I mean $X$ is locally compact Hausdorff space, $\mu$ is Borel regular measure, and $\mu(\{x\})=0$. For any subset $A$ with finite measure. How to prove for any $0, we always can find a Borel subset $B$ of $A$, such that $\mu(B)=b$?
$X$ is locally compact Hausdorff space, $\mu$ is Borel regular measure. How to prove $\mu$ is cover $[0,\mu(A)]$
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real-analysis
measure-theory
compactness
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1If I understand correctly, $X$ is locally compact and Hausdorff, $\mu$ is a regular Borel measure that vanishes on singletons, and you want to prove that if $0, there is a Borel set $B\subseteq A$ such that $\mu(B)=b$. Is that correct? – 2012-10-31
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0yes, that is what I want to prove. – 2012-10-31
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0When $X=\mathbb{R}^n$ you can solve this by looking at the function $x\mapsto \mu(A\cap [-x,x))$. Maybe this can be used in a similar fashion in this setup. – 2012-10-31