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I know that the composition of a continuous function with a measurable function is measurable, whereas the composition of a measurable function with a measurable function is not necessarily measurable. Now I want to prove that the composition of a Borel measurable function $ g:(X,\mathscr B)\to(Y,\hat\tau)$ with a measurable function $f:(\Omega,\mathscr M) \to (X, \tau)$, i.e. $g \circ f$, is measurable. Proof: let $A$ be open. Then $g^{-1}(A)$ is either $G{\delta}$ (countable intersection of open sets) or $F{\sigma}$ (countable union of closed sets). Suppose the former case. Then

$f^{-1}(\bigcap_{i \mathop \in N} {A_i}) = \bigcap_{i \mathop \in N} (f^{-1}({A_i})) $ where $A_i$ open, $f^{-1}(A_i) \in \mathscr M $ and so under countable intersection. How can I prove the same result in the case of $F{\sigma}$? Indeed the counterimage of a closed set is not necessarily measurable. Here I am exploiting the definition: a functions is measurable if the counterimage of a open set is measurable. Can you give me a hint, please?

Federico Lavatori
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A measurable space is a pair $\newcommand\cA{\mathcal{A}}\newcommand\cB{\mathcal{B}}\newcommand\cC{\mathcal{C}}(X,\cA)$ of a set and a $\sigma$-algebra on that set. We say that a function between measurable spaces $f:(X,\cA)\to(Y,\cB)$ is measurable if $\newcommand{\inv}{^{-1}}f\inv(B)\in \cA$ for all $B\in\cB$. It is fairly easy to see with this definition that if $g:(Y,\cB)\to(Z,\cC)$ and $f:(X,\cA)\to (Y,\cB)$ are measurable, then $g\circ f : (X,\cA)\to (Z,\cC)$ is measurable, since $g\inv(C)\in\cB$ for $C\in\cC$, so $(g\circ f)\inv(C)=f\inv(g\inv(C))\in \cA$ for $C\in\cC$.

Now we just need to relate this definition to the notion of measurability you've given for functions from measurable spaces to topological spaces. Well, topological spaces $(X,\tau_X)$ come equipped with a natural $\sigma$-algebra, the Borel $\sigma$-algebra, generated by the open sets. So we'll show that $f:(X,\cA)\to (Y,\tau_Y)$ is measurable with the definition you've given if and only if $f:(X,\cA)\to (Y,\cB(Y))$ is measurable with the definition above, where $\cB(Y)$ is the Borel $\sigma$-algebra, $\sigma(\tau_Y)$.

It turns out that to check measurability as a function between measure spaces, it suffices to check the definition on a generating set for the Borel $\sigma$-algebra, i.e. the open sets. There are several strategies to prove this. The key idea is that you can define a preimage $\sigma$-algebra, which is generated by the preimages of the generators, and if the preimages are in $\cA$, the $\sigma$-algebra they generate must be contained in $\cA$. See this question and answer for the relevant lemma needed. Or this answer for another proof and application of the lemma to prove a similar statement to the one desired here. I won't go into the details here, the linked questions and answers should provide enough detail along with what I've said already.

jgon
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