I suspect that proving what I wish to prove is not so very tough, but I haven't succeeded yet. I am pushing around sets between three spaces using two functions and their composition.
I did some poking around old questions, feeling that this must have been answered already, but I didn't find much besides this (which I do use) and this other one (which I do not think will help).
Proposition.
Let $\phi:X \to Y$, and let $(X,\Sigma)$ be a measurable space. We can obtain a related measurable space $(Y,T)$ by defining $T=\{F\subset Y:\phi^{-1}F\in \Sigma\}.$ Also fix a function $g:\text{dom}(g)\to\mathbb{R}$ with $\text{dom}(g)\subset Y.$
Show that $g$ is $T$-measurable iff $g\phi=g \circ \phi$ is $\Sigma$-measurable.
I have already showed the "only if" part, which is straight-forward:
Proof.
Fix any half-line $H=H_a=\{t:t<a\}$ for $a$ real. We need to show $(g\phi)^{-1}H$ is an element of the subspace sigma-algebra: $$\Sigma_{\phi^{-1}\text{dom}(g)} = \Sigma_{\text{dom}(g\phi)}.$$ We note that $(g\phi)^{-1}H={\phi}^{-1}(g^{-1}H)$, and that $g^{-1}H$ belongs (by hypothesis) to $$T_{\text{dom}(g)},$$ and hence $g^{-1}H=\text{dom}(g) \cap A$, for some $A$ in $T$. Finally, $$(g\phi)^{-1}H={\phi}^{-1}(\text{dom}(g) \cap A)={\phi}^{-1}\text{dom}(g) \;\cap\; {\phi}^{-1}A=\text{dom}(g\phi) \;\cap\; {\phi}^{-1}A,$$ with ${\phi}^{-1}A\in\Sigma.$ This means the composition is a measurable function.
We needed to swap the operations $\phi^{-1}$ and $\cap$ to make it work, but apart from that the proof is mainly applying the definition. But for the converse statement, I stalled with trying this approach. You can aim to prove $g^{-1}H \in T_{\text{dom}(g)}$, knowing that $A$ belongs to $\Sigma$ and $$\phi^{-1}(g^{-1}H)=(\phi^{-1}\text{dom}(g))\cap A,$$
but the forward mapping $\phi$ on sets only cooperates fully with union.
Some of the ideas I have tried:
We can also pull back a sigma-algebra under a mapping. For example, if $\mathcal{B}$ is the Borel algebra of real sets, we can pull it back to a sigma-algebra of subsets of the domain of $g$; equally well we can pull it back to a sigma-algebra of subsets of the domain of $g\phi$.
We aim for $g^{-1}H \in T_{\text{dom}(g)}$ for any half-line. We can note the set of all real sets satisfying this is a sigma-algebra: $\{E\subset\mathbb{R}: g^{-1}E \in T_{\text{dom}(g)}\}$
It might simplify matters to consider the restriction of $\phi$ to the domain of $g\phi$, say $\phi_0: X_0 \to Y_0.$
Appendix:
My attempts to find an existing answer: