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Endowing $R$ with a finite borel measure. How to find a closed set with its total measure and every closed subset of it has minor measure?

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This closed set is called the "support" of the measure. – GEdgar Jul 18 '12 at 13:46
It is interesting that in a non-Lindelof space, it could happen that there is no such set. – GEdgar Jul 18 '12 at 13:47
up vote 0 down vote accepted

Let $\mathcal O:=\{O\subset \Bbb R, m(O)=0, O\mbox{ open}\}$ and $S':=\bigcup_{O\in\mathcal O}O$. $S'$ is open hence separable, and by Lindelöf property we can extract a countable subcover, which proves that $S'$ is open and of measure $0$. Let $S:=\Bbb R\setminus S'$ (support of $m$). It has full measure.

Let $F\subset S$ a closed subset and assume that $m(S)=m(F)$. We have $S'=S^c\subset F^c$ and since $m$ is finite, $m(F^c)=m(S^c)=0$. So $F^c\in\mathcal O$ and $F^c\subset S'$ which gives $S\subset F$. So if a closed subset of $S$ has full measure it should be $S$.

Note that it works for a Lindelöf topological space, since it's the only feature of the real line we have used.

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Why is $S'$ open? – Chris Eagle Jul 18 '12 at 13:41
@ChrisEagle I should have writtent that $\mathcal O$ is the collection of open subsets of measure $0$. I will fix it, thanks! – Davide Giraudo Jul 18 '12 at 13:44

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