If $f=h\circ g$, then there's a measurable function $\tilde h$ such that $f=\tilde h\circ g$

Question

Let $X,Y,Z$ be three standard measurable spaces and $f:X\to Z$ and $g:X\to Y$ two measurable functions. Suppose that there's a function $h:Y\to Z$ such that $f=h\circ g$. How can I show that there's a measurable function $\tilde h:Y\to Z$ such that $f=\tilde h \circ g$?

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According to Joscha Gillessen's answer to Minimal sufficient statistic: a measurability issue in a well-known theorem that proposition is true and, if I haven't misunderstood, the proposition is a consequence of the following theorem:

Blackwell: Let $(X,\mathcal{A})$ be a standard measurable space and $\mathcal{A}_0$ a $\sigma$-algebra on $X$ generated by a countable subset of $\mathcal{A}$. Then a subset of $X$ belongs to $\mathcal{A}_0$ if and only if it belongs to $\mathcal{A}$ and is the union of a family of atoms of $\mathcal{A}_0$.

However I don't know how to use this theorem to obtain that result.

Note: A measurable space $X$ is said to be standard if there is a polish space $Z$ such that $X$ and $Z$ are isomorphic.

Answer 1

I try to provide a direct proof here using the above-mentioned theorem of Blackwell and Kuratowski's extension theorem.

In the following consider the subspace $W:=g(X) \subseteq Y$ endowed with the subspace-$\sigma$-algebra: $$\Sigma_W := \{ B \cap W \mid B \in \Sigma_Y \}.$$

Kuratowski's extension theorem states that every measurable map $ h:W \to Z$ to a standard measurable space $Z$ can be extended to a measurable map $\tilde h:Y \to Z$.

This means that we only need to consider the surjective measurable map $g: X \to W$ and a map $h: W \to Z$ with $h \circ g = f$. We are thus left to show that such an $h$ is measurable.

Since $Y$ is a standard measurable space, $\Sigma_Y$ is countably generated and separated, which implies the same for $\Sigma_W$. So there exists a countable algebra $F$ that separates the points of $W$ such that $\Sigma_W = \sigma(F)$. In particular, the atoms of $\Sigma_W$ are the singletons $\{w\}$ for $w \in W$.

Now consider the two sub-$\sigma$-algebras of $\Sigma_X$: $$ \Sigma_1 := \sigma(\{ g^{-1}(B) \mid B \in F \}), \qquad \Sigma_2 := \{ g^{-1}(B) \mid B \subseteq W, g^{-1}(B) \in \Sigma_X \}. $$ Since $g$ is measurable, it is clear that we have the inclusion: $$\Sigma_1 \subseteq \Sigma_2 \subseteq \Sigma_X.$$ Furthermore, $\Sigma_1$ is countably generated and its atoms are of the form $g^{-1}(w)$ for $w \in W$. Now Blackwell's theorem also implies the reverse inclusion $\Sigma_2 \subseteq \Sigma_1$. Indeed, if $A = g^{-1}(B) \in \Sigma_2$ with $g^{-1}(B) \in \Sigma_X$ and $B \subseteq W$, then $A = \bigcup_{w \in B} g^{-1}(w)$ and thus $A \in \Sigma_1$ by Blackwell's theorem. We thus get the equality: $$\Sigma_1 = \Sigma_2.$$ We now focus on the push-forward $\sigma$-algebras: $$ g_*\Sigma := \{ B \subseteq W \mid g^{-1}(B) \in \Sigma \}.$$

By the surjectivity of $g$ we always have $g(g^{-1}(B))=B$ for every $B \subseteq W$. We thus get the chain of equalities: $$ g_*\Sigma_X = g_*\Sigma_2 = g_*\Sigma_1 = \sigma(F)=\Sigma_W. $$

Now let $C \in \Sigma_Z$. By the measurability of $f$ we get: $$\Sigma_X \ni f^{-1}(C) = g^{-1}(h^{-1}(C)).$$ This shows that: $$h^{-1}(C) \in g_*\Sigma_X = \Sigma_W,$$ which shows that $h$ is measurable. This concludes the proof.

Edit: A short argument to see Kuratowski's extension theorem is to consider $Z=[0,\infty]$, which one wlog can always assume if $Z$ is a standard measurable space. Then the map $h:W \to Z$ is the limit of simple functions of the form: $$ \sum_n c_n \cdot 1\!\!1_{B_n \cap W}.$$ Then an extension $\tilde h: Y \to Z$ is just given by taking the same limit of simple functions of the form: $$ \sum_n c_n \cdot 1\!\!1_{B_n}.$$