Here is a result asserting that, under some technical conditions, we can replace a Borel almost everywhere homomorphism by a Borel homomorphism. It is taken from Zimmer's book "Ergodic Theory and Semisimple Lie Groups" (pp.197, Appendix B), but I'm not interested in the result itself.
The main point is that I want to understand how its proof uses Fubini's theorem (click here for the statement), because I havent't got it.
Basic definitions: By Borel structure we mean the $\sigma$-algebra generated by the topology. For Haar measure, click here.
Proposition Suppose $J\subseteq [0,1]$ is a topological group whose Borel structure is countably generated (i.e. there exists a countable family of Borel sets which separates points and generates the $\sigma$-algebra). Let $G$ be a locally compact group with probability Haar measure $m$. Let $\pi\colon G\to J$ be a Borel function s.t. $\pi(xy)=\pi(x)\pi(y)$ for almost all $(x,y)\in G\times G$. Then there is a Borel homomorphism $\pi_0\colon G\to J$ s.t. $\pi=\pi_0$ almost everywhere (a.e.).
The proof starts as follows: Consider the set $$H\{y\in G\mid \text{the function } x\mapsto \pi(x)^{-1}\pi(xy) \text{ is constant a.e.}\}.$$
By Fubini, $H$ is conull (i.e. the complement has measure $0$) [and I don't understand why].
Later on, he uses Fubini's theorem once again, this time to prove that $\pi_0$ is Borel [How?]: he argues that it follows from the fact that $\pi_0(y)=\int \pi(x)^{-1}\pi(xy)\,dm$.
Question: As I stated above, what I look for is a clear (possibly detailed) explanation of how the author applied Fubini's Theorem in both cases.
The set $M=\{(x,y)\in G\times G:\pi(x)^{-1}\pi(xy)=\pi(y)\}$ is co-null by assumption. By Fubini, for almost all $y\in G$, the section $M_y$ of $M$ at level $y$ is (measurable and) co-null. That section is $M_y=\{x\in G:\pi(x)^{-1}\pi(xy)=\pi(y)\}$, so, whenever $M_y$ is co-null. the function $x\mapsto \pi(x)^{-1}\pi(xy)$ is constant a.e. (specifically, constant with value $\pi(y)$ on $M_y$). Since this happens for a.e. $y$, we have that the $H$ in the question is co-null.