In generalizing Fubini's theorem to functions on Banach space, I encounter a result that I'm unable to prove. Could you shed some light on this issue?
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 $f \in \mathcal S (X \times Y, \lambda, E)$ and $F' := L_1 (Y, \nu, E)$ be the quotient space of $\mathcal L_1 (Y, \nu, E)$. Then $\Phi : x \mapsto [f_x]$ is defined $\mu$-a.e. and belongs to $\mathcal L_1 (X, \mu, F)$, i.e., $\Phi$ is integrable. Here $[f_x] \in F'$ is the equivalence class containing $f_x \in \mathcal L_1 (Y, \nu, E)$.
My attempt: I can only prove a very slight result.
Lemma 1: Let $f \in \mathcal S (X \times Y, \lambda, E)$ and $f_x : Y \to E, y \mapsto f(x, y)$. Then $f_x \in\mathcal S (Y, \nu, E)$ for $\mu$-a.e. $x \in X$.
Proof of Lemma 1: Assume $f = \sum_{k=1}^n e_k 1_{G_k}$ where $0 \neq e_k \in E$ and $(G_k)$ is a finite sequence of pairwise disjoint (p.w.d.) sets with finite measure in $\mathcal C$. Then $f_x =\sum_{k=1}^n e_k 1_{G_{k,x}}$ where $G_{k, x} \subseteq Y$ is the $x$-section of $G_k$. It is proved here that $G_{k, x} \in \mathcal B$ for all $x \in X$. By Tonelli's theorem, $$\int_X \nu( 1_{G_{k, x}}) \mathrm d \mu (x) = \lambda(G_{k}) < \infty.$$
It follows that $\nu( 1_{G_{k, x}}) < \infty$ for $\mu$-a.e. $x \in X$. This means there exist null sets $A_k \in \mathcal A$ such that $\nu( 1_{G_{k, x}}) < \infty$ for all $x \in A_k^c$. Let $A := \cup_{k=1}^n A_k$. Then $A \in \mathcal A$ is a null set such that $\nu( 1_{G_{k, x}}) < \infty$ for all $k=1, \ldots, n$ and for all $x \in A^c$. It follows that $f_x \in \mathcal S (Y, \nu, E)$ for all $x \in A^c$.
Update: After many trials and errors, It seems I found the proof. First, we need below cornerstone lemma whose proof is given here.
Lemma 2: 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 $\Phi: x \mapsto [1_{G_x}]$ 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 $1_{G_x} \in \mathcal L_1 (Y, \nu, \mathbb R)$.
By the Lemma 1, $f_x \in\mathcal S (Y, \nu, E) \subseteq \mathcal L_1 (Y, \nu, E)$ and thus $[f_x] \in F$ for $\mu$-a.e. $x \in X$. Assume $f = \sum_{k=1}^n e_k 1_{G_k}$ where $0 \neq e_k \in E$ and $(G_k)$ is a finite sequence of p.w.d. sets with finite measure in $\mathcal C$. Then $f_x =\sum_{k=1}^n e_k 1_{G_{k,x}}$. There exists a null set $A \in \mathcal A$ such that $\nu(G_{k, x}) < \infty$ for all $k=1, \ldots,n$ and $x \in A$. Let $\Phi_k' : x \mapsto [1_{G_{k, x}}]$ and $\Phi_k: x \mapsto [e_k1_{G_{k, x}}]$. By Lemma 2, $\Phi_k' \in \mathcal L_0 ( X, \mu, F )$, so $\Phi_k \in \mathcal L_0 ( X, \mu, F' )$, i.e., $\Phi_k$ is measurable. We have \begin{align} \|e_k1_{G_{k, x}}\|_1 &= \int_Y |e_k1_{G_{k, x}}| (y)\mathrm d\nu(y) &&= \int_Y |e_k| 1_{G_{k, x}} (y)\mathrm d\nu(y)\\ &= |e_k| \int_Y 1_{G_{k, x}} (y)\mathrm d\nu(y) &&= |e_k| \nu(G_{k, x}). \end{align}
Hence $$\int_X \|\Phi_k(x)\|_1 \mathrm d \mu (x) = \int_X \|e_k1_{G_{k, x}}\|_1 \mathrm d \mu (x) = |e_k| \int_X \nu(G_{k, x}) \mathrm d \mu (x) =|e_k| \lambda(G_k) < \infty.$$ So $\Phi_k$ and thus $\Phi = \sum_{k=1}^n \Phi_k$ are integrable. This completes the proof.
