The proof of theorem 10.4.1 (Fubini) in the text "Real Analysis" by P.A. Loeb says:
Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ two complete measure spaces [...] Given a non negative integrable function $f$ on $X \times Y$ there is an increasing sequence of simple non negative functions $\phi_n$ with limit $f$. [...] For $\mu$-almost all $x \in X$ each function $\phi_n(x,\cdot)$ is measurable on $Y$. [...]
I can't see why $\phi_n(x,\cdot)$ are measurable on $Y$ only for $\mu$-almost all $x \in X$. I would say: since $\phi_n$ is a measurable function for every $\alpha \in \mathbb{R}$ the set $\{\phi_n>\alpha\}$ is in $\mathcal{A} \otimes \mathcal{B}$. I found here that (for $\sigma$-finite measure spaces, but I doesn't seem to me that this is necessary) a cross section $E_x:=\{y\in Y :(x,y) \in E\}$ where $E \in \mathcal{A} \otimes \mathcal{B}$ is in $Y$. Therefore, since $\{\phi_n(x,\cdot)>\alpha\} = \{\phi_n>\alpha\}_x$, I would conclude that $\phi_n(x,\cdot)$ is measurable for every $x$.
I must have missed something. Any help would be greatly appreciated.
I think that the point is the following.
If your measure spaces are complete, then the $\sigma$-algebra on $X\times Y$ is often implicitly assumed to be the completion of the product $\sigma$-algebra $\mathcal{A}\times \mathcal{B}$. If this is the case, then the measurability of the sections holds only for almost every section.
For details you can see Rudin, R&CA, pp. 167-170 ("Completion of Product Measures").