Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces,
- $(E, | \cdot |)$ a Banach space,
- $S (X)$ the space of $\mu$-simple functions from $X$ to $E$,
- $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$,
- $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$,
- $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
- $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.
Let $f:X \times Y \to E$ be $\lambda$-measurable. Then there is a sequence $(f_n) \subset S(X \times Y)$ such that $f_n \to f$ $\lambda$-a.e. Let $N$ be a $\lambda$-null set on which $f_n \not \to f$. For $x \in X$, let $N_x := \{y \in Y : (x, y) \in N\}$ be the $x$-section of $N$. Then $N_x \in \mathcal B$ for all $x \in X$. By Tonelli's theorem, $$ 0=\lambda(N_x) = \int_X \nu (N_x) \mathrm d \mu (x). $$
Then for $\mu$-a.e. $x\in X$ we have $\nu(N_x)=0$ and thus $f_n(x, \cdot) \xrightarrow{n \to \infty} f(x, \cdot)$ $\nu$-a.e.
Could you confirm if my above understanding is fine?
