Countable generation of $\sigma$-algebras

Question

I feel that a positive answer to the following question would be helpful in solving some exercises in introductory measure theory:

Suppose $\cal A$ is a collection of subsets of a set $X$ and let $\cal \Sigma$ be the smallest $\sigma$-algebra containing $\cal A$, that is to say, the intersection of all $\sigma$-algebras containing $\cal A$.

Suppose $S \in \cal \Sigma$. Then, can we state that there is a countable collection of sets $\cal B \subset \cal A$ such that, $S$ can be obtained from $\cal B$ with countably many operations of complement, intersection and union?

Does the above query have a positive answer and if so where can a proof be found?

Thank you! I'd been thinking about why $C([0,\infty))$ is not a measurable subset of $\mathbb{R}^{[0,\infty)}$ and this solves it. — snar, Aug 12 '14 at 05:55

score 4 · Accepted Answer · answered Aug 12 '14 at 03:34

4

Hint: Let $\Sigma'$ be the collection of sets that can be obtained from countable subsets of $\mathcal A$ by countably many operations. Show that $\Sigma'$ is a $\sigma$-algebra.

answered Aug 12 '14 at 03:34

Robert Israel

448,999

That simple... oh! Thanks a lot! – Aug 12 '14 at 05:51
The downvote boggles the mind. Also +1 for something that'd been bothering me for a while. – snar Aug 12 '14 at 05:54

Countable generation of $\sigma$-algebras

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