Understanding product $\sigma$-algebra 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 $M_\alpha$ is a $\sigma$-algebra on $X_\alpha$ for each $\alpha$, what is the product $\sigma$-algebra on $X$?
I am reading a book, and it says that the product $\sigma$-algebra on $X$ is the $\sigma$-algebra generated by 
$$
\{\pi_\alpha^{-1}(E_\alpha): E_\alpha \in M_\alpha, \alpha \in A\}
$$
I was wondering:


*

*Why isn't it simply $\prod _{\alpha \in A}M_\alpha$?

*I am completely confused with this statement of the book, can anyone give a simple example to clarify what the book is trying to say?

 A: 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:  

Let $(X,\Sigma)$ be a mensurable space and $f: X \to  \prod _{\alpha \in A}X_\alpha$ then: $f$ is measurable IF AND ONLY IF , for all $\alpha \in A$, $\pi_\alpha \circ f$ is measurable. 

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.
