A step in the proof of Fubini theorem (Theorem 2.36, Folland) This is a first case of the proof of the Fubini-Tonelli theorem, given in Folland's Real Analysis. I'm confused with the line underlined in blue at the end (namely, 'the preceding argument applies to' part):$\newcommand{\blueunderline}[1]{\color{blue}{\underline{\color{black}{\text{#1}}}}}$

2.36 Theorem. Suppose $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are $\sigma$ -finite measure spaces. If $E \in \mathcal{M} \otimes \mathcal{N},$ then the functions $x \mapsto \nu\left(E_{x}\right)$ and $y \mapsto \mu\left(E^{y}\right)$ are measurable on $X$ and $Y,$ respectively, and
$$
\mu \times \nu(E)=\int \nu\left(E_{x}\right) d \mu(x)=\int \mu\left(E^{y}\right) d \nu(y)
$$
Proof. First suppose that $\mu$ and $\nu$ are finite, and let $\mathcal{C}$ be the set of all $E \in$ $\mathcal{M} \otimes \mathcal{N}$ for which the conclusions of the theorem are true. If $E=A \times B$, then $\nu\left(E_{x}\right)=\chi_{A}(x) \nu(B)$ and $\mu\left(E^{y}\right)=\mu(A) \chi_{B}(y),$ so clearly $E \in \mathcal{C} .$ By additivity
it follows that finite disjoint unions of rectangles are in $\mathcal{C},$ so by Lemma 2.35 it will suffice to show that $\mathcal{C}$ is a monotone class. If $\left\{E_{n}\right\}$ is an increasing sequence in $\mathcal{C}$ and $E=\bigcup_{1}^{\infty} E_{n},$ then the functions $f_{n}(y)=\mu\left(\left(E_{n}\right)^{y}\right)$ are measurable and increase pointwise to $f(y)=\mu\left(E^{y}\right) .$ Hence $f$ is measurable, and by the monotone convergence theorem,
$$
\int \mu\left(E^{y}\right) d \nu(y)=\lim \int \mu\left(\left(E_{n}\right)^{y}\right) d \nu(y)=\lim \mu \times \nu\left(E_{n}\right)=\mu \times \nu(E).
$$
Likewise $\mu \times \nu(E)=\int \nu\left(E_{x}\right) d \mu(x),$ so $E \in \mathcal{C} .$ Similarly, if $\left\{E_{n}\right\}$ is a decreasing sequence in $\mathcal C$ and $\bigcap_{1}^{\infty} E_{n},$ the function $y \mapsto \mu\left(\left(E_{1}\right)^{y}\right)$ is in $L^{1}(\nu)$ because $\mu\left(\left(E_{1}\right)^{y}\right) \leq \mu(X)<\infty$ and $\nu(Y)<\infty,$ so the dominated convergence theorem can be applied to show that $E \in \mathcal{C}$. Thus $\mathcal{C}$ is a monotone class, and the proof is complete for the case of finite measure spaces.
Finally, if $\mu$ and $\nu$ are $\sigma$ -finite, we can write $X \times Y$ as the union of an increasing sequence $\left\{X_{j} \times Y_{j}\right\}$ of rectangles of finite measure. $\blueunderline{If $E \in \mathcal{M} \otimes \mathcal{N},$}$ $\blueunderline{the preceding argument}$ $\blueunderline{applies to}$ $\blueunderline{$E \cap\left(X_{j} \times Y_{j}\right)$}$ for each $j$ to give $$\mu \times \nu\left(E \cap\left(X_{j} \times Y_{j}\right)\right)=\int \chi_{X_{j}}(x) \nu\left(E_{x} \cap Y_{j}\right) d \mu(x)=\int \chi_{Y_{j}}(y) \mu\left(E^{y} \cap X_{j}\right) d \nu(y)$$
and a final application of the monotone convergence theorem then yields the desired result. $\blacksquare$

Transcribed from this screenshot
This was my initial thinking: Assume that $\nu,\mu$ are $\sigma$-finite. As $X,Y$ have exhausting sequences formed by elements of $\mathcal{M}, \mathcal{N}$, we can write $X \times Y$ as the union of an increasing sequence $X_j \times Y_j$ of rectangles of finite measure when measured by $\mu \times \nu$. Let $E \in \mathcal{M} \times \mathcal{N}$. We know that the restriction to $X_j \times Y_j \in \mathcal{M} \times \mathcal{N}$ is still a sigma algebra; in other words, from $(X \times Y, \mathcal{M} \otimes \mathcal{N}, \mu \times \nu)$ is a measure space, we know that the restriction $(X_j \times Y_j, (\mathcal{M} \otimes \mathcal{N}) \cap (X_j \times Y_j), (\mu\times\nu)|_{(X_j \times Y_j)}(\cdot) = \mu \times \nu (\cdot \cap X_j \times Y_j))$ is a measure space. This space is clearly a finite measure space. We apply the preceding result and conclude.
However, I realized there is no reason for an integral in a restricted measure space to equal in integral in a larger measure space. Furthermore, thinking in this way causes some issues when checking for the conditions to use the monotone convergence theorem. Can anyone help clarify how the 'preceding argument applies to $E \cap (X_i \times Y_i)$' to give the result? Thanks.
 A: I believe that it's a non trivial result. Here is an answer to this question:-
Let's start with a lemma which we'll use to prove the result in case $\mu, \nu$ are $\sigma$- finite.  The case when $\mu, \nu$ are finite has already been taken care of.
For each $i\in \mathbb N$,
Define $M_i=\{A\cap X_i: A\in M\}, N_i=\{B\cap Y_i: B\in N\}$ and note that $(X_i, M_i, \mu|_{X_i}) $ and $(Y_i, N_i, \nu|_{Y_i}) $ are measure spaces on $X$, where $\mu|_{X_i}, \nu|_{Y_i}$ denote restriction of the measures to $X_i$ and $Y_i$. Note that $\mu|_{X_i}$ and $ \nu|_{Y_i}$ are finite measures.
Lemma:  For each $i\in \mathbb N$ and for each $E\in M\otimes N,  E_i:= E\cap (X_i\times Y_i)\in M_i\otimes N_i$
Proof: Fix $i$. Let $\scr C$ be the set of all elements $S$ of $M\otimes N$ which satisfy $S\cap (X_i\times Y_i)\in M_i\otimes N_i $. Then it can be shown that $\scr C$ contains all the rectangles (i.e., elements of $M\times N$) and their finite disjoint unions. Note that $\scr C$ is a monotone class so by monotone class theorem, it follows that $\scr C$=$M\otimes N.$ This proves the lemma.
For each $i\in \mathbb N$ and for every $x\in X_i, (E_i)_x\in N_i$ (by the lemma) and $(E_i)_x \uparrow E_x$. By continuity of $\nu$, it follows that $\nu ((E_i)_x)\uparrow \nu(E_x)$ for each $x\in X$. Note that $\nu((E_i)_x)=\nu((E_i)_x)1_{X_i}$ for every $x\in X$.
By using conclusion(s) of finite measures case, we get: $$\int_{X_i}\nu ((E_i)_x)d\mu|_{X_i}(x)=\nu(E_i)\implies \int \nu((E_i)_x) 1_{X_i}(x)d\mu(x)=\nu(E_i)$$
By monotone convergence theorem, $\int \nu(E_x)d\mu(x)=\nu(E)$.
A: The idea not that complicated:  $X$ and $Y$ can be decomposed as the union of pairwise disjoint sets $(A_n:n\in\mathbb{N})$ and $(B_m:n\in\mathbb{N})$ respectively and such that $0<\mu(A_n), \nu(B_m)<\infty$. Then
$$X=\bigcup_{n,m}A_n\times B_m$$
The sets $\{A_n\times B_m: (n,m)\in\mathbb{N}\times\mathbb{N}\}$ are pairwise disjoint.  Now, as $\mathbb{N}\times\mathbb{N}$ is countable, the sets $\{A_n\times B_m: (n,m)\in\mathbb{N}\times\mathbb{N}\}$ can be put into a list $\{A'_k\times B'_k:k\in\mathbb{N}\}$. Then we can define
$$G_n=\bigcup^n_{k=1}(A'_k\times B'_k)=\Big(\bigcup^n_{k=1}A'_k\Big)\times\Big(\bigcup^n_{k=1}B'_k\Big)=: X_n\times Y_n$$
This is an increasing sequence of measurable sets with finite $\nu\times\mu$ measure where you can proceed as the books says (notice that $X_n\nearrow X$ and $Y_n\nearrow Y$). The measures $\mu_n(\cdot)=\mu_n(\cdot\cap X_n)$ and $\nu_n(\cdot)=\nu(\cdot\cap Y_n)$ are finite. Thus, for any measurable set $E\subset X\times Y$,
$$(\mu\times\nu)(E)=\lim_n(\mu\times\nu)\big(E\cap(X_n\times Y_n)\big)=\lim_n(\mu_n\times\nu_n)(E)$$
\begin{align}
(\mu_n\times \nu_n)(E)&=\int_{X\times Y}\mathbb{1}_E\,d\mu_n\times\nu_n=\int_{X}\Big(\int_{Y}\mathbb{1}_{ E}(x,y)\,\nu_n(dy)\Big) \,\mu_n(dx)\\
&=\int_{X}\nu_n(E_x)\,\mu_n(dx)=\int_{X_n}\nu(Y_n\cap E_x)\,\mu(dx)
\end{align}
Similarly
\begin{align}
(\mu_n\times \nu_n)(E)&=\int_{X\times Y}\mathbb{1}_E\,d\mu_n\times\nu_n=\int_{Y}\Big(\int_{X}\mathbb{1}_{ E}(x,y)\,\mu_n(dx)\Big) \,\nu_n(dy)\\
&=\int_{Y}\mu_n(E^y)\,\nu_n(dy)=\int_{Y_n}\nu(X_n\cap E^y)\,\nu(dy)
\end{align}
A: Based on Matematleta's helpful comments, here's what I got: (this is an incomplete proof)
We apply the dominated convergence theorem to obtain, as above, $$\int \mu(E^y) d \nu(y) = \int \lim_{n \to \infty} \mu(E_n^y) d \nu(y) = \lim_{n \to \infty} \int \mu(E_n^y) d \nu(y) = \lim_{n \to \infty} \mu \times \nu (E_n) = \mu \times \nu (E_n). $$
Thus $\mathcal{C}$ is a monotone class, and the proof is complete for the case of finite measure spaces. Now, assume that $\nu,\mu$ are $\sigma$-finite. As $X,Y$ have exhausting sequences formed by elements of $\mathcal{M}, \mathcal{N}$ (denoted $X_j,Y_j$), we can write $X \times Y$ as the union of an increasing sequence $X_j \times Y_j$ of rectangles of finite measure when measured by $\mu \times \nu$. We know that the restriction to $X_j \times Y_j \in \mathcal{M} \times \mathcal{N}$ is still a sigma algebra; in other words, from $(X \times Y, \mathcal{M} \otimes \mathcal{N}, \mu \times \nu)$ is a measure space, we know from results we proved before that the restriction $(X_j \times Y_j, (\mathcal{M} \otimes \mathcal{N}) \cap (X_j \times Y_j), (\mu\times\nu)_{(X_j \times Y_j)}(\cdot) = \mu \times \nu (\cdot \cap X_j \times Y_j))$ is a measure space. This space is clearly a finite measure space. Similarly, we deal with measure spaces $(X_j, \mathcal{M} \cap X_j, \mu_{X_j}(\cdot) = \mu(\cdot \cap X_j))$, $(Y_j, \mathcal{N} \cap Y_j, \nu_{Y_j}(\cdot) = \mu(\cdot \cap Y_j))$ where $\mu_{X_j}$ is the restriction of $\mu$ to $X_j$, and same for $\nu$. I claim that the measure space generated from these two measures are equal to $(X_j \times Y_j, (\mathcal{M} \otimes \mathcal{N}) \cap (X_j \times Y_j), (\mu\times\nu)_{(X_j \times Y_j)}(\cdot))$.
(I am unable to prove this claim. Matematleta, would you know of any way to avoid using this? I'm only doing this because in the previous instance, we proved the theorem for finite measures. Now in this case, $\mu,\nu$ are not finite, even if they are finite measured on $\cdot \cap (X_j\times Y_j)$. Hence the question Product measure space generated by two sigma finite measure spaces)
Now, as $\mu_{X_j}, \nu_{Y_j}$ are finite, we can apply the previous argument in the case of finite measures to obtain, for $E \in \mathcal{M} \otimes \mathcal{N}$: $$(\mu\times\nu)_{X_j \times Y_j}(E \cap (X_j \times Y_j)) = \int \nu((E \cap (X_j \times Y_j))_x) d \mu_{X_j}(x) = \int \mu((E \cap (X_j \times Y_j))^y) d \nu_{Y_j}(y).$$
Note that $\nu((E \cap (X_j \times Y_j))_x) = \chi_{X_j}(x) \nu(E_x \cap Y_j)$, as $\nu((E \cap (X_j \times Y_j))_x) = 0$ if $x \notin X_j$, and if $x \in X_j$, $(E \cap (X_j \times Y_j))_x = \{y \in Y : y \in E_x \text{ and } y \in Y_j\} = E_x \cap Y_j$. Similarly, $\mu((E \cap (X_j \times Y_j))^y) = \chi_{Y_j}(y) \mu(E^y \cap X_j)$. Thus, $$(\mu\times\nu)_{X_j \times Y_j}(E \cap (X_j \times Y_j)) = \int \chi_{X_j}(x) \nu(E_x \cap Y_j) d \mu_{X_j}(x) = \int \chi_{Y_j}(y) \mu(E^y \cap X_j) d \nu_{Y_j}(y).$$
We make a few observations. Firstly, $(\mu\times\nu)_{X_j \times Y_j}(E \cap (X_j \times Y_j)) = (\mu\times\nu)(E \cap (X_j \times Y_j))$, by definition. Secondly, $g_j(x) =  \chi_{X_j}(x) \nu(E_x \cap Y_j) $ is increasing with respect to $j$, as $X_j, Y_j$ are increasing. The same applies to $h_j = \chi_{Y_j}(y) \mu(E^y \cap X_j)$. Thirdly, I claim that $\int \chi_{X_j}(x) \nu(E_x \cap Y_j) d \mu(x)  = \int \chi_{X_j}(x) \nu(E_x \cap Y_j) d \mu_{X_j}(x)$, and $\int \chi_{Y_j}(y) \mu(E^y \cap X_j) d \nu(y) = \int \chi_{Y_j}(y) \mu(E^y \cap X_j) d \nu_{Y_j}(y)$. If this is the case, we conclude that, for each $j$, $$\mu \times \nu\left(E \cap\left(X_{j} \times Y_{j}\right)\right)=\int \chi_{X_{j}}(x) \nu\left(E_{x} \cap Y_{j}\right) d \mu(x)=\int \chi_{Y_{j}}(y) \mu\left(E^{y} \cap X_{j}\right) d \nu(y).$$
As $g_j, h_j$ are increasing functions of $L^+$ (I think it is possible to justify this quickly, using previous theorems in Folland's), we use the monotone convergence theorem and continuity from below of $\mu \times \nu$. $$ \lim_{j \to \infty} \mu \times \nu\left(E \cap\left(X_{j} \times Y_{j}\right)\right) =\mu \times \nu\left(E \right) = \lim_{j \to \infty}\int \chi_{X_{j}}(x) \nu\left(E_{x} \cap Y_{j}\right) d \mu(x) $$ $$=\int \lim_{j \to \infty}\chi_{X_{j}}(x) \nu\left(E_{x} \cap Y_{j}\right) d \mu(x)=\int \chi_{X}(x) \nu\left(E_{x}\right) d \mu(x)$$ $$= \lim_{j \to \infty}\int \chi_{Y_{j}}(y) \mu\left(E^{y} \cap X_{j}\right) d \nu(y) =\int \lim_{j \to \infty}\chi_{Y_{j}}(y) \mu\left(E^{y} \cap X_{j}\right) d \nu(y) = \int \chi_{Y}(y) \mu\left(E^{y}\right) d \nu(y) .$$
In summary, $$\mu \times \nu\left(E \right) =\int  \nu\left(E_{x}\right) d \mu(x) = \int  \mu\left(E^{y}\right) d \nu(y).$$
