Uniqueness of product measure with sigma-finite measure spaces I'm trying to understand the proof for problem 5A.10 in Axler's Measure Theory book. I don't understand the last paragraph of this proof. In my attempted proof, I'm not using $\sigma$-finiteness, so I know I'm missing something.
Problem

Partial Proof
Define
$$\mathcal M := \{ E \in \mathcal S \otimes \mathcal T : \omega(E) = \mu \times \nu(E) \}$$
We claim that $\mathcal M = \mathcal S \otimes \mathcal T$. By construction $\subset$ is obvious, so it remains to show the reverse inclusion. Clearly $\mathcal M$ contains the algebra of finite unions of measurable rectangles in $\mathcal S \otimes \mathcal T$ (since each of those unions can be written as a finite disjoint union of measurable rectangles). Now it suffices to show that $\mathcal M$ is a monotone class, and then apply the Monotone Class Theorem, which would imply that $\mathcal S \otimes \mathcal T \subset M$.
Suppose $E_1 \subset E_2 \subset \ldots \in \mathcal M$. Then $\omega(E_k) = \mu \times \nu(E_k)$ for each $k \in \mathbb N$. Now reformulate the sets to be disjoint:
$$
\tilde E_k = E_k - \bigcup_{j=1}^{k-1} E_j
$$
Then
\begin{align*}
\omega(\cup E_k) &= \omega (\cup \tilde E_k) \\
& = \sum \omega(\tilde E_k) \\
&= \sum \mu \times \nu(\tilde E_k) \\
&= \mu \times \nu(\cup \tilde E_k) \\
&= \mu \times \nu(\cup E_k)
\end{align*}
Hence, $\cup E_k \in \mathcal M$. Similarly, let $E_1 \supset E_2 \supset \ldots \in \mathcal M$. Because $X \times Y$ is a $\sigma$-finite space, let $\{F_k\}_{k \in \mathbb N} \subset \mathcal S \times \mathcal T$ be such that $X \times Y = \cup F_k$ and $\omega(F_k) < \infty$.  Define
$$
\hat E_{j, k} = E_j \cap F_k
$$
Then
\begin{align*}
\omega(\cap E_k) &= \omega (\bigcap_j \bigcup_k \hat E_{j, k}) \\
&= \text{???}
\end{align*}
(not proven: So $\cap E_k \in \mathcal M$, and $\mathcal M$ is a monotone class. Therefore $\mathcal M = \mathcal S \otimes \mathcal T$. $\square$)
Questions

*

*Is my partial proof correct -- have a shown that $\mathcal M = \mathcal S \otimes \mathcal T$?

*Why is this not sufficient to show that the product measure is unique? What am I missing? Again, the explanation here (final paragraph) confused me, so I need a bit more explanation.

Update 1

*

*I believe I fixed the first part of my proof that $\mathcal M$ is a monotone class.


*To show that $\omega (\cap E_k) = \lim_{k \to \infty} E_k$, I can't quite figure out how to use $\sigma$-finiteness here. Help on this last part would be much appreciated!
Update 2
Submitted a possible answer.
Update 3
Axler's Proof of Tonelli's Theorem (5.28, p. 129) provides a shorter, more elegant way of showing that $\mathcal M$ is a monotone class, using the monotone convergence theorem.
 A: Define
$$\mathcal M := \{ E \in \mathcal S \otimes \mathcal T : \omega(E) = \mu \times \nu(E) \}$$
We claim that $\mathcal M = \mathcal S \otimes \mathcal T$. By construction $\subset$ is obvious, so it remains to show the reverse inclusion. Clearly $\mathcal M$ contains the algebra of finite unions of measurable rectangles in $\mathcal S \otimes \mathcal T$ (since each of those unions can be written as a finite disjoint union of measurable rectangles). Now it suffices to show that $\mathcal M$ is a monotone class, and then apply the Monotone Class Theorem, which would imply that $\mathcal S \otimes \mathcal T \subset M$.
Suppose $E_1 \subset E_2 \subset \ldots \in \mathcal M$. Then $\omega(E_k) = \mu \times \nu(E_k)$ for each $k \in \mathbb N$. Now reformulate the sets to be disjoint (let $E_o = \emptyset)$:
$$
\tilde E_k = E_k - \bigcup_{j=1}^{k-1} E_j
$$
Then
\begin{align*}
\omega(\cup E_k) &= \omega (\sqcup \tilde E_k) \\
& = \sum \omega(\tilde E_k) \\
&= \sum \mu \times \nu(\tilde E_k) \\
&= \mu \times \nu(\sqcup \tilde E_k) \\
&= \mu \times \nu(\cup E_k)
\end{align*}
Hence, $\cup E_k \in \mathcal M$. Similarly, let $E_1 \supset E_2 \supset \ldots \in \mathcal M$. Because $X \times Y$ is a $\sigma$-finite space, let $\{F_k\}_{k \in \mathbb N} \subset \mathcal S \times \mathcal T$ be such that $X \times Y = \cup F_k$ and $\omega(F_k) < \infty$.  First note that
$$
\bigcap_k E_k = \bigcup_j \left( \bigcap_k E_k \right) \cap F_k = \bigcup_j \bigcap_k (E_k \cap F_j)
$$
So define
\begin{align*}
A_{j,k} &= E_k \cap F_j \\
\hat A_i &= \bigcup_{j \in \mathbb N - \{i\} } \bigcap_k A_{j,k}
\end{align*}
Note that
\begin{align*}
\bigcap_k E_k &= \bigcup_j \bigcap_k A_{j,k} \\
&= \bigcup_j \left( \bigcap_k A_{j,k} \right) - \hat A_j \\
&= \bigsqcup_j \bigcap_k (A_{j,k} - \hat A_j)
\end{align*}
Now observe that for each $j \in \mathbb N$, $\{ A_{j,k} - \hat A_j \}_{k \in \mathbb N}$ is a decreasing sequence that is finite at $k = 1$.
Then
\begin{align*}
\omega(\cap E_k) &= \omega (\bigsqcup_j \bigcap_k (A_{j,k} - \hat A_j)) \\
&= \sum_j \omega (\bigcap_k (A_{j,k} - \hat A_j)) \\
&= \sum_j \mu \times \nu(\bigcap_k (A_{j,k} - \hat A_j)) \\
&= \mu \times \nu(\bigsqcup_j \bigcap_k (A_{j,k} - \hat A_j)) \\
&= \mu \times \nu(\cap E_k)
\end{align*}
So $\cap E_k \in \mathcal M$, and $\mathcal M$ is a monotone class. Therefore $\mathcal M = \mathcal S \otimes \mathcal T$. $\square$
