Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
$\mathbb R$ 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
1  
You may also want to read this and that to learn about accepting answers. –  Asaf Karagila Jun 7 '12 at 19:07

1 Answer 1

up vote 1 down vote accepted

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.

Hopefully that's enough to get you started.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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