Recall the definition of a $\sigma$-algebra: it is a collection of subsets of some particular $X$, such that:
- It is closed under countable unions;
- It is closed under taking complement (relative to $X$).
From this, countable intersections follow by DeMorgan's Laws.
Now suppose that $X_i\in\mathcal F$ for $i\in\mathbb N$. So for all $\alpha\in A$ we have $X_i\in\mathcal F_\alpha$. Since $\mathcal F_\alpha$ is a $\sigma$-algebra we have that $\bigcap X_i\in\mathcal F_\alpha$ for all $\alpha\in A$ and therefore $\bigcap X_i\in\bigcap\mathcal F_\alpha=\mathcal F$.
For complements, the principle is the same.
Furthermore, the same idea works to show that $\mathcal F$ contains $\mathcal C$. Lastly we need to show that this is indeed the smallest:
Suppose that $\mathcal S$ is a $\sigma$-algebra which contains $\mathcal C$, then for some $\alpha\in A$ we have $\mathcal S=\mathcal F_\alpha$, so it took part in the intersection which generated $\mathcal F$, therefore $\mathcal F\subseteq\mathcal S$.
Further Reading:
- The $\sigma$-algebra of subsets of $X$ generated by a set $\mathcal{A}$ is the smallest sigma algebra including $\mathcal{A}$
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.
Best Answer
Please note that, even if $A$ is finite, an element of $\prod _{\alpha \in A}M_\alpha$ is not a subset of $\prod _{\alpha \in A}X_\alpha$. So it can not be (or generate) a $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$. I think you meant to write: the $\sigma$-algebra generated by $\{\prod _{\alpha \in A} E_\alpha : E_\alpha \in M_\alpha, \alpha \in A\}$.
Let $M$ be the $\sigma$-algebra generated by $$ \{\pi_\alpha^{-1}(E_\alpha): E_\alpha \in M_\alpha, \alpha \in A\} $$ and $N$ be the $\sigma$-algebra generated by $$ \{\prod _{\alpha \in A} E_\alpha : E_\alpha \in M_\alpha, \alpha \in A\} $$ First, as Tyr mentioned, if $A$ is (at most) countable, then $M=N$. In the general case, we have $M\subseteq N$. In all cases, the product $\sigma$-algebra is the SMALLEST $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables, and that is $M$. It is similar to the way we define product topology.
Why we want the product $\sigma$-algebra to be the SMALLEST $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables ?
Because the smallest $\sigma$-algebra that makes all the coordinate maps $\pi _\alpha$ measurables is precisely the only $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ which makes the following proposition hold:
Any $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ larger than $M$ (the product $\sigma$-algebra) will result in the existence of $f$ which is not measurable but, for all $\alpha\in A$, $\pi_\alpha \circ f$ is measurable.
Any $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$ smaller than $M$ (the product $\sigma$-algebra) will result in the existence of a measurable $f$ such that there is $\alpha\in A$, $\pi_\alpha \circ f$ is not measurable.
In fact, let $S$ be a $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$.
If $S \varsupsetneq M$, take $X= \prod _{\alpha \in A}X_\alpha$ and $\Sigma=M$, then $Id: (\prod _{\alpha \in A}X_\alpha,M) \to (\prod _{\alpha \in A}X_\alpha, S) $ is NOT measurable, but, for all $\alpha\in A$, $\pi_\alpha \circ Id$ is measurable.
If $S \varsubsetneq M$, take $X= \prod _{\alpha \in A}X_\alpha$ and $\Sigma=S$, then $Id: (\prod _{\alpha \in A}X_\alpha,S) \to (\prod _{\alpha \in A}X_\alpha, S) $ is measurable, but there is $\alpha\in A$ such that $\pi_\alpha \circ Id$ is not measurable.