# 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.

# 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 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.

• 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. Aug 27, 2022 at 22:11
• 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! Aug 27, 2022 at 22:32

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