This is [cross-posted](https://math.stackexchange.com/questions/2173639/does-the-doob-dynkin-lemma-hold-for-any-separable-measure-space) from math.se, where I got no responses.


Let's say that a measurable space $(Z, \mathcal Z)$ has the "[Doob-Dynkin](https://en.wikipedia.org/wiki/Doob%E2%80%93Dynkin_lemma) 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](https://math.stackexchange.com/questions/508203/failure-of-doob-dynkin-lemma-in-general-measurable-spaces) 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 $h^{-1}(A)=B$.



Does the Doob-Dynkin lemma hold for any measurable space that separates points?
A characterization of spaces with the Doob-Dynkin property was given by N. Pintacuda in his paper *Sul lemma di misurabilità di Doob* (1989). Unfortunately, the paper is in Italian and it doesn't show up on Google. 


L. Pratelli gave another proof of Pintacuda's result in his paper *Sur le lemme de mesurabilité de Doob* (1990). You can find this paper (in French) [here](http://www.numdam.org/article/SPS_1990__24__46_0.pdf). His main result is the following (the translation from French is my own, and I changed the notation and terminology to match yours): 



> 
> **Theorem**: A measurable space $(Z, \mathcal{Z})$ has the Doob-Dynkin property if and only if it is separated and injective.
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> 
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A measure space $(Z, \mathcal{Z})$ is said *injective* if for any space $(Y, \mathcal{Y})$ and any subset $A \subset Y$ (measurable or not), any measurable map from $A$ (with the subspace sigma algebra) to $Z$ can be extended to a measurable map from $Y$ to $Z$. 


The proof of the "if" part 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.


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 $f$ injective. It is an easy check that this is equivalent to $Z$ being separated.


[1] [MR1008597](http://www.ams.org/mathscinet-getitem?mr=1008597) Pintacuda, Nicolò. On Doob's measurability lemma. (Italian. English summary) *Boll. Un. Mat. Ital. A* (7) 3 (1989), no. 2, 237–241. 


[2] [MR1071531](http://www.ams.org/mathscinet-getitem?mr=1071531) 
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. 


