I have been studying Fubini theorem and its proof on "Probability and Stochastics" by Erhan Cinlar. Premise: a measure $\mu$ on a measurable space $\big( E,\mathcal{E} \big)$ is said to be $\Sigma$-finite if there is a countable collection of finite measures $\big( \mu_n \big)_{n=1}^{\infty}$ such that $\mu=\sum_{n=1}^{\infty}\mu_n$. It can be shown that for any $\mathcal{E}$-measurable function $f$, it holds $\mu f = \sum_{n=1}^{\infty} \mu_n f$.
Statement Let $\mu$ and $\nu$ be $\Sigma$-finite measures on $\big( E, \mathcal{E} \big)$ ad $\big( F, \mathcal{F} \big)$, respectively. Then, there exists a unique $\Sigma$-finite measure $\pi$ on $\big( E \times F, \mathcal{E} \otimes \mathcal{F} \big)$ such that, for every positive $f$ in $\mathcal{E} \otimes \mathcal{F}$,
\begin{equation} \pi f= \int_E \mu(dx) \int_F \nu(dy) f(x,y) = \int_F \nu(dy) \int_E \mu(dx) f(x,y) \end{equation}
Issue with proof The proof starts by considering the first double integral, and observes that, under the assumption that both $\mu$ and $\nu$ are $\Sigma$-finite, it can be expressed as
\begin{equation} \pi f= \int_E \mu(dx) \int_F \nu(dy) f(x,y) = \\ = \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\int_E \mu_i(dx) \int_F \nu_j(dy) f(x,y) = \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} (\mu_i \times \nu_j) f \tag{1} \end{equation}
Then, it defines the function
\begin{align*} \hat{f} \colon F \times E & \longrightarrow \mathbb{R_+}\\ (y,x) &\longmapsto f(x,y) \end{align*}
and considers the second iterated integral:
\begin{equation} \hat{\pi} \hat{f}= \int_F \nu(dy) \int_E \mu(dx) \hat{f}(y,x) = \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} \int_F \nu_j(dy) \int_E \mu_i(dx) \hat{f}(y,x) = \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} (\nu_j \times \mu_i) \hat{f} \end{equation}
where $\hat{\pi}(B \times A) = \nu(B) \mu(A)$.
The goal is obviously to show that $ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} (\mu_i \times \nu_j) f = \pi f = \hat{\pi} \hat{f} = \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} (\nu_j \times \mu_i) \hat{f}$
Since all the integrals inside the summations are non-negative numbers, the order of summation can be exchanged. The goal is then to show that the order of integration can be exchanged. The proof now rightly states that it suffices to show that the integrals can be exchanged for some arbitrarily fixed $i$ and $j$, i.e.
\begin{equation} \int_E \mu_i(dx) \int_F \nu_j(dy) f(x,y) = \int_F \nu_j(dy) \int_E \mu_i(dx) f(x,y) \end{equation}
The advantage is that now $\mu_i$ and $\nu_j$ are finite, i.e., the proof for $\Sigma$-finite measures boils down to proving the exchangeability of integrals for finite measures. Hence, in the following we will get rid of the indexes $i$ and $j$ and pretend that we are working under the assumption that $\mu$ and $\nu$ are finite. The textbook now considers a measurable rectangle $A \times B$, and defines the function
\begin{align*} h \colon E \times F & \longrightarrow F \times E\\ (x,y) &\longmapsto (y,x) \end{align*}
Also, it observes that $f = \hat{f} \circ h$.
Here is the core of the proof, that is conducted under the assumption of finiteness of $\mu$ and $\nu$, which is not needed in my view. The book observes that
\begin{equation} \pi \circ h^{-1}(B \times A)=\pi(h^{-1}(B \times A)) = \pi(A \times B) = \\ = \mu(A) \nu(B) = \nu(B) \mu(A) = \hat{\pi} (B \times A) \tag{2} \end{equation}
Hence, upon noticing that
\begin{equation} f(x,y)=(\hat{f} \circ h)(x,y)=\hat{f}(h(x,y))=\hat{f}(y,x)=f(x,y) \end{equation}
it holds
\begin{equation} \hat{\pi} \hat{f} = \pi (\hat{f} \circ h) = \pi f \end{equation}
which proves the statement. My problem is: I believe that equation 2 can be applied also to non-$\Sigma$-finite measures. In this case, I would just do that on the measure $\pi$ in equation 1 and get the exchangeability of the integrals for any product measure, without the need for the $\Sigma$-finiteness assumption.
The way I would proceed I wondered if the actual measure of the rectangle $A \times B$ requires assumption on $\mu$ and $\nu$ in order to allow the exchangeability. In other words, I wondered if the assumption of $\Sigma$-finiteness of both is required for
\begin{equation} \pi(A \times B) = \int_A \mu(dx) \int_B \nu(dy) = \mu(A) \nu(B) = \nu(B) \mu(A) = \int_B \nu(dy) \int_A \mu(dx) = \hat{\pi}(B \times A) \end{equation}
and I would say that $\mu(A) \nu(B) = \nu(B) \mu(A)$ holds irrespective of any $\Sigma$-finiteness assumption.