Why do we need countability of A to prove proposition 1.3 in Folland Real Analysis?

Question

1.3 Proposition If A is countable, then $\oplus_{\alpha\in A}\mathcal{M}$ is the $\sigma$-algebra generated by $\{\prod_{\alpha\in A}E_\alpha:E_\alpha\in\mathcal{M}_\alpha \}$

I understand the proof, but where did the proof make use of countability of A? Is it because if A is not countable, then $\prod_{\alpha\in A}E_\alpha=\bigcap_{\alpha\in A}\pi^{-1}(E_\alpha)$ is not necessarily true since $\sigma$-algebra is only closed under countable intersections?

Proof here: If $E_\alpha\in\mathcal{M}_\alpha$,then $\pi^{-1}(E_\alpha)=\prod_{\beta\in A}E_\beta$ where $E_\beta=X_\beta$ for $\beta\neq\alpha$; on the other hand, $\prod_{\alpha\in A}E_\alpha=\bigcap_{\alpha\in A}\pi^{-1}(E_\alpha)$. The result therefore follows from Lemma 1.1.

Lemma 1.1 If $\mathcal{E}\subset\mathcal{M}(\mathcal{F})$ then $\mathcal{M}(\mathcal{E})\subset\mathcal{M}(\mathcal{F})$.

Answer 1

Suppose that $A$ is not countable. So, a $\sigma$-algebra $M$ is $\oplus_{a \in A} M_a$. We could choose a subfamily $B$ of $A$ which is uncountable. Then, $\bigcap_{b \in B} \oplus_{a\in A} M_a = \oplus_{b \in B} M_b$ exists. So, infinite intersections always exist, for every set $B$ smaller than $A$! Contradiction.

Answer 2

By definition, $\otimes_{\alpha \in A}\mathcal{M}_\alpha = \mathcal{M}(\{\pi^{-1}_\alpha(E_\alpha): E_\alpha \in \mathcal{M}_\alpha, \alpha \in A \}) $ (the $\sigma$-algebra generated by..). If $A$ is countable, we have to show that this is equal to $\mathcal{M}(\{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \})$. $$ \pi^{-1}_\alpha(E_\alpha)=\Pi_{\beta \in A}E_\beta, E_\beta = X_\beta \text{ if } \beta \ne \alpha \\ \implies \pi^{-1}_\alpha(E_\alpha) \subset \mathcal{M}(\{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \}) \\ \implies \mathcal{M}(\{\pi^{-1}_\alpha(E_\alpha): E_\alpha \in \mathcal{M}_\alpha, \alpha \in A \}) \subset \mathcal{M}(\{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \}) \\ \implies \otimes_{\alpha \in A}\mathcal{M}_\alpha \subset \mathcal{M}(\{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \}) $$ Now we have to show the opposite inclusion. $$ \Pi_{\alpha \in A} E_\alpha = \bigcap_{\alpha \in A}\pi^{-1}_\alpha(E_\alpha) $$ Now, since each $\pi^{-1}_\alpha(E_\alpha) \in \otimes_{\alpha \in A}\mathcal{M}_\alpha $ and $\sigma$-algebras are closed under COUNTABLE intersections, we have that $\Pi_{\alpha \in A} E_\alpha \in \otimes_{\alpha \in A}\mathcal{M}_\alpha $. Therefore: $$ \{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \} \subset \otimes_{\alpha \in A}\mathcal{M}_\alpha \\ \implies \mathcal{M}(\{ \Pi_{\alpha \in A} E_\alpha: E_\alpha \in \mathcal{M}_\alpha \}) \subset \otimes_{\alpha \in A}\mathcal{M}_\alpha $$