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?

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

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.

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  • $\begingroup$ Hi Daniel. I just started studying measure theory, and I am deeply confused about $\sigma$ algebra. I would really appreciate if you can explicitly provide $S_1, S_2$ such that $S_1 \cap S_2 = E_1 \times E_2$? $\endgroup$ – James Chung Nov 5 '19 at 4:11
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    $\begingroup$ Take $S_k = \pi_k^{-1}(E_k)$ for $k \in \{1,2\}$, @JamesChung. Then $S_1 = \{(x_1,x_2) \in X_1 \times X_2 : x_1 \in E_1\} = E_1 \times X_2$ and $S_2 = \{(x_1,x_2) \in X_1 \times X_2 : x_2 \in E_2\} = X_1 \times E_2$, thus $S_1 \cap S_2 = (E_1 \times X_2) \cap (X_1 \times E_2) = E_1 \times E_2$. $\endgroup$ – Daniel Fischer Nov 5 '19 at 9:34
  • $\begingroup$ Thank you Daniel! $\endgroup$ – James Chung Nov 5 '19 at 18:27

$$E_1\times E_2=\pi_1^{-1}(E_1)\cap\pi_2^{-1}(E_2)$$ 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.

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