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!