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.


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.

  • $\begingroup$ Ok..I can do that. Does it mean that $E$ plays no part in the proof? $\endgroup$ – Joe Sep 3 '11 at 17:18
  • $\begingroup$ @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". $\endgroup$ – Dylan Moreland Sep 3 '11 at 17:26
  • $\begingroup$ I was about to ask that..i.e...why it suffices to show that $H=\sigma(C)$ $\endgroup$ – Nana Sep 3 '11 at 17:28
  • $\begingroup$ @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. $\endgroup$ – Dylan Moreland Sep 3 '11 at 17:48
  • $\begingroup$ @Dylan Oh ok...thanks $\endgroup$ – Joe Sep 3 '11 at 17:56

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