Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete 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 $\nu$.
I would like to verify that
Theorem Let $f \in L^0(X \times Y)$. Then for $\mu$-a.e. $x \in X$ we have $f(x, \cdot) \in L^0(Y)$.
Could you have a check on my below attempt? Thank you so much for your help!
There is a sequence $(f_n) \subset S(X \times Y)$ such that $f_n \to f$ $\lambda$-a.e. Lemma 1 in this shows that for a fixed $n$: for $\mu$-a.e. $x \in X$ we have $f_n (x, \cdot) \in S (Y)$. Countable union of null sets is a null set, so there is a $\mu$-null set $N_1$ such that $$ f_n(x, \cdot) \in S(Y) \quad \forall n \in \mathbb N, x\in N_1^c. $$
I already proved here there is a $\mu$-null set $N_2$ such that for every $x \in N_2^c$ we have $f_n(x, \cdot) \xrightarrow{n \to \infty} f(x, \cdot)$ $\nu$-a.e.
Let $N := N_1 \cup N_2$. Then $N$ is a $\mu$-null set such that for every $x\in N^c$: $$ f_n(x, \cdot) \in S(Y) \quad \forall n \in \mathbb N, \\ f_n(x, \cdot) \xrightarrow{n \to \infty} f(x, \cdot) \quad \nu \text{-a.e.} $$
By definition of Bochner measurability, $f(x, \cdot) \in L^0(Y)$ for all $x \in N^c$. The claim then follows.
