Here is a restatement of a problem in a textbook I encountered. I'm well beyond the age of doing homework and this is purely for self-study.
Exercise: Let $f : (X,\Sigma_1) \to (Y, \Sigma_2)$ and $h:(X,\Sigma_1) \to (\mathbb R, \mathcal B)$ be measurable maps where in the latter case $\mathcal B$ denotes the Borel $\sigma$-algebra over $\mathbb R$. Let $\Sigma_f = \sigma(f)$. Show that $h$ is $\Sigma_f$-measurable if and only if there exists $g : (Y,\Sigma_2) \to (\mathbb R, \mathcal B)$ such that $h(x) = g(f(x))$ for all $x \in X$.
One direction of the proof is easy. Suppose such a $g$ exists. Then, for all $B \in \mathcal B$, $h^{-1}(B) = f^{-1}( g^{-1}(B) )$ and so $h^{-1} \in \Sigma_f$.
There seem to be some holes in the opposite direction which I can't quite fill.
For all $z \in \mathbb R$, I defined
$ A_z = \{x: h(x) = z\}. $
Then $A_z \in \Sigma_1$ since the singletons $\{z\}$ are Borel-measurable. Also, for $z \neq z'$, it is true that $A_z \cap A_{z'} = \emptyset$. Now, if $h$ is $\Sigma_f$-measurable, then $A_z = f^{-1}(B_z)$ for some $B_z \in \Sigma_2$. But then, for $z \neq z'$, we have that $B_z \cap B_{z'} = \emptyset$ as well, so the $\{B_z\}_{z \in \mathbb R}$ sets partitions $Y$ modulo the portion not in the image of $f$.
Now, set $g(y) = z$ on $B_z$ and set $g(y) = 0$ on $y \in N_0 := Y \setminus \cup_{z \in \mathbb R} B_z$. It seems reasonable to claim that $N_0$ is a measurable set by considering that $N_0 = Y \setminus \cup_n C_n$ where $f^{-1}(C_n) = h^{-1}((-\infty,n))$ and $C_n \in \Sigma_2$ by assumption.
But, this only seems to show that we can construct a well-defined $g$. It doesn't seem to prove that it is measurable! To get measurability we need to show something in addition to this, like $\{y: g(y) \leq z\} \in \Sigma_2$ for all $z \in \mathbb R$.
For z < 0 it seems we should be able to get a correspondence between $\{y: g(y) \leq z\}$ and $C_z$ where $C_z \in \Sigma_2$ satisfies $f^{-1}(C_z) = h^{-1}((-\infty,z))$. For $z \geq 0$, I think it would be something like $\{y : g(y) \leq z\} = C_z \cup N_0$, I think.
I can't quite seem to make the argument go through.
Questions:
- Is this on the right track? If so, how do we finish it off? (It seems a little "too constructive" for a typical measure-theoretic argument.)
- Is there some other more clever or direct argument? If so, what is it?