9
$\begingroup$

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.

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.