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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

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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

  1. Is my partial proof correct -- have a shown that $\mathcal M = \mathcal S \otimes \mathcal T$?
  2. 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

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

  2. 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.

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    $\begingroup$ Your proof is not correct. $\omega(\bigcup_{k} E_k) = \sum_{k} \omega(E_k)$ doesn't hold (in general) for an increasing union of sets. Furthermore, the downward convergence $\omega(\bigcap{k} E_k) = \lim_{k \to \infty} \omega(E_k)$ doesn't hold in infinite measure spaces, hence the need for sigma-finiteness. $\endgroup$
    – unwissen
    Aug 27, 2022 at 22:11
  • $\begingroup$ I see that in each case I need to rewrite the $E_k$'s as a subset of some finite union of the finite sets that cover the space. I'll think on this, then update the proof. Thanks! $\endgroup$
    – IsaacR24
    Aug 27, 2022 at 22:32

1 Answer 1

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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 $\mathcal M = \mathcal S \otimes \mathcal T$. $\square$

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