Below we use [Bochner measurability](https://www.wikiwand.com/en/Bochner_measurable_function) and [Bochner integral](https://www.wikiwand.com/en/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 $E$,
- $Z := X \times Y$,
- $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
- $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$,
- $(Z, \overline{\mathcal C}, \overline{\nu})$ the completion of $(Z, \mathcal C, \nu)$.

We are going to prove
>**Theorem** Let $p \in [1, \infty)$ and $f \in L^0 (X, L^p(Y))$. Then $f \in L^0 (Z)$.

---

1. First, we consider the case $\mu (X) + \nu(Y) < \infty$. There is a sequence $(f_n) \subset S(X, L^p(Y))$ such that $f_n \to f$ $\mu$-a.e. in $L^p(Y)$. Let
$$
f_n (x) = \sum_{k=1}^{\varphi_n} 1_{A_{n, k}} (x) h_{n, k},
$$
where $h_{n, k} \in L^p (Y)$ and $A_{n, k} \in \cal A$ with $\mu(A_{n, k}) < \infty$. Let
$$
F_n: X \times Y \to E, (x, y) \mapsto f_n(x)(y).
$$

Clearly, $F_n \in L^0 (X \times Y)$. For $\mu$-a.e. $x \in X$,
$$
\|F_n(x, \cdot) - f(x) \|_{L^p(Y)} \xrightarrow{n \to \infty} 0,
$$
which implies $(F_n(x, \cdot))_n$ is a Cauchy sequence in $L^p(Y)$.

>Let $\rho_Z$ be a [pseudometric](https://www.wikiwand.com/en/Pseudometric_space#Definition)  on $L^0(Z)$ defined by
$$
\rho_Z (g_1, g_2) := \int_Z \min\{|g_1 - g_2|, 1\} \, \mathrm d \lambda
\quad \forall g_1, g_2 \in L^0 (Z).
$$
>If $\mu (X) + \nu(Y) < \infty$, then
>- [Lemma 1](https://math.stackexchange.com/questions/4754912/if-for-a-e-x-in-x-the-sequence-f-nx-cdot-n-is-cauchy-in-l1-y-t) Let $f_n \in L^0(Z)$ for all $n \in \mathbb N$. Assume that for $\mu$-a.e. $x \in X$, the sequence $(f_n(x, \cdot))_n$ is a Cauchy sequence in $(L^1 (Y), \| \cdot\|_{L^1 (Y)})$. Then $(f_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$.
>- [Lemma 2](https://math.stackexchange.com/questions/4753822/the-metric-space-l0-x-rho-of-mu-measurable-functions-is-complete) $\rho_Z$ is a complete metric on $L^0 (Z)$.
>- [Lemma 3](https://math.stackexchange.com/questions/4753838/a-characterization-of-convergence-in-the-metric-space-l0-x-rho-of-mu) $(f_n)$ converges to $f$ in $(L^0 (Z), \rho_Z)$ IFF every subsequence of $(f_n)$ has in turn a further subsequence that converges to $f$ $\lambda$-a.e.

By **Lemma 1**, $(F_n)$ is a Cauchy sequence in $(L^0 (Z), \rho_Z)$. By **Lemma 2**, there is $F\in L^0(Z)$ such that $F_n \to F$ in $\rho_Z$. By **Lemma 3**, there is subsequence (also denoted by $(F_n)$ for simplicity) such that $F_{n} \xrightarrow{n \to \infty} F$ $\lambda$-a.e. I showed that for $\mu$-a.e $x \in X$ we have [(S1)](https://math.stackexchange.com/questions/4753844/let-f-in-l0x-times-y-then-for-mu-a-e-x-in-x-we-have-fx-cdot?noredirect=1&lq=1) $F_n(x, \cdot) \in L^0 (Y)$ and [(S2)](https://math.stackexchange.com/questions/4753759/almost-everywhere-convergence-in-product-measure-and-that-in-coordinate-ones) $F_n(x, \cdot) \to F(x, \cdot)$ $\nu$-a.e.

We have $f_n \to f$ $\mu$-a.e. This means for $\mu$-a.e. $x \in X$ we have $\|f_n (x) - f(x)\|_{L^p (Y)} \xrightarrow{k \to \infty} 0$. Convergence in $L^p$ [implies](https://math.stackexchange.com/questions/4301365/convergence-in-mathcal-l-1-implies-mu-a-e-convergence-for-a-subsequence) a.e. convergence of a subsequence. For  $\mu$-a.e. $x \in X$ there is a subsequence $\varphi_x$ of $\mathbb N$ such that $f_{\varphi_x (n)} (x) \xrightarrow{n \to \infty} f(x)$ $\nu$-a.e. and thus $F_{\varphi_x (n)} (x, \cdot) \xrightarrow{n \to \infty}f(x)$ $\nu$-a.e.

It follows that for $\mu$-a.e. $x \in X$ we have $F(x, \cdot) = f(x)$ $\nu$-a.e. and thus $\|F(x, \cdot) - f(x)\|_{L^p (Y)}=0$.

2. We consider the case $\mu (X) < \infty$. There is a countable measurable partition $(Y_n)$ of $Y$ such that $\sup_n \nu(Y_n) < \infty$. We define
$$
f_n:X \to L^p (Y_n), x \mapsto f(x) 1_{Y_n}.
$$

Then $f_n \in L^0(X, L^p(Y_n))$. We apply part (1) for $f_n$ and get $f_n \in L^0 (X \times Y_n)$. We have $L^p (Y_n)$ can be considered a closed subspace of $L^p (Y)$. So $L^0(X, L^p(Y_n))$ can be considered a closed subspace of $L^0(X, L^p(Y))$ w.r.t. $\rho_X$. Similarly, $L^0 (X \times Y_n)$ can be considered a closed subspace of $L^0(X \times Y)$ w.r.t $\rho_{X \times Y}$. Then $f = \lim_n \sum_{k=1}^n f_k \in L^0 (X \times Y)$.

3. We also drop the assumption $\mu (X) < \infty$. There is a countable measurable partition $(X_n)$ of $X$ such that $\sup_n \mu (X_n) < \infty$. We define
$$
f_n:X_n \to L^p (Y), x \mapsto f(x).
$$

Then $f_n \in L^0(X_n, L^p(Y))$. We apply part (2) for $f_n$ and get $f_n \in L^0 (X_n \times Y)$. We have $L^0(X_n, L^p(Y))$ can be considered a closed subspace of $L^0 (X, L^p (Y))$ w.r.t. $\rho_X$.  Similarly, $L^0 (X \times Y_n)$ can be considered a closed subspace of $L^0(X \times Y)$ w.r.t. $\rho_{X \times Y}$. Then $f = \lim_n \sum_{k=1}^n f_k \in L^0 (X \times Y)$.