• Tidak ada hasil yang ditemukan

Conditional Measures

Dalam dokumen Ergodic Theory (Halaman 152-164)

Conditional Measures and Algebras

5.3 Conditional Measures

situations null sets with respect to the original measure become more trou- blesome, since the conditional measures will often be singular with respect to the original measureμ.

Thus we need to pay more attention to null sets—and in particular we will need to specify in a more concrete way with respect to which measure a given set has measure zero. When we say N is a null set we mean that μ(N) = 0; in contrast for x X we will say N is a μAx -null set if μAx(N) = 0.

Similarly, we will need to make a distinction between the notion of “almost everywhere” (true off aμ-null set) and “μAx -almost everywhere”. From now on we will also distinguish more carefully between the space Lp(X,B, μ) of genuine functions and the more familiar spaceLp(X,B, μ) of equivalence classes of functions; in particular L(X,B) denotes the space of bounded measurable functions and will be writtenLif the underlying measure space is clear.

We next formalize our prevailing assumption about the measure spaces we deal with. A probability space is any triple (X,B, μ) where μ is a measure on the σ-algebra B with μ(X) = 1. It turns out that this definition is too permissive for some—but by no means all—of the natural developments in ergodic theory.

Example 5.11.LetX ={0,1}R, with the product topology and theσ-algebra of Borel sets. The product measureμ of the (12,12) measure on each of the sets {0,1} makes X into a probability space with the property that there is an uncountable collection {As}sR of measurable sets with the property thatμ(As) =12 for eachs∈Rand the sets are all mutually independent:

μ(As1∩ · · · ∩Asn) = 21n

for any n distinct reals s1, . . . , sn. The next definition gives a collection of probability spaces that precludes the possibility of uncountably many inde- pendent sets.

Definition 5.12.LetX be a Borel subset of a compact metric space with the restriction of the Borelσ-algebraBtoX. Then the pair (X,B) is aBorel space.

Definition 5.13.Let X be a dense Borel subset of a compact metric space X, with a probability measure μ defined on the restriction of the Borelσ-algebraBtoX. The resulting probability space (X,B, μ) is aBorel probability space.

For a compact metric space X, the space M(X) of Borel probability measures on X itself carries the structure of a compact metric space with respect to the weak*-topology. In particular, we can define the Borel σ- algebraBM(X) on the spaceM(X) in the usual way. IfX is a Borel subset of a compact metric spaceX, then we define

5.3 Conditional Measures 135

M(X) ={μ∈M(X)(XX) = 0},

and we will see in Lemma5.23thatM(X) is a Borel subset ofM(X).

We are now in a position to state and prove the main result of this chapter.

A set is calledconullif it is the complement of a null set. Forσ-algebrasC,C the relation

C

μ C

means that for anyA∈C there is a setA∈C withμ(A A) = 0. We also define

C =

μ C to mean thatC

μ C andC

μ C.

A σ-algebra A on X is countably-generated if there exists a countable set{A1, A2, . . .}of subsets ofXwith the property thatA =σ({A1, A2, . . .}) is the smallest σ-algebra (that is, the intersection of every) σ-algebra con- taining the setsA1, A2, . . ..

Theorem 5.14.Let(X,B, μ)be a Borel probability space, andA ⊆Baσ- algebra. Then there exists an A-measurable conull set X X and a sys- tem Ax |x∈X} of measures onX, referred to as conditional measures, with the following properties.

(1)μAx is a probability measure onX with E(fA)(x) =

f(y) dμAx (y) (5.3) almost everywhere for allf ∈L1(X,B, μ). In other words, for any func- tion f L1(X,B, μ) we have that

f(y) dμAx (y) exists for all x be- longing to a conull set in A, that on this set

x→

f(y) dμAx(y) depends A-measurably on x, and that

A

f(y) dμAx(y) dμ(x) =

A

fdμ for allA∈A.

(2)If A is countably-generated, thenμAx ([x]A) = 1 for allx∈X, where [x]A =

xA∈A

A

Notice that we are forced to work with genuine functions inL1in order that the right- hand side of (5.3) is defined. As we said before,μAx may be singular toμ.

is the atom ofA containingx; moreover μAx =μAy forx, y∈X when- ever[x]A = [y]A.

(3)Property(1)uniquely determinesμAx for a.e.x∈X. In fact, property(1) for a dense countable set of functions in C(X)uniquely determines μAx for a.e.x∈X.

(4)If Ais anyσ-algebra with A =

μ

A, thenμAx =μAxealmost everywhere.

Remark 5.15.Theorem5.14is rather technical but quite powerful, so we as- semble here some comments that will be useful both in the proof and in situations where the results are applied.

(a) For a countably generated σ-algebra A = σ({A1, A2, . . .}) the atom in (2) is given by

[x]A =

xAi

Ai

x /Ai

XAi (5.4)

and hence isA-measurable (see Exercise5.3.1). In fact [x]A is the small- est element ofA containingx.

(b) IfN ⊆Xis a null set forμ, thenμAx (N) = 0 almost everywhere. In other words, for a μ-null set N, the set N is also a μAx-null set for μ-almost everyx. This follows from property (1) applied to the function f =χN. In many interesting cases, the atoms [x]A are null sets with respect toμ, and so μAx is singular toμ.

(c) The conditional measures constructed in Theorem 5.14 are sometimes said to give adisintegration of the measure μ.

(d) Notice that the uniqueness in property (3) (and similarly for (4)) may require switching to smaller conull sets. That is, ifμAx forx∈X ⊆X andμAx forx∈X⊆X are two systems of measures as in (1), then the claim is that there exists a conull subsetX ⊆X∩X with μAx =μAx for allx∈X.

(e) We only ever talk about atoms for countably generated σ-algebras. The first reason for this is that for a generalσ-algebra the expression defined in Theorem 5.14(2) by an uncountable intersection may not be mea- surable (let alone A-measurable). Moreover, even in those cases where the expression happens to be A-measurable, the definition cannot be used to prove the stated assertions. We also note that it is not true that any sub-σ-algebra of a countably-generatedσ-algebra is countably gen- erated (but see Lemma5.17for a more positive statement). For example, the σ-algebra of null sets in Twith respect to Lebesgue measure is not countably-generated (but there are more interesting examples, see Exer- cise 6.1.2).

Example 5.16.LetX = [0,1]2andA =B× {,[0,1]}as in Example5.3. In this case Theorem5.14claims that any Borel probability measureμonX can

5.3 Conditional Measures 137

be decomposed into “vertical components”: the conditional measuresμA(x

1,x2)

are defined on the vertical line segments {x1} ×[0,1], and these sets are precisely the atoms ofA. Moreover,

μ(B) =

X

μA(x1,x2)(B) dμ(x1, x2). (5.5) In this exampleμA(x

1,x2)=νx1does not depend onx2, so (5.5) may be written as

μ(B) =

[0,1]

νx1(B) dμ(x1) (5.6)

whereμ=πμis the measure on [0,1] obtained by the projection π: [0,1]2 −→[0,1]

(x1, x2)−→x1.

While (5.6) looks simpler than (5.5), in order to arrive at it a quotient space and a quotient measure has to be constructed (see Sect.5.4). For simplicity we will often work with expressions like (5.5) in the general context.

Onceμis known explicitly, the measuresμA(x

1,x2)can often be computed.

For example, ifμis defined by

fdμ=13

f(s, s) ds+ 1

0

s 0

f(s, t) dtds, then

μA(x

1,x2)= 1

√x1+ 1/3δx1×1

3δx1+m[0,x1] .

To see that this equation holds, the reader should use Theorem5.14(3). How- ever, the real force of Theorem5.14lies in the fact that it allows an unknown measure to be decomposed into components which are often easier to work with.

Proof of Theorem 5.14. By assumption, X is contained in a compact metric spaceX, which is automatically separable. We note that the statement of the theorem for the ambient compact metric spaceX implies the theorem for X by Remark 5.15(b). Hence we may assume that X = X is itself a compact metric space.

Suppose first thatx}andx}are families of measures defined for almost everyxthat both satisfy (5.3) for a countable dense subset{fn}nNinC(X).

Then for eachn1 and almost everyx,

fndρx=E(fnA) =

fndνx. (5.7)

So there is a common null setNwith the property that (5.7) holds for alln 1 and x /∈ N. By uniform approximation and the dominated convergence theorem (Theorem A.18), this easily extends to show that

fdρx=

fdνx

for allf ∈C(X) andx /∈N. Henceρx=νxforx /∈N, which shows that the conditional measures—if they exist—must be unique as claimed in (3).

Now let

A=

μ A

and write A for the smallest σ-algebra containing both Aand A. Then for any f C(X), g = E(fA) (or E(f A)) satisfies the characterizing properties ofE(fA), so they are equal almost everywhere. Noting this for a countable dense subset of C(X) shows (as in the proof of uniqueness) thatμAx =μAxealmost everywhere, showing (4).

Turning to existence, let

F ={f01, f1, f2, . . .} ⊆C(X)

be a vector space overQthat is denseinC(X). For everyi1, choose anA- measurable function gi∈Lμ1withgirepresentingE(fiA). Defineg0to be the constant function 1. Then

gi(x)0 almost everywhere iffi0;

|gi(x)|fi almost everywhere;

• iffi=αfj+βfk withα, β∈Q, thengi(x) =αgj(x) +βgk(x) for almost allx.

LetN ∈A be the union of all the null sets on the complement of which the properties above hold; since this is a countable union,N is a null set.

For x /∈ N, define Λx(fi) to be gi(x). Then by the properties aboveΛx is a Q-linear map fromF to R withΛx 1. It follows that Λx extends uniquely to a continuous positive linear functional

Λx:C(X)R.

By the Riesz representation theorem, there is a measureμAx onX character- ized by the property that

SinceX is separable we may find a set{h01, h1, h2, . . .}that is dense inC(X). The vector space overQspanned by this set is dense and countable, and may be written in the form{f01, f1, f2, . . .}.

Notice that this is a genuine function rather than an equivalence class of functions, so there is a choice involved despite Theorem5.1(1).

5.3 Conditional Measures 139

Λx(f) =

fdμAx

for allf ∈C(X); moreoverΛx(1) = 1, soμAx is a probability measure.

By our choice of the setF, for any f C(X) there is a sequence (fni) withfni−→f uniformly. We have already established that

x→

fnidμAx isA-measurable (by Theorem5.14(1)), and that

A

fnidμAx(x) =

A

fnidμ

for allA∈A. So, by the dominated convergence theorem (Theorem A.18),

fnidμAx

fdμAx (5.8)

isA-measurable as a function ofx, and

A

fdμAx(x) =

A

fdμ (5.9)

for allA∈A. For any open setO there is a sequence (fni) withfni χO, so by the monotone convergence theorem (5.8) and (5.9) hold forχO. Thus we have (5.8) and (5.9) for the indicator function of any closed A⊆X, by taking complements. Similarly, these equations extend to anyGδ-set Gand anyFσ-setF. Define

M ={B∈B|f =χB satisfies (5.8) and (5.9)}.

By the monotone convergence theorem (Theorem A.16), if B1, B2, . . . ∈M with

B1⊆B2⊆ · · · , then

n1Bn∈M and ifC1, C2, . . .∈M with C1⊇C2⊇ · · ·, then

n1Cn M. Thus M is a monotone class (see Definition A.3 and Theorem A.4). Define

R= n

i=1

Oi∩Ai|Oi⊆X is open andAi⊆X is closed

forn∈N. We claim thatRis an algebra (that is,Ris closed under comple- ments, finite intersections and finite unions). To see this, notice that theσ- algebraC generated by finitely many open and closed sets has the property that every element ofC is a disjoint union of atoms of the partition generated by the same open and closed sets, all of which are precisely of the formO∩A.

Since any setO∩Ais aGδ-set and (5.8) and (5.9) are linear conditions, it follows that (5.8) and (5.9) also hold for functions of the form

χR= n i=1

χOiAi

for all

R= n i=1

Oi∩Ai∈R.

By the monotone class theorem (Theorem A.4),B =σ(R)⊆M. In other words, for any Borel measurable setB B, the characteristic function χB

satisfies (5.8) and (5.9). By considering simple functions and applying the monotone convergence theorem, it follows that (5.8) and (5.9) also hold for anyB-measurable functionf 0.

Finally, given anyB-measurable integrable functionf, we may write f =f+−f

withf+, fnon-negative, measurable, and integrable functions. Then, by (5.9),

f+dμAx ,

fdμAx <∞

almost everywhere. In particular,f isμAx-integrable for almost everyx, and where it isμAx -integrable,

fdμAx is anA-measurable function ofx. Finally, (5.9) holds, proving (1).

Suppose now thatA =σ({A1, A2, . . .}) is countably-generated. Then E(χAiA)(x) =χAi(x)

=μAx(Ai)

almost everywhere, for anyi 1. Collecting all the null sets arising into a single null setN gives

μAx(Ai) =

1 ifx∈AiN;

0 ifx∈X(Ai∪N).

SinceμAx is a measure, it follows by (5.4) that μAx ([x]A) = 1

5.3 Conditional Measures 141

ifx /∈N. WritingX forXN, recall that the map X x−→

fdμAx isA-measurable for anyf ∈C(X). Thus

fdμAx =

fdμAy

ifx, y∈X and [x]A = [y]A, so that [x]A = [y]A implies thatμAx =μAy . One of the many desirable properties of Borel probability spaces is that there is a constraint on the complexity of their sub-σ-algebras.

Lemma 5.17.If (X,B, μ)is a Borel probability space and A ⊆B is a σ- algebra then there is a countably-generatedσ-algebraAwithA =

μ

A.

Proof. Recall that C(X) is separable for any compact metric space X (see Lemma B.8). SinceC(X) is mapped continuously to a dense subspace ofL1(X,B, μ), the same holds forL1(X,B, μ). Since subsets of a separable space are separable, it follows that the space

A|A∈A} ⊆L1(X,A, μ)⊆L1(X,B, μ)

is separable. Thus there is a set{A1, A2, . . .} ⊆A such that for anyε >0 andA∈A there is somenwith

μ(A An) =χA−χAn1< ε.

LetA=σ({A1, A2, . . .}), so thatA⊆A andA|A∈A}is dense in A|A∈A}

with respect to theL1μ norm. GivenA∈A, we can find a sequence (nk) for which

χA−χAnk1< 1 k

fork1. Then the sequence (χAnk) is Cauchy inL1(X,A, μ)⊆L1(X,A, μ), so has a limit f L1(X,A, μ). We must have f = χA almost everywhere since the limit is unique, so there is someA∈Awithμ(A A) = 0. It follows thatA =

μ

Aas required.

In the remainder of this section we give extensions and reformulations of Theorem5.14.

Lemma 5.18.Let(X,B, μ)be a Borel probability space and letA ⊆Bbe a countably-generatedσ-algebra. Iff ∈L(X,B)is constant on atoms ofA, thenf|X isA-measurable, where X is as in Theorem5.14.

A setB (or a functionf) isA-measurable moduloμif, after removing aμ- null set,B(orf) becomesA-measurable. Thus the conclusion of Lemma5.18 is thatf isA-measurable moduloμ.

Proof of Lemma5.18.By Theorem5.14(2), onX we have

fdμAx =f(x)

sinceμAx ([x]A) = 1 and, by assumption, f is constant (and equal to f(x)) on the set [x]A. By Theorem5.14(1) we know thatf|X isA-measurable.

In Theorem5.14the conditional measure was characterized in terms of the conditional expectation. The following proposition gives a more geometrical characterization.

Proposition 5.19.Let(X,B, μ)be a Borel probability space and let A be a countably-generated sub-σ-algebra ofB. Suppose that there is a set X ∈B with μ(X) = 1, and a collection x|x∈X} of probability measures with the property that

x→νx is measurable, that is for anyf ∈L we have that x→ fdνx

is measurable,

νx=νy for[x]A = [y]A andx, y∈X,

νx([x]A) = 1, and

μ=

νxdμ(x)in the sense that

fdμ= fdνxdμ(x)for all f ∈L. Thenνx=μAx for a.e.x. The same is true if the properties hold for a dense countable set of functions in C(X).

Proof.First notice that we may assume that μAx and νx are defined on a common conull setX. Moreover, we may replaceX byX and simultane- ously replace A by A|X = {A∩X | A A}. After this replacement, Lemma 5.18says that any function f which is constant on A-atoms is A- measurable. In order to apply Theorem5.14(3) we need to check that

fdνx=E fA

(x) (5.10)

almost everywhere, for allf in a countable dense subset of C(X).

That x→

fdνx is measurable is the first assumption on the family of measures in the proposition. Together with Lemma5.18, the second property shows thatx→

fdνx is actually A-measurable. This is the first require- ment in the direction of showing that (5.10) holds.

5.3 Conditional Measures 143

To show (5.10), we also need to calculate

A

fdνxdμ(x) for anyA∈A, as in Theorem5.1(1). We know thatχA(x) is constantνx-almost everywhere for any A A, by the third property. In fact χA(x) equals 1 νx-almost everywhere ifx∈Aand equals 0 otherwise. Therefore, by the fourth property applied to the functionχAf, we get

A

f(z) dνx(z) dμ(x) =

χA(x)

[x]A

f(z) dνx(z) dμ(x)

=

χA(z)f(z) dνx(z) dμ(x)

=

χA(z)f(z) dμ(z) =

A

fdμ

as required. By Theorem5.14(3) it follows thatνx=μAx almost everywhere.

It remains to prove the last claim of the proposition. So suppose we only assume the first and fourth properties for all functions in a dense countable subset ofC(X). Using dominated convergence, monotone convergence, and the monotone class theorem (Theorems A.18, A.16 and A.4) just as in the proof of Theorem5.14on p.138, we can extend the first and fourth properties in turn to allf ∈C(X), allf =χBforBany open set, any closed set, anyGδ, any Fσ, any Borel set, and finally to anyf ∈L(X). This implies the last

claim.

Proposition 5.20.Let(X,B, μ)be a Borel probability space, and let A ⊆A ⊆B

be countably-generated sub-σ-algebras. Then [z]A [z]A for z X, and for almost every z X the conditional measures for the measureμAz with respect toA are given forμAz -almost every x∈[z]A by (μAz)Ax =μAx .

The proof of this result will reveal that it is a reformulation of Theo- rem5.1(4).

Proof of Proposition5.20.We will show that the mapx→μAx satisfies all the assumptions in Proposition5.19with respect to the measureμAz for almost everyz∈X. LetμAz be defined onXA ∈A and letμAx be defined on XA ∈A with all the properties in Theorem 5.14. By Remark 5.15(b), we have μAz (XA ) = 1 for μ-almost every z. Now fix some z XA

withμAz (XA ) = 1. Forx, y∈XA we know thatμAx =μAy if [x]A = [y]A and that μAx([x]A) = 1 by Theorem 5.14(2). Also, if f L, we know that

fdμAx is measurable by Theorem5.14(1). Thus we have shown the first three assumptions of Proposition5.19on the complement of a singleμAz -null set.

It remains to check that

μAz =

μAx dμAz (x) (5.11)

for almost everyz. Letf ∈C(X). By Theorem5.14(1), for almost everyz,

fdμAx dμAz (x) =

E(fA)(x) dμAz (x) =E

E(fA)A (z), which by Theorem5.1(1) is equal to

E(fA)(z) =

fdμAz

for almost everyz. Using a dense subset{f1, f2, . . .} ⊆C(X), and collecting the countably many null sets arising in these two statements for eachninto a single null set, we obtain equality in (5.11) on a conull setZ. In other words, we have checked all the requirements of Proposition 5.19 for the family of measuresνx=μAx (and therefore they are equal almost everywhere to μAx) and for the measureμAz forz∈Z∩XA withμAz (XA ) = 1.

Theorem5.14(3) and the more geometric discussion above highlights the significance of the countably-generated hypothesis on theσ-algebraA, for in that case the conditional measuresμAx can be related to the atoms [x]A. In a Borel probability space it is safe to assume thatσ-algebras are countably- generated by Lemma5.17.

By combining the increasing and decreasing martingale theorems (Theo- rems5.5and5.8) with the characterizing properties of the conditional mea- sures we get the following corollary (see Exercise5.3.5).

Corollary 5.21.If An A or An A then μAxn −→ μAx in the weak*- topology forμ-almost every x.

This gives an alternative construction ofμAx for a countably-generatedσ- algebra. More concretely, ifA =σ({A1, A2, . . .}) andAn=σ({A1, . . . , An}) is the finiteσ-algebra generated by the firstngenerators ofA, thenμAxn is readily defined, andμAxn→μAx.

Exercises for Sect. 5.3

Exercise 5.3.1.Prove the equality claimed in (5.4).

Exercise 5.3.2.(59)Let (X,B, μ) be an aperiodic measure-preserving trans- formation on a Borel probability space (see Exercise 2.9.2 for the definition of aperiodic). Prove that for anyk1 there is a setA∈Bwith μ(A)>0 andμ(Tk(A)∩A) = 0.

Dalam dokumen Ergodic Theory (Halaman 152-164)