Question about a proposition on sigma algebras To define our terms: Let $\{ X_{\alpha} \}_{\alpha \in A } $ be any collection and $X = \prod_{\alpha \in A} X_{\alpha} $. Let $\pi_{\alpha}: X \to X_{\alpha} $ be coordinate maps and let $\mathcal{M}_{\alpha} $ be sigma algebras on $X_{\alpha} $. The product sigma algebra on $X$ is generated then by the  $\{ \pi_{\alpha}^{-1}(E_{\alpha} ) : E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A\}$, and we denote it by $\bigotimes_{\alpha \in A} \mathcal{M}_{\alpha} $.
PROP:
Suppose $\mathcal{M}_{\alpha} $ is generated by $\mathcal{E}_{\alpha} $, and let $\mathcal{F} = \{ \pi_{\alpha}^{-1}(E_{\alpha} ) : E_{\alpha} \in \mathcal{E}_{\alpha} \} $. Im trying to understand the proof to show $\bigotimes_{\alpha \in A} \mathcal{M}_{\alpha} \subseteq \sigma( \mathcal{F} ) $
The proof is as follows: Consider the collection $\mathcal{G} = \{ V \subset X_{\alpha} : \pi_{\alpha}^{-1}(V) \in \sigma( \mathcal{F} ) \} $. I understand and have proved that this is a $\sigma-$algebra. What I don't understand is
Why does $\mathcal{G} $ contain $\mathcal{E}_{\alpha} $ and $\mathcal{M}_{\alpha} $ ??
 A: This is clear just from sorting through the definitions. Including every detail I can imagine, hoping to hit the one you're missing:
We want to show that $\mathcal E_\alpha\subset\mathcal G$. The definition of $A\subset B$ is that every element of $A$ is an element of $B$. So we assume that $E\in\mathcal E_\alpha$ and now we need to show that $E\in\mathcal G$.
The definition of $\mathcal F$ shows that $\pi_\alpha^{-1}(E)\in\mathcal F$. Since $\mathcal F\subset\sigma(\mathcal F)$ this shows that $\pi_\alpha^{-1}(E)\in\sigma(\mathcal F)$. Now the definition of $\mathcal G$ shows that $E\in\mathcal G$, which is what we needed to prove.
So $\mathcal E_\alpha\subset \mathcal G$. It follows that $\mathcal M_\alpha\subset \mathcal G$: We know that $\mathcal M_\alpha$ is the sigma-algebra generated by $\mathcal E_\alpha$. This means that $\mathcal M_\alpha$ is the smallest sigma-algebra containing $\mathcal E_\alpha$. That says precisely that $\mathcal M_\alpha\subset\mathcal G$, since $\mathcal G$ is a sigma-algebra containing $\mathcal E_\alpha$.
