TITLE: Product $\sigma$-algebra in countable case (Proposition 1.3 in Folland)
QUESTION [5 upvotes]: In the real analysis, by Folland, p. 23:  


I know $\prod_{\alpha\in A}E_{\alpha}=\bigcap_{\alpha\in A}\pi_{\alpha}^{-1}(E_{\alpha})$. But I cannot figure out why the product $\sigma$-algebra in the countable case should be defined in $\bigcap_{\alpha\in A}\pi_{\alpha}^{-1}(E_{\alpha})$. And I have no idea of "The result therefore follows from Lemma 1.1".    

The following is the general definition of product $\sigma$-algebra (on p.22) (It does not say anything about the intersection):  
 
The following is the Lemma 1.1:  


I can understand this Lemma, however, what does this Lemma relate to that proposition?

REPLY [6 votes]: Let $\mathscr M$ be the $\sigma$-algebra generated by $\{\prod_{\alpha\in A}E_\alpha:E_\alpha\in\mathscr M_\alpha, \alpha\in A\}$.
Since, $\pi_\alpha^{-1}(E_\alpha)=\prod_{\beta\in A}E_\beta$, then $\{\pi_\alpha^{-1}(E_\alpha)\}\subseteq\mathscr M$.  So, by Lemma 1.1, $\bigotimes_{\alpha\in A}\mathscr M_\alpha\subseteq\mathscr M.$ 
On the other hand, $\prod_{\alpha\in A}E_\alpha=\bigcap_{\alpha\in A}\pi_\alpha^{-1}(E_\alpha)$ implies that
$\{\prod_\alpha E_\alpha:E_\alpha\in\mathscr M_\alpha, \alpha\in A\}\subseteq\bigotimes_{\alpha\in A}\mathscr M_\alpha$, and again by Lemma 1.1, $\mathscr M\subseteq\bigotimes_{\alpha\in A}\mathscr M_\alpha$.