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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?

Andrés E. Caicedo
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möbius
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    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? – Ziyuan Dec 01 '17 at 22:11
  • @ziyuang Not condition, it's case. Did you read the proposition and the Wikipedia article? – Integral Dec 02 '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? – Ziyuan Dec 02 '17 at 17:07

3 Answers3

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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.

Daniel Fischer
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  • 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 $S_1 \cap S_2 = E_1 \times E_2$? – James C Nov 05 '19 at 04:11
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    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 $S_1 \cap S_2 = (E_1 \times X_2) \cap (X_1 \times E_2) = E_1 \times E_2$. – Daniel Fischer Nov 05 '19 at 09:34
  • Thank you Daniel! – James C Nov 05 '19 at 18:27
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$$E_1\times E_2=\pi_1^{-1}(E_1)\cap\pi_2^{-1}(E_2)$$ 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.

drhab
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This is a very good question. The Folland definition depends on cylinders in the product space, while the Wikipedia definition refers to rectangles in the product space. It's unusual to define cylinders in terms of the projections function, but it works.

Cylinders are subsets of a product which are delimited in one dimension, but infinite - or better, equal to the base domain - in every other dimension. In a sense, the cylinders are infinite rectangles with one finite side. The accepted answer makes the correct observation that every rectangle is the intersection of two cylinders. The Folland definition is perhaps more abstract than it needs to be, but it's certainly elegant.

I'm adding this answer because my own project depends on cylinders, and I appreciate seeing the topic brought up in this context.

Ken Presting
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