In proving this result, I need to prove this result. I'm not very sure whether part (iv) (which is important and combines results I proved above) of my proof contains subtle mistakes. Could you please have a check on it?
Related definitions of Bochner integrals can be found here. Let
$(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces, and $(E, | \cdot |)$ a Banach space.
$\Sigma := \mathcal A \times \mathcal B$.
$\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$.
$\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.
$\mathcal S (X \times Y, \lambda, E)$ the space of $\lambda$-simple functions from $X \times Y$ to $E$.
$\mathcal L_0 (Y, \nu, E)$ the space of $\nu$-measurable functions from $Y$ to $E$.
$\mathcal L_1 (Y, \nu, E)$ the space of $\nu$-integrable functions from $Y$ to $E$.
Theorem: Let $G \in \mathcal C$ have finite measure and $F := L_1 (Y, \nu, \mathbb R)$ be the quotient space of $\mathcal L_1 (Y, \nu, \mathbb R)$. Then $\Phi: x \mapsto [1_{G_x}]$ is defined $\mu$-a.e. and belongs to $\mathcal L_0 (X, \mu, F)$. Here $G_x$ is the $x$-section of $G$ and $[1_{G_x}] \in F$ is the equivalence class containing $1_{G_x} \in \mathcal L_1 (Y, \nu, \mathbb R)$.
My attempt: We proceed by approximation. But we first need a lemma.
Lemma: Let $(f_n)$ be a sequence in $\mathcal L_0(X, \mu, E)$ that converges $\mu$-a.e. to $f \in E^X$. Then $f \in \mathcal L_0(X, \mu, E)$. [A proof can be found here]
(i) Let $G = A \times B \in \Sigma$ such that $\mu(A), \nu(B) < \infty$. Then $\Phi(x) = 1_B$ if $x \in A$ and $1_\emptyset$ otherwise, i.e., $\Phi (x) = 1_B 1_A (x)$. It follows from $\nu(B) < \infty$ that $1_B \in F$. This combining with $\mu(A) < \infty$ implies $\Phi \in \mathcal S (X, \mu, F) \subseteq \mathcal L_0 (X, \mu, F)$.
(ii) Let $G = A \times B \in \Sigma$ such that $\lambda(G) < \infty$. Because $\mu$ and $\nu$ are $\sigma$-finite, $A$ and $B$ can be partitioned into collections $(A_k)$ and $(B_h)$ of pairwise disjoint (p.w.d.) sets with finite measure in $\mathcal A$ and $\mathcal B$ respectively. We relabel $(A_k \times B_h)$ to obtain $(A_k \times B_k)$. Let $$G_k: = A_k \times B_k \qquad G_{k, x} := (G_k)_x \qquad \phi_k := x \mapsto 1_{G_{k,x}} \qquad \Phi_n := \sum_{k = 0}^n \phi_{k}.$$
Then $1_G = \sum_k 1_{G_k}$ and $1_{G_x} = \sum_k 1_{G_{k, x}}$. Also, $\phi_k$ and thus $\Phi_n$ belong to $\mathcal S(X, \mu, F)$ by (i). We have \begin{align} \|\Phi_n (x) - \Phi (x)\|_F &= \int_Y | \Phi_n (x) (y) - \Phi (x) (y)| \mathrm d \nu (y) &&= \int_Y \left | \sum_{k = 0}^n \phi_{k} (x) (y) - \Phi (x) (y) \right | \mathrm d \nu (y) \\ &= \int_Y \left | \sum_{k = 0}^n 1_{G_{k,x}} (y) - 1_{G_{x}} (y) \right | \mathrm d \nu (y) &&= \int_Y \sum_{k = n+1}^\infty 1_{G_{k,x}} (y) \mathrm d \nu (y) \\ &= \sum_{k = n+1}^\infty \int_Y 1_{G_{k,x}} (y) \mathrm d \nu (y) &&= \sum_{k = n+1}^\infty \nu (G_{k,x}). \end{align}
From $\lambda(G) < \infty$ and Tonelli's theorem, $\nu (G_x) = \sum_k \nu (G_{k,x}) < \infty$ for $\mu$-a.e. $x \in X$. Hence $\|\Phi_n (x) - \Phi (x)\|_F \to 0$ for $\mu$-a.e. $x \in X$. Hence $\Phi_n \to \Phi$ for $\mu$-a.e. and thus $\Phi \in \mathcal L_0 (X, \mu, F)$ by our Lemma.
(iii) Let $G = \bigcup_k G_k$ such that $\lambda(G) < \infty$ and that $(G_k)$ is a sequence of p.w.d. sets in $\Sigma$. Clearly, $\lambda(G_k) < \infty$ for all $k$. Let $$G_{k, x} := (G_k)_x \qquad \phi_k := x \mapsto 1_{G_{k,x}} \qquad \Phi_n := \sum_{k = 0}^n \phi_{k}.$$
Then $1_G = \sum_k 1_{G_k}$ and $1_{G_x} = \sum_k 1_{G_{k, x}}$. Also, $\phi_k$ and thus $\Phi_n$ belong to $\mathcal L_0 (X, \mu, F)$ by (ii). With a similar reasoning as above, $\Phi_n \to \Phi$ for $\mu$-a.e. and thus $\Phi \in \mathcal L_0 (X, \mu, F)$ by our Lemma.
(iv) Let $G \in \mathcal C$ with $\lambda(G) < \infty$. Then there exists a sequence $(G_m)$ such that $G_m = \bigcup_k A^{(m)}_k \times B^{(m)}_k \supseteq G$, $A^{(m)}_k \times B^{(m)}_k \in \Sigma$, and $\lambda (G_m) \searrow \lambda(G)$. Without loss of generality, we assume $\lambda(G_0) < \infty$. Let $E_n := \bigcap_{m=0}^n G_m$ and $E := \bigcap_n E_n$. Then $G \subseteq E_n$, $\lambda(E_n) \le \lambda(E_0) = \lambda(G_0) <\infty$, $\lambda(G) = \lambda(E)$, and $E_n \searrow E$. By results (a) and (b), $E_n$ can be written as a countable union of p.w.d. sets in $\Sigma$.
Let $E_{n, x} := (E_n)_x$. Then $\Phi_n: x \mapsto 1_{E_{n,x}}$ belongs to $\mathcal L_0 (X, \mu, F)$ by (iii). We have $E_x = (\bigcap_n E_n)_x =\bigcap_n E_{n, x}$, so $E_{n, x} \searrow E_x$ and thus $1_{E_{n,x}} \searrow 1_{E_x}$. This means $\Phi_n \to \Psi$ where $\Psi: x \mapsto 1_{E_x}$. Hence $\Psi \in \mathcal L_0 (X, \mu, F)$ by our Lemma.
Let $F := E \setminus G$. Then $E = G \sqcup F$ and $\lambda(F) = 0$. Let $\Lambda: x \mapsto 1_{F_x}$. It follows from $\lambda(F) = 0$ and Tonelli's theorem that $\nu(F_x) = 0$ for $\mu$-a.e. $x \in X$. This means for $\mu$-a.e. $x \in X$, $1_{F_x} (y)=0$ for $\nu$-a.e. $y \in Y$. This in turn implies $\Lambda (x) = [y \mapsto 0]$ for $\mu$-a.e. $x \in X$. Clearly, the equivalence class $[y \mapsto 0]$ belongs to $F$. Hence $\Lambda \in \mathcal L_0 (X, \mu, F)$ Then $\Phi = \Psi - \Lambda \in \mathcal L_0 (X, \mu, F)$. This completes the proof.
