# Baire sigma-algebra

Halo everyone. I would like to enquire how do I solve this question which I extract from Cohn book on Measure Theory.

Let $X$ be a compact Hausdorff space, and let $C(X)$be the set of all real-valued continuous funtions on $X$. Then $B_{o}(X)$, the Baire $\sigma$-algebra on $X$ is the smallest $\sigma$-algebra on $X$ that makes each function in $C(X)$ measurable; the sets that belong to $B_{o}(X)$ are called the Baire subsets of $X$. A Baire measure on $X$ is a finite measure on $(X,B_{o}(X))$

(i) Show that $B_{o}(X)$ is the $\sigma$-algebra generated by the closed $G_{\delta}$'s in $X$.
(ii) Show that if the compact Hausdorff space $X$ is second countable, then $B_{o}(X)=B(X)$

Note: $B(X)$ is the Borel $\sigma$-algebra.

Other question not from the text:
(i) What is the distinction between a locally compact Hausdorff space and a compact Hausdorff space?

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$\mathbb R$ is locally compact and Hausdorff. It is not compact. –  user31373 Jun 7 '12 at 18:10
Do you mean $G_\delta$ in the first part of the question? –  Asaf Karagila Jun 7 '12 at 18:44
Thanks Asaf. You are correct. A typo there. I will correct it. –  Sandra Jun 7 '12 at 19:05
You may also want to read this and that to learn about accepting answers. –  Asaf Karagila Jun 7 '12 at 19:07

For the first part, you must show two things: (1) that every closed $G_\delta$ subset of $X$ is an element of $B_0(X)$, and (2) that if any $\sigma$-algebra $A$ on $X$ has every closed $G_\delta$ subset of $X$ as an element, then $B_0(X)\subseteq A$. For the latter, it suffices that every continuous function is $A$-measurable, by definition of $B_0(X)$.
For the second part, note that any $G_\delta$ subset of $X$ is an element of $B(X)$--since it is generated by the open sets, and so every countable intersection of open sets is $B(X)$-measurable--so in particular, every closed $G_\delta$ subset of $X$ is an element of $B(X)$, and so $B_0(X)\subseteq B(X)$ by the first part. Note that we didn't even use the fact that $X$ is compact, here, so in general we can say $B_0(X)\subseteq B(X)$. To show the other containment, we must show that every closed subset of the second countable compact Hausdorff space $X$ is an element of $B_0(X)$--again, since $B(X)$ is generated by the closed subsets of $X$--which (as Asaf points out in the comments) follows the fact that closed sets are $G_\delta$ in this context.
For the second part actually note that every $G_\delta$ is Borel so $B_0\subseteq B$ always, and in the second countable case every closed set is $G_\delta$, so $B_0(X)$ is actually generated by the closed sets, so by minimality of both $B(X)$ and $B_0(X)$ we have equality. –  Asaf Karagila Jun 7 '12 at 19:06
@AsafKaragila: True, since we're dealing with $\sigma$-algebras and a $G_\delta$ is a countable intersection of open (so $B(X)$-measurable) sets. I will make suitable alterations. –  Cameron Buie Jun 7 '12 at 19:36