# Proving that if $A$ is countable then $\otimes_{\alpha}M_{\alpha}$ is created by sets of form ${\displaystyle \prod_{\alpha\in A}E_{\alpha}}$

I am given an exercise, the following is the first part of the exercise:

let $X_{\alpha}$ be a measureable space with $\sigma-algebra$ $M_{\alpha}$ , mark $$X\triangleq{\displaystyle \prod_{\alpha\in A}X_{\alpha}}$$ and $\pi_{\alpha}:X\rightarrow X_{\alpha}$ the projection to the i-th coordinate.

define $\otimes_{\alpha\in A}M_{\alpha}$ as the $\sigma-algebra$ that is created from sets of form $\pi_{\alpha}^{-1}\left(E_{\alpha}\right)$ where $E_{\alpha}\in M_{\alpha}$

Prove that if $A$ is countable then $\otimes_{\alpha}M_{\alpha}$ is created by sets of form ${\displaystyle \prod_{\alpha\in A}E_{\alpha}}$

I don't know where to start, I numbered the elements of $A$ and I wanted to prove by showing two containments, but I don't know how to start any of them.

Can someone please give some hint on how to start ?

Edit: I have managed to prove the containment that what created by the sets of form ${\displaystyle \prod_{\alpha\in A}E_{\alpha}}$ are in whats created from sets of form $\pi_{\alpha}^{-1}\left(E_{\alpha}\right)$

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Hint: $\prod_{\alpha \in A} E_\alpha = \bigcap_{\alpha \in A} \pi_\alpha^{-1}(E_\alpha)$, so $\bigotimes_{\alpha \in A} M_\alpha$ contains these sets.
@Belgi On the other side $\pi_\beta^{-1}(E_\beta) = \prod_{\alpha \in A} E_\alpha'$ with $E'_\alpha = X_\alpha$ for $\alpha \ne \beta$ and $E'_\beta = E_\beta$. –  martini Nov 7 '12 at 15:07