Doob-Dynkin Lemma for probability space

Let f, g be measurable functions from a probability space (\Omega, \mathcal F, \mathbb P) to measurable spaces (X, \mathcal A) and (Y, \mathcal B) respectively. Consider the following three conditions:
(1) \sigma(f) \supset \sigma(g) \bmod \mathbb P (in other words, each element in \sigma(g) is equivalent (up to null sets) to some element in \sigma(f). The notation \bmod \mathbb P is used when an author wants to make explicit the practice of treating sub-sigma algebras ignoring null sets.)
(2) There is a measurable H: (X, \mathcal A) \to (Y, \mathcal B) with g = H \circ f a.e.
(3) There is a measure preserving map H: (X, \mathcal A, \mu) \to (Y, \mathcal B, \nu) with g = H \circ f a.e. where \mu, \nu are the pushforward measures obtained from \mathbb P.

(2) implies (3) trivially. (3) implies (2) trivially (even if the measure preserving map is only a.e. defined.)

(2) implies (1) trivially. The Doob-Dynkin lemma for probability space is that (1) implies (2) if (Y, \mathcal B) is a standard Borel space. In this post, we present some proofs of this lemma and some corollaries and others. Compare with the Doob-Dynkin lemma for measurable spaces (the previous article). As we’ll see, Doob-Dynkin lemma for probability space is a special case of that for measurable spaces. The one for probability spaces suffices for most applications in probability theory and ergodic theory.

1. Proof 1

In this proof, we simply show that the Doob-Dynkin lemma for probability space follows from that for measurable spaces. It is enough to show that the condition \sigma(f) \supset \sigma(g) \bmod \mathbb P can be improved to \sigma(f) \supset \sigma(g) after discarding all points in some null set from \Omega.

For each B \in \mathcal B, there is A_B \in \mathcal A such that the symmetric difference E_B := g^{-1}B \cap f^{-1}A_B is a \mathbb P-null set. It would be nice to be able to discard all points in the union \bigcup_{B \in \mathcal B} E_B but is this a null set? Perhaps not.

Exercise 10. Show that it is enough to discard all points in the countable union \bigcup_{n} E_{B_n} where (B_n)_n is a sequence that generates \mathcal B (Existence of such a sequence is guaranteed by the assumption that (Y, \mathcal B) is a standard Borel space).

2. Proof 2

In this proof, we adapt an argument in the previous article to our setting of ignoring null sets. We divide into two cases.

Case I: when the image of g countable

In this case, there is A_i \in \mathcal A such that g^{-1}y_i = f^{-1}A_i a.e. for each i.

Notice that f^{-1}A_i form a countable partition modulo \mathbb P, in other words, the intersection of f^{-1}A_i and f^{-1}A_j is a null set when i \ne j and their union over all i has measure 1, or equivalently that for a.e. \omega, there exists unique i for which \omega \in f^{-1}A_i.

Exercise 20. Show that A_i form a countable partition mod \mu. (the general idea is that the measure algebra homomorphism A \mapsto f^{-1}A behaves like an inclusion map)

Let H be any (everywhere-defined) measurable function from X to Y such that for \mu-a.e. x and for each i we have x \in A_i \implies H(x) = y_i. It is easy to define such an H.

Now it only remains to show g = H \circ f a.e. For a.e. \omega \in \Omega, we have:
there is i for which f(\omega) is in A_i;
for that i, H(f(\omega)) = y_i (by property of H);
but that y_i = g(\omega) (by how A_i chosen);
so H\circ f = g on this \omega.

Case II: when the image of g is not countable.

WLOG assume (Y, \mathcal B) is the unit interval.

Let g_n := \lfloor 2^n g \rfloor / 2^n. There is a measurable H_n: (X, \mathcal A) \to (Y, \mathcal B) with g_n = H_n \circ f a.e..

Notice that g_n \nearrow g holds everywhere. It would be nice to take H = \lim H_n but we only know H_n converges at least on f(\Omega_0) for some subset \Omega_0 of measure 1, and we don’t know if this image is measurable.

A work around is that we let H be any (everywhere-defined) measurable such that H(x) = \lim H_n(x) for all x for which the limit exists. Such H exists and with this H we can proceed to show g = H \circ f a.e..

Another work around is to notice that the subset [H_n \nearrow] (alternatively also [\exists \lim H_n]) is measurable and so one can show that its measure is 1. Then we set H to be a \mu-a.e. limit of H_n. Then H \circ f is a P-a.e. limit of H_n \circ f. So we can proceed with this H too.

3. Proof 3

In this proof, we show that (1) implies (3) using the fact that (Y, \mathcal B, \nu) is a Lebesgue space (which follows because (Y, \mathcal B) is a standard Borel space) and the fact that a Lebesgue space can be approximated by a sequence of partitions. This proof is probably equivalent to the previous one, but let’s do it for illustration.

The idea in this proof is that points in X (resp. Y) more or less should correspond to appropriate nested sequence of elements in \mathcal A (resp. \mathcal B), so in order to build H, we only need establish an appropriate procedure to transform an appropriate nested sequence of elements in \mathcal A to that of \mathcal B, but that is done for free by observing that with abuse of notation we may safely pretend that \mathcal B \subset \mathcal A \subset \mathcal F \bmod \mathbb P (if we identify \sigma(f), \sigma(g) with \mathcal A, \mathcal B). We’ll proceed without abuse of notation.

Notice that WLOG we may replace (X, \mathcal A, \mu) and (Y, \mathcal B, \nu) with any other almost-isomorphic probability spaces at the minor cost of losing everywhere definedness of f, g.

WLOG there is a non-decreasing sequence of countable partitions \beta_n \subset B such that for any non-increasing choice B_n \in \beta_n the intersection \cap_n B_n is a single point and that (\beta_n)_n generates \mathcal B. This is possible because we may assume that the probability space Y is a disjoint union of the Cantor space and atoms, which follows from the result that any Lebesgue space is almost-isomorphic to a disjoint union of an interval and atoms).

Now we want to pick a good sequence \alpha_n such that for all n we have g^{-1}(\beta_n) = f^{-1}(\alpha_n) \bmod \mathbb P. We can assume that for each n all elements of \beta_n (and hence also those of g^{-1}(\beta_n) too) have positive measure. So we can ensure that all elements of \alpha_n have positive measure. For each n, \alpha_n is a countable partition modulo \mu and the sequence (\alpha_n)_n is non-decreasing modulo \mu. After discarding only countably many \mu-null sets from X, we can ensure that \alpha_n form a non-decreasing sequence of countable partitions. The reason we ensure elements of \alpha_n and \beta_n only have elements of positive measure is because we want to ensure the natural bijection between \alpha_n and \beta_n.

For each x \in X, let \alpha_n(x) be the unique element in \alpha_n containing x, and let \beta_n(x) be the unique element in \beta_n which corresponds to \alpha_n(x) in the sense that g^{-1}(\beta_n(x)) = f^{-1}(\alpha_n(x)) \mod \mathbb P. Define H by \{H(x)\} = \cap_n \beta_n(x) which is a well defined map from X to Y (which is because \beta_n(x) is non-decreasing (for a fixed x) because its counterpart in \Omega is non-decreasing up to null sets because its counterpart in Y is.).

(Measurability of H) For each B_n \in \beta_n, it is easy to show that H^{-1}(B_n) is measurable, moreover, equal to A_n (where A_n \in \alpha_n is the element corresponding to B_n).

(On g = H \circ f a.e.) Notice that H is measurable and everywhere defined, so g' := H\circ f is an almost-everywhere defined measurable function. So we are comparing two a.e. defined measurable functions from (\Omega, \mathcal F, \mathbb P) to (Y, \mathcal B). Inverse images of B_n \in \beta_n under the two functions are equal up to null sets because of H^{-1}(B_n) = A_n. Now notice that the set [g \neq g'] is a subset of \bigcup_{n, B, B': B, B' \in \beta_n, B \neq B'} (g^{-1}B \cap g'^{-1}B') where the terms in the latter are null sets because of the previous sentence. We have shown g = g' a.e..

(On measure preserving) It is easy to show that H is measure preserving from the fact that g = g' a.e.. Alternatively, it also follows from H^{-1}(B_n) = A_n.

4. Uniqueness of H

If H and H’ satisfy (2) , then H = H' holds \mu-a.e.. This follows from the following two observations:

  • H \circ f = H' \circ f holds \mathbb P-a.e.
  • [H = H'] is measurable (because (Y, \mathcal B) is a standard Borel space).

5. applications

Exercise 25. Let f, g be measurable functions from a probability space (\Omega, \mathcal F, \mathbb P) to measurable spaces (X, \mathcal A) and (Y, \mathcal B) respectively. Show that \sigma(f) = \sigma(g) \bmod \mathbb P if and only if there are X_0 \in \mathcal A with \mu(X_0) = 1 and Y_0 \in \mathcal B with \nu(Y_0) = 1 and a bi-measurable bijection H: X_0 \to Y_0 such that g = H \circ f a.e..

Exercise 30. Given two measure-theoretic dynamical systems (X, \mathcal A, \mu, T) and (Y, \mathcal B, \nu, S) (measure preserving transformations on Lebesgue spaces), show that the latter is a factor of the former if and only if there is a joining \lambda (a T\times S-invariant probability measure on (X \times Y)) such that \mathcal B \subset \mathcal A mod \lambda (with abuse of notation).

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