Let $(X, \mathcal X, \mu)$ and $(Y, \mathcal Y, \nu)$ be $\sigma$-finite measure spaces. Let $\overline{\mathbb R} := \mathbb R \cup \{\pm \infty\}$.
- $f:X \to \overline{\mathbb R}$ is called $\mu$-simple if it has the form $f = \sum_{i=1}^n a_i 1_{A_i}$ where $(a_i) \subset \mathbb R \setminus \{0\}$ and $(A_i) \subset \mathcal X$ is pairwise disjoint such that $\mu(A_i) < \infty$ for all $i = 1, \ldots,n$.
- $f:X \to \overline{\mathbb R}$ is called $\mu$-measurable if it is a $\mu$-a.e. pointwise limit of a sequence of $\mu$-simple functions.
Let $\mathcal X \otimes \mathcal Y$ the product $\sigma$-algebra on $X \times Y$. Let $\pi^X:X \times Y \to X$ and $\pi^Y:X \times Y \to Y$ be the projection maps. Let $\gamma$ be a measure on $\mathcal X \otimes \mathcal Y$ with marginals $\mu$ on $X$ and $\nu$ on $Y$, i.e., $(\pi^X)_\sharp \gamma = \mu$ and $(\pi^Y)_\sharp \gamma = \nu$. s
Is $c_x$ is $\nu$-measurable for $\mu$-a.e. $x \in X$? If not, is there $x \in X$ such that $c_x$ is $\nu$-measurable?
The answer is affirmative for a very particular case below.
Theorem If $\mu(X)=\nu(Y)=1$ and $\gamma = \mu \otimes \nu$, then $c_x$ is $\nu$-measurable for $\mu$-a.e. $x \in X$.
Proof Let $(c_n)$ be a sequence of $\gamma$-simple functions such that $c_n (x, y) \to c(x, y)$ as $n \to \infty$ for $\gamma$-a.e. $(x, y) \in X \times Y$. This implies there is $N \in \mathcal X \otimes \mathcal Y$ such that $\gamma(N)=0$ and $$ c_n (x, y) \to c(x, y) \quad \forall (x,y) \in S := (X \times Y) \setminus N. $$
For $x \in X$, let $S_x := \{y \in Y : (x, y) \in S\}$ be the $x$-section of $S$. Then $S_x \in \mathcal Y$. By Fubini's theorem, $\nu (S_x) =1$ for $\mu$-a.e. $x \in X$. WLOG, we assume $c_n (x, y) = \sum_{i=1}^{\varphi_n} a_{ni} 1_{A_{ni}} (x, y)$. Let $A_{ni,x}$ be the $x$-section of $A_{ni}$. Then $A_{ni, x} \in \mathcal Y$. Let $$ c_{n, x} : Y \to \overline{\mathbb R}, y \mapsto \sum_{i=1}^{\varphi_n} a_{ni} 1_{A_{ni, x}} (y). $$
Then $c_{n, x}$ is $\nu$-simple for every $(n,x) \in \mathcal N \times X$. Clearly, $c_{n, x} \to c_x$ on $S_x$ for $\mu$-a.e. $x\in X$. This completes the proof.
