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?
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The answer is affirmative for a very particular case below.




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> 
> **Theorem** If $\mu(X)=\nu(Y)=1$ and $\gamma = \mu \otimes \nu$, then $c\_x$ is $\nu$-measurable for $\mu$-a.e. $x \in X$.
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*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.



Let $c: X \times Y \to \overline{\mathbb R}$ be $\gamma$-measurable. Is $c_x:Y \to \overline{\mathbb R}, y \mapsto c(x, y)$ $\nu$-measurable?
No to both. Let $\gamma$ be the uniform distribution on $\Delta=\{(x,x)\mid x\in [0,1]\}$, the diagonal of $[0,1]^2$. The marginals are simply the uniform distribution on $[0,1]$. Fix some function $g:[0,1]\to\mathbb{R}$ that is not Lebesgue measurable. Define $c$ by $c(x,y)=g(y)$ if $x\neq y$ and $c(x,y)=0$ for $x=y$. Then $c$ is $\gamma$-measurable, but each section agrees almost everwhere with $g$ and is, therefore, not Lebesgue-measurable. Here, this is the same as not being $\nu$-measurable.


