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Given a collection of sets $\mathcal{C}$ and $E$ an element in the $\sigma$-algebra generated by $\mathcal{C}$, how do I show that $\exists$ a countable subcollection $\mathcal{C_0} \subset \mathcal{C}$ such that $E$ is an element of the $\sigma$-algebra, $\mathcal{A}$ generated by $\mathcal{C_0}$?

The hint says to let $H$ be the union of all $\sigma$-algebras generated by countable subsets of $\mathcal{C}$....although I don't know why.

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A good strategy would be to prove that $H$ is equal to $\sigma(\mathcal C)$, the $\sigma$-algebra generated by $\mathcal C$. It should be clear that $H \subset \sigma(\mathcal C)$. For the reverse inclusion, it is enough to show that $H$ is, in fact, a $\sigma$-algebra. Check each of the axioms! At some point, you will have to use the fact that a countable union of countable sets is countable.

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    Ok..I can do that. Does it mean that $E$ plays no part in the proof?2011-09-03
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    @Joe To be careful, you might want to show why $H = \sigma(C)$ proves the statement about this $E$. But I don't see how to write a proof that goes, "give me an $E$, and I'll find a $\mathcal C_0$ for you".2011-09-03
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    I was about to ask that..i.e...why it suffices to show that $H=\sigma(C)$2011-09-03
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    @Nana You two have me worried! Here is my reasoning: we will know that $E \in H$, and $H$ is just a union of $\sigma(\mathcal D)$ as $\mathcal D$ runs over the countable subsets of $\mathcal C$, so $E$ is in one of them.2011-09-03
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    @Dylan Oh ok...thanks2011-09-03
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    @Dylan Sorry I had you worried. Thanks for the explanation though...:)2011-09-03
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    @Dyland Morland. It just so happen that I am answering the same question that was asked by Joe. Can you help me on understanding how $H$ being a sigma algebra implies the reverse inclusion holds? Thanks a lot.2018-10-02