Measurability of the measure of a "marginal" set (Lemma for Fubini's theorem) My quesion is the following:
I have $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{S}, \nu)$ as measured spaces. $\nu$ is $\sigma$-finite. $E\in \mathcal{A} \otimes \mathcal{S}$ and for $x \in X$, I define $E_{x,.} := \{y\in Y | (x,y) \in E\}$

Demonstrate that $[x\rightarrow \nu (E_{x,.})] \in \mathcal{L}^0(X,\mathcal{A},\mu)$

I have no clue whatsoever.
 A: @drhab has addressed how to show that each $E_x \in \mathcal{S}$ in the comments, so I will focus on showing that the maps $f_E$ defined by $f_E(x) = \nu(E_x)$ are measurable.
To this end, define
$$
\mathcal{E}
= \{E \in \mathcal{A} \otimes \mathcal{S}: f_E \text{ is measurable}\}.
$$
We will show that $\mathcal{E} = \mathcal{A} \otimes \mathcal{S}$ using the $\pi$-$\lambda$ theorem, first by noticing that $G = \{A \times S :A \in \mathcal{A}, S \in \mathcal{S}\}$ is a $\pi$-system that generates $\mathcal{A} \otimes \mathcal{S}$.
Next, we show that $\mathcal{E}$ is a $\lambda$-system.
Checking the axioms one-by-one,

*

*$f_{X \times Y}(x) = \nu((X \times Y)_x) = \nu(Y)$ is constant and so measurable, rendering  $X \times Y \in \mathcal{E}$.


*If $E, F \in \mathcal{E}$ with $E \subseteq F$ (since $\nu$ is $\sigma$-finite, we can assume wlog that $\nu(F) < \infty$), then, for any $B \in \mathcal{B}_{\mathbf{R}}$, we have that $(F\setminus E)_x = F_x \setminus E_x$, rendering
\begin{align*}
f_{F\setminus E}^{-1}((a, \infty))
&= \{x : \nu((F\setminus E)_x) > a\} \\
&= \{x : \nu(E_x) < \nu(F_x) - a\} \\
&= f_E^{-1}([0, \nu(F_x) - a)),
\end{align*}
which is in $\mathcal{S}$ since $E \in \mathcal{E}$.
Hence, $f_{F\setminus E}$ is measurable, and so $F\setminus E \in \mathcal{E}$.


*If $\{E_n\}_n$ is a sequence of increasing sets in $\mathcal{E}$ with $E = \bigcup_{n=1}^\infty E_n$, then
\begin{align*}
f_E^{-1}((a, \infty))
&= \Bigl\{x : \nu\Bigl(\bigcup_{n=1}^\infty E_n\Bigr) > a\Bigr\} \\
&= \{x : \lim_{n\rightarrow\infty}\nu(E_n) > a\} \\
&= \bigcup_{n=1}^\infty \{x : \nu(E_x) > a\} \\
&= \bigcup_{n=1}^\infty f_{E_n}^{-1}((a, \infty)),
\end{align*}
which is in $\mathcal{S}$ since each $E_n \in \mathcal{E}$.
Hence, $f_E$ is measurable and so $E \in \mathcal{E}$.
Thus, $\mathcal{E}$ is a $\lambda$-system, and we conclude from the $\pi$-$\lambda$ theorem that
$$
\mathcal{A}\otimes \mathcal{S}
= \sigma(G)
\subseteq \mathcal{E}
\subseteq \mathcal{A}\otimes \mathcal{S},
$$
from which we conclude that all such $f_E$ are measurable.
