# When can a measurable mapping be factorized?

Problem 13.3 of Probability and Measure by Billingsley states:

$(\Omega, \mathcal{F})$ and $(\Omega', \mathcal{F}')$ are two measurable spaces. Suppose that $f: \Omega \rightarrow \mathbb{R}^1$ and $T:\Omega \rightarrow \Omega'$. Show that $f$ is measurable $T^{-1}\mathcal{F}':= \{ T^{-1}A': A' \in \mathcal{F}' \}$ if and only if there exists a map $\phi: \Omega' \rightarrow \mathbb{R}^1$ such that $\phi$ is measurable $\mathcal{F}'$ and $f= \phi T$.

Hint: First consider simple functions and then use Theorem 13.5.

where Theorem 13.5 states

If $f$ is real and measurable $\mathcal{F}$, there exists a sequence $\{f_n\}$ of simple functions, each measurable $\mathcal{F}$, such that $$0 \leq f_n(\omega)\uparrow f(\omega) \text{ if }f(\omega) \geq 0$$ and $$0 \geq f_n(\omega) \downarrow f(\omega) \text{ if }f(\omega) \leq 0.$$

I would like to consider a general case of Problem 13.3 where the codomain of $f$ is a general measurable space $(X, \mathcal{N})$ rather than $(\mathbb{R}^1, \mathcal{B}(\mathbb{R}^1))$, i.e. when it can be true that

$(\Omega, \mathcal{F})$ and $(\Omega', \mathcal{F}')$ are two measurable spaces. Suppose that $f: \Omega \rightarrow X$ and $T:\Omega \rightarrow \Omega'$. $f$ is measurable $T^{-1}\mathcal{F}'/\mathcal{N}$ if and only if there exists a map $\phi: \Omega' \rightarrow X$ such that $\phi$ is measurable $\mathcal{F}'/\mathcal{N}$ and $f= \phi T$.

noticing that Theorem 13.5 does not apply here.

Thanks and regards!

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In order to make Billingsley's outlined argument work you need that the $\sigma$-algebra on $X$ is countably generated and separated, so you could take $X$ to be any separable metrizable space, e.g. $\mathbb{R}^n$ or some subset of it. But where does this problem arise? What do you need that statement for? –  t.b. Nov 3 '11 at 6:58
t.b.: Thanks! I may seem aimless, but I just wanted to know what makes the case when the codomain of $f$ is $\mathbb{R}$ with its Borel $\sigma$-algebra so special that factorization holds. –  Tim Nov 3 '11 at 7:02
Exactly what I mentioned in my previous comment: that the $\sigma$-algebra is countably generated and separated (i.e., that you work with a (complete) separable metric space as range). You can use a technique similar to the one given by André in his answer for a generalization (and for many results that deal with the reals as range of measurable functions). –  t.b. Nov 3 '11 at 7:06
@t.b.: Thanks! Not sure if you are familiar with the first two quoted results. I wonder if the sequence of simple functions in Theorem 13.5 (the second quote) is not unique (Billingsley constructed one in the same way as Rudin did in his Real and Complex Analysis), and if the measurable mapping $\phi$ is not unique either in Problem 13.3 (the first quote)? –  Tim Nov 24 '11 at 20:47