In generalizing Fubini's theorem to functions on Banach space, I encounter below result that takes me a great deal of time to prove. The proof is delicate for me because we play with $\mathcal L_1$ as well as its quotient space $L_1$. So it's likely that there are subtle mistakes that I'm unable to recognize.
I really hope that somebody helps me check it out.
Can we have a stronger result, i.e., $(f_{n, x})_n$ is a Cauchy sequence that converges to $f_x$ $\nu$-a.e.?
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 $(f_n)$ be a Cauchy sequence in $\mathcal S (X \times Y, \lambda, E)$ that converges $\lambda$-a.e. to $f$. We define $f_x,f_{n,x}: Y \to E$ by $f_x(y) := f (x, y)$ and $f_{n,x} (y) := f_n (x, y)$. Then there is a subsequence $\varphi$ of $(0, 1, 2, \ldots, )$ such that for $\mu$-a.e. $x \in X$, $(f_{\varphi (n), x})_n$ is a Cauchy sequence in $\mathcal S (Y, \nu, E)$ and converges to $f_x$ both in $\mathcal L_1 (Y, \nu, E)$ and $\nu$-a.e.
My attempt: First we need some lemmas.
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. $x \in X$. [A proof can be found here]
Lemma 2: 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 $f_x \in \mathcal L_1 (Y, \nu, E)$. [A proof can be found here]
Lemma 3: Let $(h_n)$ be a sequence in $\mathcal L_1 (Y, \nu, E)$ that converges to $h$ in $\mathcal L_1 (Y, \nu, E)$. Then there exists a subsequence $(h_{\varphi(n)})$ that converges $\nu$-a.e. to $h$. [A proof can be found here]
Let $C \in \mathcal C$ be a set on which $f_n \not \to f$. Let $C_x$ and $C^y$ be the $x$-section and $y$-section of $C$ respectively. By Tonelli's theorem, $$0 = \lambda (C)=\int_X \nu (C_x) \mathrm d\mu(x) = \int_Y \mu (C^y) \mathrm d \nu(y).$$ This means for $\mu$-a.e. $x \in X$, $\nu(C_x) =0$. Then there is a null set $A \in \mathcal A$ such that for each $x \in A^c$, there is a null set $B_x \in \mathcal B$ such that $f_{n,x} (y) \to f_x (y)$ for all $y \in B_x^c$. By Lemma 1, there is a null set $A \in \mathcal A$ such that $f_x,f_{n, x} \in \mathcal S (Y, \nu, E)$ for all $x \in A^c$ and for all $n \in \mathbb N$. It follows that for each $x \in A^c$, $(f_{n,x})_n$ is a sequence in $\mathcal S (Y, \nu, E)$ that converges to $f_x \in \mathcal S (Y, \nu, E)$ on $B_x^c$.
Let $F' := L_1 (Y, \nu, E)$ be the quotient space of $\mathcal L_1 (Y, \nu, E)$. By Lemma 2, $\Phi_n: x \mapsto [f_{n, x}]$ belongs to $\mathcal L_1 (X, \mu, F')$. Here $[f_{n, x}] \in F'$ is the equivalence class containing $f_{n, x} \in \mathcal L_1 (Y, \nu, E)$. We have \begin{align} \| \Phi_n - \Phi_m \|_1 &= \int_X \|\Phi_n (x) - \Phi_m (x) \|_{F'} \mathrm d \mu(x) \\ &= \int_X \int_Y |f_{n, x} (y) - f_{m, x} (y)| \mathrm d \nu (y) \mathrm d \mu (x) \\ &= \int_X \int_Y |f_{n} (x, y) - f_{m} (x, y)| \mathrm d \nu (y) \mathrm d \mu (x) \\ &= \int_{X \times Y} |f_n - f_m| \mathrm d \lambda, \quad \text{ by Tonelli's theorem} \\ &= \|f_n - f_m \|_1. \end{align}
It follows that $(\Phi_n)$ is a Cauchy sequence in $\mathcal L_1 (X, \mu, F')$. Because $\mathcal L_1 (X, \mu, F')$ is complete, there exists $\Phi \in \mathcal L_1 (X, \mu, F')$ such that $\Phi_n \to \Phi$ in $\mathcal L_1 (X, \mu, F')$.
By Lemma 3, there is a subsequence $\varphi$ of $(0, 1, \ldots,)$ such that $\Phi_{\varphi(n)} \to \Phi$ $\mu$-a.e. Let $\Phi: x \mapsto [\hat f_x]$ with $\hat f_x \in \mathcal L_1 (Y, \nu, E)$. Then for $\mu$-a.e. $x \in X$, $[f_{\varphi(n), x}] \xrightarrow{n \to \infty} [\hat f_{x}]$ in $F'$ and thus $f_{\varphi(n), x} \xrightarrow{n \to \infty} \hat f_{x}$ in $\mathcal L_1 (Y, \nu, E)$.
Notice that $\mathcal L_1 (Y, \nu, E)$ is complete (w.r.t. the semi-norm $\| \cdot \|_1$), so $(f_{\varphi(n), x})_n$ is a Cauchy sequence in $\mathcal L_1 (Y, \nu, E)$ for $\mu$-a.e. $x \in X$.
By Lemma 3 again, for $\mu$-a.e. $x \in X$, there is a subsequence $\tau_x$ of $\varphi$ such that $f_{\tau_x(n), x} \xrightarrow{n \to \infty} \hat f_{x}$ $\nu$-a.e. We have already proved that for $\mu$-a.e. $x \in X$, $f_{n,x} \xrightarrow{n \to \infty} f_x$ $\nu$-a.e. Hence $\hat f_x = f_x$ $\nu$-a.e.
As such, for $\mu$-a.e. $x \in X$, $(f_{\varphi (n), x})_n$ is a Cauchy sequence in $\mathcal L_1 (Y, \nu, E)$ that converges to $f_x$ both in $\mathcal L_1 (Y, \nu, E)$ and $\nu$-a.e. This completes the proof.
