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The following is a proposition and its proof from chapter 1 of Folland's Real Analysis.

1.3 Proposition. If $A$ is countable, then $\bigotimes_{\alpha\in A}\mathcal M_\alpha$ is the $\sigma$-algebra generated by $\left\{\prod_{\alpha\in A}E_\alpha:E_\alpha\in\mathcal M_\alpha\right\}$.

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

Where is the countability of $A$ invoked? I.e., how does this argument fail for $A$ uncountable?

Lemma 1.1 reads:

1.1 Lemma. If $\mathcal E\subset\mathcal M\left(\mathcal F\right)$ then $\mathcal M\left(\mathcal E\right)\subset\mathcal M\left(\mathcal F\right)$.

Here, $\mathcal M\left(\mathcal E\right)$ stands for the $\sigma$-algebra generated by $\mathcal E$.

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  • What is Lemma 1.1? Perhaps the countability is used there. – User8128 Jun 15 '17 at 00:22
  • @User8128, I have edited in the lemma. – wjmolina Jun 15 '17 at 00:28
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    $\bigcap_{\alpha\in A}\pi^{-1}\alpha(E\alpha)$ this is the intersection of countably many sets, since $A$ is assumed countable. – Mirko Jun 15 '17 at 00:33
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    @Mirko, I see. So, $\prod_{\alpha\in A}E_\alpha$ need not necessarily be in $\bigotimes_{\alpha\in A}\mathcal M_\alpha$. – wjmolina Jun 15 '17 at 01:12
  • The answer can be found in https://math.stackexchange.com/questions/1497667/product-sigma-algebra-in-countable-case-proposition-1-3-in-folland – MMH Oct 22 '19 at 16:18

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