Consider three measurable spaces $(X, \mathcal{X})$, $(Y, \mathcal{Y})$, $(Z, \mathcal{Z})$. Let $S$ and $T$ be maps from $X$ to $Y$ and from $Y$ to $Z$, respectively. Assume $S$ is $\mathcal{X}/\mathcal{Y}$-measurable and that the composition map $T \circ S$ is $\mathcal{X}/\mathcal{Z}$-measurable. Is then necessarily the case that $T$ is $\mathcal{Y}/\mathcal{Z}$-measurable?
Of course, for any set $B \in \mathcal{Z}$ we have $(T \circ S)^{-1}(B)=S^{-1}(T^{-1}(B)) \in \mathcal{X}$. Are we then allowed to conclude that $T^{-1}(B) \in \mathcal{X}$ because of the measurability of $S$?
Not true. Let $X=Y=Z=\mathbb R$ each with the Borel sigma algebra.Let $T=I_A$ where $A$ is a subset of $(0,\infty)$ which is not a Borel set. Let $Sx=-x^{2}$. Then $T\circ S=0$, hence $T\circ S$ is measurable and $S$ is measurable but $T$ is not.