusin’s separation theorem 22

Question: How to characterize analytic sets that are Borel?

Theorem

Let X be Polish and A, A1 Ď X are analytic with AXA1 “ H.Then A and A1 are Borel separated in the sense that

DB P BpXq : A Ď B ^ A1 Ď Bc

In particular,

@A Ď X analytic : A P BpXq ô Ac is analytic

Proof of Lusin’s separation theorem 23

Key idea: Divide and conquer

Let A :“ class of analytic sets. Note: If tEiuiě1 Ď A andtE1iuiě1 Ď A obey

´

ď

iě1Ei¯

X´

ď

jě1E1j¯

“ H.

then@i, j ě 1 DBi,j P BpXq : Ei Ď Bi,j ^ E1j Ď Bci,j

impliesŤ

iě1 Ei ĎŤ

iě1Ş

jě1 Bi,j and

ď

jě1E1j Ď

ď

jě1

č

iě1Bci,j Ď

č

iě1

ď

jě1Bci,j “

´

ď

iě1

č

jě1Bi,j

¯c

Conclusion: IfŤ

iě1 Ei andŤ

jě1 E1i are NOT Borel separated,

then at least one pair of sets Ei and E1j is NOT Borel separated

Proof of Lusin’s separation theorem continued . . . 24

Assume A, A1 disjoint analytic but NOT Borel separated.

Projection map: f : XˆNN Ñ X. Then DC, C1 Ď XˆNN closed,disjoint with A “ f pCq and A1 “ f pC1q

Apply “divide and conquer” for intersections with closed ballsof radii 2´k´1, k ě 1, to recursively define tCkukě1 and tC1kukě1closed non-empty such that (for all k ě 1)

C1 :“ C ^ C11 :“ C1

Ck`1 Ď Ck ^ C1k`1 Ď C1k

diampCkq ď 2´k and diampC1kq ď 2´k

f pCkq, f pC1kq analytic but NOT Borel separatedBut then

Ş

kě1 Ck “ txu andŞ

kě1 C1k “ tx

1uwith f pxq P A andf px1q P A1. So x ‰ x1 and Ck and C1k are separated by disjointopen balls. Hence, f pCkq and f pC1kq ARE Borel separated, acontradiction!

Universal sets 25

Next question: Are there analytic sets that are not Borel?

Answer: YES but we need to work a bit

Definition (Universal set)

Let X be a set and A a collection of its subsets. We say that a setE Ď XˆNN is universal for A if

@A P A Dn̄ P NN :

x P X : px, n̄q P E(

“ A.

In order words, each A P A appears as a section of E of constantsecond coordinate.

An example 26

Lemma

Let pX, T q be a second-countable topological space. Then there existsan open set E Ď XˆNN that is universal for T .

Proof:Let tUnuně1 be open sets generating T by unions. Abbreviatingn̄ “ tniuiě1, let

E :“!

px, n̄q P XˆNN : x Pď

iě1Uni

)

Open because tpx, n̄q P XˆNN : x P Uniu is open. Let O P T .Then there is n̄1 “ tn1iuiě1 P NN such that O “

Ť

iě1 Un1i . So

x P X : px, n̄1q P E(

“ O

and O is thus a section of E.

Universal analytic set 27

Proposition

Let X be a Polish space. Then there is an analytic set E Ď XˆNNthat is universal for the class of analytic subsets of X.

Proof:Let F Ď pXˆNNq ˆNN universal for open sets. Then D :“ Fcuniversal for closed sets in XˆNNNext: A Ď X analytic is a projection of a closed C Ď XˆNNThen C is a section of D for constant 3rd coordinate so letting

E :“ projection of D on 1st and 3rd coordinate

we get that A is a section of E.

Remark on cardinality 28

Corollary

The class of analytic (and thus also Borel) subsets of a Polish space isat most of the cardinality of the continuum. In any infinite Polishspace X, the cardinality of BpXq is that of the continuum.

Proof:Number of sections ď |NN| “ |R|X infiniteñ BpXq contains all subsets of an infinite set. So|BpXq| ě |t0, 1uN| “ |R|

Note: This proves that BpRq Ĺ LpRq

Analytic non-Borel set in NN 29

Lemma

The Baire space NN contains an analytic set that is not Borel

Proof: Let E Ď NN ˆNN be universal for analytic sets in NN.Define

A :“

n̄ P NN : pn̄, n̄q P E(

.

Then A analytic. If Ac analytic, then

Dn̄1 P NN : Ac “

n̄ P NN : pn̄, n̄1q P E(

.

As n̄1 P A is equivalent to n̄1 P Ac. So Ac is NOT analytic and soA R BpNNq.

Analytic non-Borel sets 30

Theorem

Every uncountable Polish space X contains an analytic set A suchthat A R BpXq

Proof: X embeds t0, 1uN which embeds NN, with both mapshomeomorphic and thus Borel-bimeasurable. The maps movethe analytic non-Borel set in NN to X.

Co-analytic sets 31

The above motivates:

Definition (Co-analytic set)

A subset A Ď X of a Polish space X is said to be co-analytic if Acis analytic.

Interesting explicit examples:Mazurkiewicz 1936: Set of differentiable functions inCpr0, 1sq co-analytic but not BorelMauldin 1979: Set of nowhere differentiable functions inCpr0, 1sq co-analytic but not Borel

µ-measurable sets and completion 32

Any measure µ extends uniquely to its completion µ̄ on:

Definition (µ-measurable sets)

Let µ be a measure on a measurable space pX,Fq. A set A Ď Xis then called µ-measurable if

DE, F P F : E Ď A Ď F ^ µpF r Eq “ 0.

We will write Fµ for the class of µ-measurable sets.

The extended measure space pX,Fµ, µ̄q is complete.

Question: Characterize sets in Fµ

Universal measurability 33

Theorem (Lusin)

Let X be a Polish space and let µ be a finite measure on pX,BpXqq.Then every analytic or co-analytic subset of X is µ-measurable.

Technical lemma 34

Outer measure:

µ‹pAq :“ inf

µpBq : B P BpXq ^ A Ď B(

.

Key step: Inner regularity (of µ‹)

Lemma

Let A Ď X be analytic with A ‰ H. Then

@e ą 0 DK Ď X : K compact ^ K Ď A ^ µ‹pAq ď µpKq ` e.

Proof of Lemma 35

Note: µ‹ is regular so each set admits measurable cover. Hence:Monotonce Convergence (MCT) holds for µ‹

Pick A Ď X and find f : NN Ñ X continuous with f pNNq “ ALet

Rkpr1, . . . , rkq :“!

tniuiě1 P NN : p@i ď k : ni ď riq)

Note Rk`1pr1, . . . , rk, rq Ò Rkpr1, . . . , rkq as r Ñ8. So, by MCT,

µ‹`

f pR1prqq˘

ÝÑrÑ8

µ‹pAqand

µ‹`

f pRk`1pr1, . . . , rk, rqq˘

ÝÑrÑ8

µ‹`

f pRkpr1, . . . , rkqq˘

.

So, given e ą 0, there is triuiě1 P NN s.t.

@k ě 1 : µ‹`

f pRkpr1, . . . , rkqq˘

ě µ‹pAq ´ e

Now define . . .

Proof of Lemma continued . . . 36

... define

rK :“!

tniuiě1 P NN : p@i ě 1 : ni ď riq)

Then K :“ f prKq is compact with K Ď f pNNq “ A. NTS:

K “č

kě1f`

Rkpr1, . . . , rkq˘

as that implies µpKq ě µ‹pAq ´ e by MCT.Ď immediate. For Ě pick x in intersection. Then

@k ě 1 Dn̄pkq P Rkpr1, . . . , rkq : f pn̄pkqq “ x

But then Dkj Ñ8 s.t. n̄pkjq Ñ n̄ P rK. Continuity:x “ f pn̄q P f prKq “ K

Proof of Lusin’s theorem 37

As BpXqµ σ-algebra, suffices to consider A Ď X analytic.

Finite outer measure:

@e ą 0 DBe P BpXq : A Ď Be ^ µpBq ď µ‹pAq ` e

Lemma:

@e ą 0 DKe P BpXq : Ke Ď A ^ µpKeq ě µ‹pAq ´ e

TakeE :“

ď

ně1K1{n ^ F :“

č

ně1B1{n

to get E Ď A Ď F with µpF r Eq “ 0.

Universal measurability 38

Definition (Universal measurability)

A Ď X is universally measurable if it belongs to the σ-algebra

F‹ :“č

µ : finitemeasure on pX,Fq

Fµ (1)

of universally measurable sets on pX,Fq.

Lusin’s Theorem: For pX,BpXqqwith X uncountable we have

BpXq Ď BpXq‹

Important e.g. in stochastic analysis

Universally measurable functions 39

Definition

f : X Ñ X is said to be universally measurable if f´1pF‹q Ď F‹.

LemmaLet pX,Fq be measurable space. Then every F{F -measurablef : X Ñ X is universally measurable.

Conclusion 40

Analytic sets play important in logic, set up for certainprobability models, etc

More info:A.S. Kechris’ “Classical descriptive set theory”D.L. Cohn’s “Measure theory”