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In the real analysis, by Folland, p. 23:

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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):

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The following is the Lemma 1.1:

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

Michael Hardy
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sleeve chen
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    The correct spelling is $\sigma$-algebra (with a hyphen), not $\sigma-$algebra (with a minus sign). I changed it. ${}\qquad{}$ – Michael Hardy Oct 26 '15 at 00:24

1 Answers1

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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, $\mathscr M\subseteq\bigotimes_{\alpha\in A}\mathscr M_\alpha$.

Tim Raczkowski
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  • Short question: can I say $(\prod_{\alpha\in A}E_\alpha)^c = {\prod_{\beta\in A, \beta \neq \alpha}E_\beta : E_{\beta}\in \mathcal{M}_{\beta}, \beta \in A}$ – sleeve chen Oct 27 '15 at 09:02
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    No, in fact $\prod_{\beta\in A,\beta\ne\alpha}E_\beta\nsubseteq\prod_{\alpha\in A}$ – Tim Raczkowski Oct 27 '15 at 13:52
  • Then how about $(\prod_{\alpha\in A}E_\alpha)^c=\prod_{\alpha\in A}(X_{\alpha}\setminus E_\alpha)$ where $E_\alpha\in\mathscr M_\alpha, \alpha\in A$? "$c$" stands for complement. – sleeve chen Oct 27 '15 at 19:01
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    Sorry, this does work either. Consider $\Bbb R^2\setminus([0,1]\times[0,1])$and $(\Bbb R\setminus [0,1])\times(\Bbb R\setminus [0,1])$.. – Tim Raczkowski Oct 28 '15 at 13:47