Does the Doob-Dynkin lemma hold for any measurable space that separates points?

Question

This is cross-posted from math.se, where I got no responses.

Let's say that a measurable space $(Z, \mathcal Z)$ has the "Doob-Dynkin property" iff for any set $X$, measurable space $(Y, \mathcal Y)$ and function $f:X\to Y$, a function $g:X\to Z$ is $f^{-1}(\mathcal Y)$ measurable if and only if it's a measurable function of $f$. This, with $\mathbb R$ playing the role of $Z$, is the Doob-Dynkin lemma as I was taught it.

It's easy to show that not every measure space has the Doob-Dynkin property. But let's say a separating measure space is one in which, given $x\ne y$, there exists a measurable set containing $x$ but not $y$.

Does every separating measure space have the Doob-Dynkin property?

I think I have a simple proof that this is the case when $f$ is surjective. Can this result be proven for a non-surjective $f$, or is there a counter-example?

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Here's an outline of my proof.

First - with the above notation, if $g$ is $f^{-1}(\mathcal Y)$ measurable, and $Z$ separates points, then $g$ is a function (not necessarily measurable) of $f$. Proof: we only need to show that $f(x)=f(y)\implies g(x)=g(y)$, so let $x$ and $y$ be such that $f(x)=f(y)$. Then every set in $f^{-1}(\mathcal Y)$ either contains $x$ and $y$, or contains neither. If $g(x)\ne g(y)$, then let $A$ be some set of $\mathcal Z$ containing $g(x)$ but not $g(y)$. Then $g^{-1}(A)$ contains $x$ but not $y$ and so is not in $f^{-1}(\mathcal Y)$, thus $g$ is no measurable. Thus $g$ is a function of $f$, say $g=h\circ f$.

If $f$ is surjective then $h$ (which is unique when $f$ is surjective) is measurable. Proof: let $A\in \mathcal Z$. Since $g$ is $f^{-1}(\mathcal Y)$ measurable, $g^{-1}(A)=f^{-1}(B)$ for some $B\in\mathcal Y$. One can check that $h^{-1}(A)=B$.

Answer 1

According to a paper of Pratelli [2], a characterization of spaces with the Doob-Dynkin property was given by Pintacuda [1]. Unfortunately, I was not able to locate Pintacuda's paper, so I cannot comment on its contents.

In [2], Pratelli gives another proof of Pintacuda's result. You can find this paper (in French) here. His main result is the following (which I translated from French):

Theorem: A measurable space $(Z, \mathcal{Z})$ has the Doob-Dynkin property if and only if it is separated and injective.

In this statement, "injective" means that $(Z,\mathcal{Z})$ is injective in the category-theoretic sense with respect to the class of morphisms given by subspace inclusions. More explicitely, $(Z, \mathcal{Z})$ is injective if for any space $(Y, \mathcal{Y})$ and any subset $A \subset Y$ equipped with the subspace $\sigma$-algebra, any measurable map $A \to Z$ can be extended to a measurable map $Y \to Z$.

The proof of the "if" part of the theorem is essentially the one you gave, and the injectivity assumption is an ad hoc way to account for the fact that $f$ might not be surjective. You first prove that there exists a map $h: f(X) \to Z$ such that $g = h \circ f$. The author shows that this map is measurable if $f(X)$ is equipped with the subspace sigma algebra. By the injectivity of $Z$, it extends to a measurable map from $Y$ to $Z$. When $f$ is assumed surjective, the proof reduces to the one you gave.

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Let's say that a space $Z$ has the weak Doob-Dynkin property if it has the Doob-Dynkin property with respect to surjective maps $f$. By the discussion above, a separated space has the weak Doob-Dynkin property. In fact, I believe that the converse also holds, i.e. a space has the weak Doob-Dynkin property if and only if it is separated.

Here is a sketch of a proof, which is essentially a minor modification of Proposition (1.1) in Pratelli's paper. Suppose $(Z, \mathcal{Z})$ has the weak Doob-Dynkin property. Take $X= Z$ and $g: Z \to Z$ the identity. Following Pratelli, consider the space $\tilde{Y}$ of maps $\mathcal{Z} \to \{ 0, 1\}$ and the map $\tilde{f}: Z \to \tilde{Y}$ that sends $x \in Z$ to the dirac measure $\delta_x: \mathcal{Z} \to \{0, 1 \}$. Equip $\tilde{Y}$ with the sigma algebra $\mathcal{A}$ generated by sets of the form $$ A_{U,V} = \{ h: \mathcal{Z} \to \{ 0, 1 \} \; : \; h(U) \in V \} $$

where $U$ ranges over $\mathcal{Z}$ and $V$ ranges over the power set of $\{0, 1 \}$. Now, let $Y \subset \tilde{Y}$ be the image of $\tilde{f}$, $f: Z \to Y$ be the restriction and $\mathcal{A}_Y$ be the subspace sigma algebra on $Y$. Then $f$ is surjective and one can check that it is measurable and strict, i.e. $f^{-1}(\mathcal{A}_Y) = \mathcal{Z}$ (this is Pratelli's terminology).

By the Doob-Dynkin property, there is a measurable map $h: Y \to Z$ with $id_Z = h \circ f$. But this implies that $f$ is injective. It is an easy check that this is equivalent to $Z$ being separated.

[1] MR1008597 Pintacuda, Nicolò. On Doob's measurability lemma. (Italian. English summary) Boll. Un. Mat. Ital. A (7) 3 (1989), no. 2, 237–241.

[2] MR1071531 Pratelli, Luca. Sur le lemme de mesurabilité de Doob. (French) [On Doob's measurability lemma] Séminaire de Probabilités, XXIV, 1988/89, 46–51, Lecture Notes in Math., 1426, Springer, Berlin, 1990.

Answer 2

Let's agree that "$g$ is a measurable function of $f$" means that there is a measurable function $h$ from $Y$ to $Z$ such that $g=h\circ f$.

The first question is answered by OP. A counter-example for the second question:

Let $Z$ be a subset of $[0,1]$ that is not analytic, so that $Z$ not a Borel measurable image of $[0,1]$ (Fremlin's Measure Theory, 423G).

$\mathcal{Z}$ is the Borel $\sigma$-algebra on $Z$, $(Y,\mathcal{Y})$ is $[0,1]$ with its Borel $\sigma$-algebra, $X=Z$, $g$ is the identity mapping from $X$ to $Z$, and $f$ is the embedding of $X$ to $Y$.

If $h$ is a mapping from $Y$ to $Z$ such that $g=h\circ f$ then $Z=h(Y)$.