Disintegration into conditional measures: Rokhlin s theorem

Transcription

1 Disintegration into conditional measures: Rokhlin s theorem Let Z be a compact metric space, µ be a Borel probability measure on Z, and P be a partition of Z into measurable subsets. Let π : Z P be the map associating to each z Z the atom P P that contains it. By definition, is a measurable subset of P if and only if π 1 ( is a measurable subset of Z. Let ˆµ be the push-forward of µ under π, in other words, ˆµ is the probability measure on P defined by ˆµ( = µ(π 1 ( for every measurable set P. Definition 1. A system of conditional measures of µ with respect to P is a family (µ P P P of probability measures on Z such that 1. µ P (P = 1 for ˆµ-almost every p P; 2. given any continuous ϕ : Z R, the function P P ϕ dµ P is measurable and ϕ dµ = ( ϕ dµ P dˆµ(p. Proposition 1. If (µ P P P is a system of conditional measures of µ relative to the partition P, then P P ψ dµ P is measurable and ψ dµ = ( ψ dµp dˆµ(p, for any µ-integrable function ψ : Z R. Proof. The class of functions that satisfy the conclusion of the lemma contains all the continuous functions, and is closed under dominated pointwise convergence. Therefore, it contains all bounded measurable functions. In particular, P µ P (E is mesurable, and µ(e = µ P (E dˆµ(p, for any measurable set E Z. Conditional measures, when they exist, are unique almost everywhere: Proposition 2. If (µ P P P and (ν P P P are two systems of conditional measures of µ with respect to P, then µ P = ν P for ˆµ-almost every P P. 1

2 Proof. Suppose otherwise, that is, there exists a measurable set 0 P with ˆµ( 0 > 0 such that µ P ν P for every P 0. Using the fact that C 0 (Z, R admits countable subsets dense with respect to the uniform norm, one gets that there exists ϕ C 0 (Z, R and a subset of 0 such that ˆµ( > 0 and (interchanging the roles of µ P and ν P, if necessary ϕ dµ P > ϕ dν P for every P. Then ( ( ϕ dµ P dˆµ(p > ϕ dν P dˆµ(p. (1 On the other hand, by Proposition 1, ( (ϕx π 1 ( dµ = (ϕx π 1 ( dµ P dˆµ(p. By assumption µ P (P = 1 for ˆµ-almost every P P. For any such P, we have (ϕx π 1 ( dµ P = X (P ϕ dµ P. Therefore, ( (ϕx π 1 ( dµ = X (P Analogously, we find (ϕx π 1 ( dµ = ( ϕ dµ P dˆµ(p = ( ϕ dν P dˆµ(p. ϕ dµ P dˆµ(p. These two last equalities contradict (1. Therefore, µ P = ν P for ˆµ-almost every P, as claimed. Definition 2. P is a measurable partition if there exist measurable subsets E 1, E 2,..., E n,... of Z such that P = {E 1, Z \ E 1 } {E 2, Z \ E 2 } {E n, Z \ E n } mod 0. In other words, there exists some full µ-measure subset F 0 Z such that, given any atom P of P we may write P F 0 = E 1 E 2 E n F 0 (2 where E j is either E j or its complement Z \ E j, for every j 1. 2

3 Example 1. Every finite or countable partition is a measurable partition. In fact, P is measurable if and only if there exists a non-decreasing sequence of finite or countable partitions P 1 P 2 P n such that P = n=1p n mod 0. Example 2. Let Z = X Y, where X and Y are compact metric spaces, and P be the partition of Z into horizontal lines X {y}, y Y. Then P is a measurable partition of Z. Theorem 3. (Rokhlin [R] If P is a measurable partition, then there exists some system of conditional measures of µ relative to P. Proof. For the purpose of the conclusion of the theorem, we may replace the space Z by any full measure subset. So, it is no restriction to suppose that the set F 0 in (2 actually coincides with Z, and we do so in all that follows. Let ψ be any bounded measurable real function on Z. For each n 1 let P n = {E 1, Z \ E 1 } {E 2, Z \ E 2 } {E n, Z \ E n } that is, P n is the partition of Z whose atoms are the sets E1 En, with Ej = E j or Ej = Z \ E j, for each 1 j n. Define ψ n : Z R as follows. If the atom P n (z of P n that contains z has positive µ-measure, then ψ n (z = 1 µ(p n (z P n(z ψ dµ. (3 Otherwise, ψ n (z = 0. Clearly, the second case in the definition of ψ n applies only to a zero µ-measure set of points. Lemma 4. Given any bounded measurable function ψ : Z R, there exists a full µ-measure subset F = F (ψ of Z such that ψ n (z, n 1, converges to some real number ψ(z, for every z F. Proof. We may always write ψ = ψ + ψ, where ψ ± are measurable, bounded, and non-negative: for instance, ψ ± = ( ψ ± ψ/2. Then ψ n = ψ n + ψ n for n 1, and so the conclusion holds for ψ if it holds for ψ + and ψ. This shows that it is no restriction to assume that ψ is non-negative. We do so from now on. For any α < β, let S(α, β be the set of points z Z such that lim inf ψ n (z < α < β < lim sup ψ n (z. 3

4 Clearly, given z Z, the sequence ψ n (z diverges if and only if z is in S(α, β for some pair of rational numbers α and β. So, the lemma will follow if we show that S = S(α, β has zero µ-measure for all α and β. For each z S fix some sequence of integers 1 a(1 < b(1 < < a(i < b(i < such that ψ a(i (z < α and ψb(i (z > β for every i 1. Define A i to be the union of the partition sets P a(i (z, and B i to be the union of the partition sets P b(i (z obtained in this way, for all the points z S. By construction, S A i+1 B i A i for every i 1. In particular, S is contained in the set S = B i F 0 = A i F 0. i=1 Given any two of the sets P a(i (z that form A i, either they are disjoint or one is contained in the other. This is because P n, n 1, is a non-decreasing sequence of partitions. Consequently, A i may be written as a two-by-two disjoint union of such sets P a(i (z. Hence, ψ dµ = ψ dµ < aµ(p a(i = αµ(a i, A i P a(i P a(i (z i=1 P a(i (z for any i 1 (the sums are over that disjoint union. Analogously, ψ dµ = ψ dµ > bµ(p b(i = βµ(b i. B i P b(i P b(i (z P b(i (z Since A i B i and we are assuming ψ 0, it follows that αµ(a i > ψ dµ A i ψ dµ > βµ(b i, B i for every i 1. Taking the limit as i, we find αµ( S βµ( S. This implies that µ( S = 0, and so S S also has zero µ-measure. 4

5 Given any bounded measurable function ψ : Z R, we shall represent as e n (ψ, e(ψ, respectively, the functions ψ n, ψ defined by (3 and Lemma 4. Let {ϕ k : k N} be some countable dense subset of C 0 (Z, R, and let F = F (ϕ k, k=1 where F (ϕ k is as given by Lemma 4. Lemma 5. Given any continuous function ϕ : Z R, the sequence e n (ϕ(z converges to e(ϕ(z as n, for every z F. Proof. The class of functions that satisfy the conclusion of the lemma is closed under uniform convergence and, by definition, it contains a dense subset of the space of all continuous functions. Let ϕ : Z R be continuous. By construction, e n (ϕ is constant on each P n P n, and so it is also constant on each atom P of P, for every n 1. Therefore, e(ϕ is constant on P F for every P P. Let e n (ϕ(p n represent the value of e n (ϕ on each P n P n. Similarly, e(ϕ(p represents the value of e(ϕ on P F whenever the latter set is non-empty. Then, since (3 defines e n (ϕ on a full µ-measure subset of Z, ϕ dµ = ϕ dµ = µ(p n e n (ϕ(p n = e n (ϕ dµ. P n µ(p n>0 µ(p n>0 Observe also that e n (ϕ sup ϕ < for every n 1. Therefore, we may use the dominated convergence theorem to conclude that ϕ dµ = e(ϕ dµ. (4 Now we are in a position to construct a system of conditional measures of µ. Let P be any atom of P such that P F is non-empty. It is easy to see that C 0 (Z, R ϕ e(ϕ(p R is a non-negative linear functional on C 0 (Z, R. By the Riesz-Markov theorem, there exists a unique probability measure µ P on Z such that ϕ dµ P = e(ϕ(p. (5 5

6 For completeness, we should define µ P also when P does not intersect F. In this case we let µ P be any probability measure on Z: since the set of all these atoms P has zero ˆµ-measure in P (in other words, their union has zero µ-measure in Z, the choice is not relevant. In view of these definitions, (4 may be rewritten as ( ϕ dµ = ϕ dµ P dˆµ(p, the fact that P P ϕ dµ P is a measurable function being a direct consequence of (5. Therefore, to conclude that (µ P P P do form a system of conditional measures of µ with respect to P we only have to prove Lemma 6. µ P (P = 1 for ˆµ-almost every P P. We use the following auxiliary result. Lemma 7. Given any bounded measurable function ψ : Z R there exists a full ˆµ-measure set F(ψ P such that the set P F is non-empty and ψ dµp = e(ψ(p, for any P F(ψ. Proof. The class of functions that satisfy the conclusion of the lemma contains all the continuous functions, and is closed under dominated pointwise convergence. Therefore, it contains all bounded measurable functions. Now we can prove Lemma 6: Proof. Define F = k,pk F(X Pk, where the intersection is over the set of all the atoms P k P k, and every k 1. Since this is a countable set, F has full ˆµ-measure. We claim that the conclusion of the lemma holds for every P F. Indeed, let k 1 and P k be the element of P k that contains P. By the definition of F µ P (P k = X Pk dµ P = e(x Pk (P. (6 For each n 1, let P n be the atom of P n that contains P. Given any z P F, e n (X Pk (z = 1 X Pk dµ. µ(p n P n 6

7 Now, for any n k we have P n P k, and then the last term is equal to 1. Therefore, e(x Pk (P = e(x Pk (z = lim n e n (X Pk (z = 1. Replacing this in (6 we get µ P (P k = 1 for every k 1. Finally, µ P (P = lim k µ P (P k = 1 because the P k, k 1, are a decreasing sequence whose intersection is P. The proof of Theorem 3 is complete. Example 3. Let Z be the 2-dimensional torus, α be some irrational number, and P be the partition of Z into the straight lines of slope m. Then P is not a measurable partition. Indeed, the Haar (Lebesgue measure on Z admits no system of conditional measures with respect to P. Example 4. (ergodic decomposition Let f : Z Z be a continuous transformation on a compact metric space Z, and B f be the subset of points z Z such that time averages are well-defined on the orbit of z: given any continuous function ϕ : Z R, the sequence 1 n 1 ϕ(f j (z n j=0 converges to some ϕ(z R when n. Let P be the partition of Z defined by (i Z \ B f is an atom of P and (ii two points z 1, 2 in B f are in the same atom of P if and only if they have the same time averages: ϕ(z 1 = ϕ(z 2 for every continuous function ϕ. Then P is a measurable partition, with respect to any probability measure µ in Z. If µ is f-invariant, then µ(z \ B f = 0 and any system of conditional measures (µ P P of µ relative to P is such that µ P is f-invariant and ergodic for ˆµ-almost every P P. References [R] V. A. Rokhlin, On the fundamental ideas of measure theory, Transl. Amer. Math. Soc., Series 1, 10 (1962,