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Let $\mathcal{A}$ be an uncountable index set (such as $\mathbb{R}$).

Let $\{A_\alpha:\alpha \in \mathcal{A}\}$ be a collection of subsets of some space $S$. Define

$$\mathcal{F}=\bigcup \sigma(\mathcal{A}_{\alpha i}:i=1,2,3,...) $$

in which the union is taken over all countable sequences $\{\alpha_1,\alpha_2,...\}\subset \mathcal{A}$. (Note this union is an uncountable union)

Show that $\mathcal{F}$ is a $\sigma$-algebra.

I know this is a well known result. However I do not know how to prove it. I found this. I know how prove it for non-decreasing sequences, but that is not the case here.

I'd appreciate any help.

Thanks

Schach21
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    I don't think this is a true result. The union of $\sigma$-algebras (even finite unions) is not in general a $\sigma$-algebra, see: https://math.stackexchange.com/questions/1122081/union-of-two-sigma-algebras-is-not-sigma-algebra. In the question you link they consider the $\sigma$-algebra generated by the union, not just the union. In fact the first answer elaborates on this, which is why they need the assumption that we have a non-decreasing sequence. – bitesizebo Mar 01 '20 at 21:50
  • Yes, in fact https://math.stackexchange.com/questions/1122081/union-of-two-sigma-algebras-is-not-sigma-algebra they provide the union of two $\sigma$-algebra that is not a $\sigma$-algebra. This is strange. Thanks. – Schach21 Mar 02 '20 at 00:13

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