After so much preparation (in proving auxiliary lemmas), I finally complete the proof of Fubini's theorem for Banach spaces. This is what I have desired after proving Tonelli's theorem :) The journey to the proof is very enriching. It solidifies my understanding in some aspects.
Why are the $\sigma$-finiteness and completeness of measure important and where do we use them?
Why does $E$ need to be complete, i.e. $E$ is a Banach space?
Where is Tonelli's theorem used?
The relation between $\mathcal L_1$ convergence and a.e. convergence.
Why is the quotient space $L_1$ of $\mathcal L_1$ useful?
Could you please have a check on my proof? It is detailed and thus easy to read. I would be grateful if any mistake is spotted.
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.
$\mathcal C :=\mathcal A \otimes \mathcal B$ 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 $E$.
Fubini's theorem: Let
$f: X \times Y \to E$ $\lambda$-integrable.
$f_x: Y \to E, \, y \mapsto f(x, y)$ for all $x \in X$.
$f_y: X \to E, \, x \mapsto f(x, y)$ for all $y \in Y$.
Then
The maps $$\Phi: X \ni x \mapsto \int_Y f_x \mathrm d \nu \quad \text{and} \quad \Psi: Y \ni y \mapsto \int_X f_y \mathrm d \mu $$ are integrable.
$$\int_X \Phi \mathrm d \mu = \int_{X \times Y} f \mathrm d \lambda = \int_Y \Psi \mathrm d \nu.$$
My proof: By symmetry, it's enough to prove for the case of $\Phi$.
Lemma 1: 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. [A proof can be found here]
Lemma 2: 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 $\varphi$ of $(0, 1, 2, \ldots, )$ such that $(h_{\varphi(n)})$ converges $\nu$-a.e. to $h$. [A proof can be found here]
(i) Let $f= 1_G$ where $G \in \mathcal C$ and $\lambda(G) < \infty$. Notice that $f$ is a characteristic function in $\mathcal S(X \times Y, \lambda, \mathbb R)$. By Tonelli's theorem, $\Phi$ is measurable and $\int_{X \times Y} f \mathrm d \lambda = \int_X \Phi \mathrm d \mu$. On the other hand, $\int_{X \times Y} f \mathrm d \lambda = \lambda(G) < \infty$. Hence $\Phi$ is also integrable and thus Fubini's theorem holds in this case.
(ii) Let $f= e1_G$ where $0 \neq e \in E$, $G \in \mathcal C$, and $\lambda(G) < \infty$. Notice that $f$ is the atomic function in $\mathcal S(X \times Y, \lambda, E)$. Let $1_{G, x} := (1_G)_x$. Then $\Phi (x) = \int_Y e 1_{G, x} \mathrm d \nu = e \int_Y 1_{G, x} \mathrm d \nu$. By (i), $x \to \int_Y 1_{G, x} \mathrm d \nu$ is integrable, so is $\Phi:x \to e \int_Y 1_{G, x} \mathrm d \nu$. Also, \begin{align} \int_X \Phi \mathrm d \mu &= \int_X \left [ e \int_Y 1_{G, x} \mathrm d \nu \right ] \mathrm d \mu (x) &&= e \int_X \int_Y 1_{G, x} \mathrm d \nu \mathrm d \mu (x) \\ &\overset{(\star)}{=} e \int_{X \times Y} 1_G \mathrm d \lambda \quad \text{by (i)} &&= \int_{X \times Y} e1_G \mathrm d \lambda \\ &= \int_{X \times Y} f \mathrm d \lambda. \end{align}
Here $(\star)$ is due to Tonelli's theorem. Hence Fubini's theorem also holds in this case. By linearity of integrals, Fubini's theorem also holds for all simple functions $f \in \mathcal S(X \times Y, \lambda, E)$.
(iii) Let $f \in \mathcal L_1 (X \times Y, \lambda, E)$. This means $f$ is $\lambda$-a.e. limit of a Cauchy sequence $(f_n)$ in $\mathcal S (X \times Y, \lambda, E)$. Let $$f_{n,x} (y) := f_n (x, y) \quad \text{and} \quad\Phi_n (x) := \int_Y f_{n, x} \mathrm d \nu.$$ By (ii), $$\Phi_n \in \mathcal L_1(X, \mu, E) \quad \text{and} \quad \int_X \Phi_n \mathrm d \mu = \int_{X \times Y} f_n \mathrm d \lambda, \quad n \in \mathbb N.$$
By Lemma 1, 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. It follows that for $\mu$-a.e. $x \in X$, $$\int_Y f_{\varphi (n), x} \mathrm d \nu \xrightarrow{n \to \infty} \int_Y f_x\mathrm d \nu.$$
This means $\Phi_{\varphi (n)} \to \Phi$ $\mu$-a.e. Let's prove that $(\Phi_{\varphi (n)})$ is indeed a Cauchy sequence in $\mathcal L_1(X, \mu, E)$. In fact, \begin{align} \| \Phi_{\varphi (n)} - \Phi_{\varphi (m)}\|_1 &= \int_X \left |\int_Y \big ( f_{\varphi (n), x}- f_{\varphi (m), x} \big ) \mathrm d \nu \right | \mathrm d \mu(x) \\ &\le \int_X \int_Y \left | f_{\varphi (n), x} - f_{\varphi (m), x} \right | \mathrm d \nu \mathrm d \mu(x) \\ &= \int_{X \times Y} | f_{\varphi (n)} - f_{\varphi (m)} | \mathrm d \lambda \quad \text{by Tonelli's theorem} \\ &= \| f_{\varphi (n)} - f_{\varphi (m)} \|_1. \end{align}
It follows that there is $\Phi' \in \mathcal L_1(X, \mu, E)$ such that $\Phi_{\varphi (n)} \to \Phi'$ in $\mathcal L_1(X, \mu, E)$. By Lemma 2, there is a subsequence $\tau$ of $\varphi$ such that $\Phi_{\tau (n)} \to \Phi'$ $\mu$-a.e. We have already proved that $\Phi_{\varphi (n)} \to \Phi$ $\mu$-a.e., so $\Phi = \Phi' \in \mathcal L_1(X, \mu, E)$ $\mu$-a.e. As such, $$\int_X \Phi_{\varphi (n)} \mathrm d \mu \xrightarrow{n \to \infty} \int_X \Phi \mathrm d \mu.$$
By definition of integrals, $$\int_{X \times Y} f_n \mathrm d \lambda \xrightarrow{n \to \infty} \int_{X \times Y} f \mathrm d \lambda.$$
Recall that $$\int_X \Phi_n \mathrm d \mu = \int_{X \times Y} f_n \mathrm d \lambda, \quad n \in \mathbb N.$$
Hence $$\int_X \Phi \mathrm d \mu = \int_{X \times Y} f \mathrm d \lambda.$$
This completes the proof.
