## 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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