# Definition of the product $\sigma$-algebra

The following is the definition of the product $\sigma$-algebra given in Gerald Folland's Real Analysis: Modern Techniques and Their Applications (pg. 22) (note that $\mathcal{M}(X)$ denotes the smallest $\sigma$-algebra generated by the set $X$):

"Let $\{X_{\alpha}\}_{\alpha \in A}$ be an indexed collection of nonempty sets, $X=\prod_{\alpha \in A} X_{\alpha}$, and $\pi_{\alpha}: X \to X_{\alpha}$ the coordinate maps. If $\mathcal{M}_{\alpha}$ is a $\sigma$-algebra on $X_{\alpha}$ for each $\alpha$, the product $\sigma$-algebra on $X$ is the $\sigma$-algebra generated by \begin{equation} \{\pi_{\alpha}^{-1}(E_{\alpha}): E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A \} \end{equation} We denote this $\sigma$-algebra by $\otimes_{\alpha \in A} \mathcal{M}_{\alpha}$."

With $A=\{1,2\}$, the wikipedia definition says that the product $\sigma$-algebra is given by: \begin{equation} \mathcal{M}_{1} \times \mathcal{M}_{2} = \mathcal{M}\left(\{E_{1} \times E_{2} : E_{1} \in \mathcal{M}_{1}, E_{2} \in \mathcal{M}_{2} \} \right) \end{equation}

I understand the Wikipedia definition. However, I am new to measure theory and am having trouble reconciling Folland's definition with the Wikipedia definition.

Assuming $A=\{1,2\}$ for simplicity, can someone show me why the two definitions of the product $\sigma$-algebra are the same?

• If you continue reading, in page 23, proposition 1.3, you can see that the countable case is this one mentioned in wikipedia. – Integral Jul 22 '15 at 14:48
• @Integral How is the countability condition used in Proposition 1.3? – ziyuang Dec 1 '17 at 22:11
• @ziyuang Not condition, it's case. Did you read the proposition and the Wikipedia article? – Integral Dec 2 '17 at 12:54
• @Integral Yes, if rephrasing my question: if $A$ were uncountable, which part of proposition 1.3's proof would go wrong? – ziyuang Dec 2 '17 at 17:07

## 2 Answers

We can write

\begin{align} \bigl\{ \pi_\alpha^{-1}(E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in \{1,2\}\bigr\} &= \{ \pi_1^{-1}(E_1) : E_1 \in \mathcal{M}_1\} \cup \{ \pi_2^{-1}(E_2) : E_2 \in \mathcal{M}_2\}\\ &= \{ E_1 \times X_2 : E_1\in \mathcal{M}_1\} \cup \{ X_1 \times E_2 : E_2 \in \mathcal{M}_2\}. \end{align}

In this form it is clear that this generating set is contained in

$$\{ E_1 \times E_2 : E_\alpha \in \mathcal{M}_\alpha\},$$

and on the other hand, every set in the latter generating family is the intersection of two members of the former, so the two families generate the same $\sigma$-algebra.

$$E_1\times E_2=\pi_1^{-1}(E_1)\cap\pi_2^{-1}(E_2)$$ so is measurable w.r.t.to the first mentioned $\sigma$-algebra.

$$\pi_1^{-1}(E_1)=E_1\times X_2$$ and

$$\pi_2^{-1}(E_2)=X_1\times E_2$$ so both sets are measurable w.r.t. the second mentioned $\sigma$-algebra.

This together proves that the mentioned $\sigma$-algebras coincide.