Meaning and Intuition behind product $\sigma$ algebras , Folland Definition I am currently studying Real Analysis using Folland's book by the same name, but I have troubles in understanding the product $\sigma$-algebras definition which is as follows:
Let $\{ X_{\alpha} \}_{\alpha \in A}$ be an indexed collection of nonempty sets, $X = \prod_{\alpha \in A}X_{\alpha}$, and $\pi_{\alpha}: X \rightarrow X_{\alpha}$ the coordinate maps. If $\mathcal{M}_{\alpha}$ is a $\sigma$-algebra on $X_{\alpha}$ for each $\alpha$, the $\textbf{product $\sigma$-algebra}$ on $X$ is the $\sigma$-algebra generated by :
$$\{ \pi^{-1}_{\alpha}(E_{\alpha}):E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A \}.$$
First I don't know what is the definition for the coordinate maps and then what is the intuition or how can I understand more practically product $\sigma$-algebras ?
Thank you!
 A: The coordinate maps are also called projection maps. They send an element of a product set into the $\alpha$-th "coordinate" space. For instance, if $X = A\times B$, then $\pi_A:X\to A$ would be defined by $\pi(a,b) = a$ for all $(a,b)\in A\times B$. Countable products can be thought of similarly, since you can write elements as infinite sequences $(a_1, a_2, \ldots)$. Uncountable products are a little bit more unwieldy.
Let's also use this example to get some feel for the product $\sigma$-algebra. For a finite product $X = A\times B$, where now $A, B = \mathbb{R}$ and each is equipped with the Borel $\sigma$-algebra. Then the product $\sigma$-algebra on $X$ is generated by
$$\{\pi_A^{-1}(E): E\in \mathscr{B}_A\}\cup\{\pi_B^{-1}(F):F\in\mathscr{B}_B\}.$$
For any $E\in\mathscr{B}_A$, we have $$\pi_A^{-1}(E) = E\times\mathbb{R}.$$ Similarly $$\pi_B^{-1}(F) = \mathbb{R}\times F.$$
Since the product $\sigma$-algebra is closed under finite intersections, this means $E\times F \in\mathcal{M}_{A\times B}$. So the product $\sigma$-algebra contains all sets of the form $E\times F$, where $E\in\mathcal{M}_A$ and $F\in\mathcal{M}_B$. (These $\sigma$-algebras are the Borel $\sigma$-algebras in this example.) In particular, taking $A$ and $B$ to be intervals (open, half-open, closed, whatever) we see that $\mathcal{M}_{A\times B}$ contains all rectangles. Since the Borel $\sigma$-algebra on $\mathbb{R}^2$ can be generated by half-open rectangles, this shows you that the product $\sigma$-algebra contains the Borel $\sigma$-algebra. This is one of the things that we expect out of the Lebesgue $\sigma$-algebra on $\mathbb{R}^2$, once we get around to defining it.
This isn't a complete description of the product $\sigma$-algebra. In general, there are many more sets in the product $\sigma$-algebra than just the Cartesian products of measurable sets. But I hope this example gives you some insight into how the product $\sigma$-algebra works - it is not as intimidating as Folland makes (arguably everything) seem.
A: The product $\sigma$-algebra is the smallest $\sigma$-algebra on $X:= \prod X_{\alpha}$ such that each projection map $\pi_{\beta} : X \to X_{\beta}$ is measurable. This is very much in analogy with the product topology on $X$, which is the smallest topology that makes the $\pi_{\beta}$ continuous.
In particular, if open sets are measurable in each $X_{\alpha}$, then continuous functions on $X$ would be measurable, which is obviously a desirable property.
A: How does the map look like when $X_1 = \{a,b\}$ and $X_2 = \{c,d\}$ and also assuming the sigma algebra on $X_1$ and $X_2$ are as follows: $$\sigma(X_1) = \mathcal P(X_1) = \{\phi,\{a\},\{b\},\{a,b\}\}$$ and $$\sigma(X_2) = \{\phi,X_2\}=\{\phi,\{c,d\}\}$$
In this case, $\pi_a^{-1}(\{a,b\})$ doesn't exist but there is still an element $\{a,b\}$ in the sigma algebra generated by $X_1$.
