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Let $X$ be a compact metric space with a Probability Borel measure $\mu$. Let $C$ be any Borel subset of $X$. Then for any small positive number $a$, we can find compact set $K$ such that $K$ is subset of C and $\mu(C\setminus K)<a$.

Why is it so?

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2 Answers 2

You can prove this, it is not that difficult if you know what to do. It is a typical $\sigma$-algebra argument.

Define $$\tag{1}\mathcal{A}=\left\{ C\subset \Omega | \forall\, \varepsilon,\ \exists U\ \text{open and}\ K\ \text{compact s.t.}\ K\subset C\subset U\ \text{and}\ \mu(U\setminus K)<\varepsilon\right\}.$$ Then you can see that any closed subset $F$ of $\Omega$ lies in $\mathcal{A}$. Indeed, fix $\varepsilon >0$. Define the $\delta$-neighborhood(*) of $F$ to be $$F_\delta=\{ x \in \Omega\ |\ \text{dist}(x, \Omega)<\delta \}.$$ This set is open. Since the sequence $F_1, F_{1/2}, F_{1/3}\ldots F_{1/n}\ldots$ is shrinking, by the theorem of continuity of measure you have $$\lim_{n \to \infty} \mu(F_{\frac{1}{n}})=\mu\left(\bigcap_{n=1}^\infty F_{\frac{1}{n}}\right)=\mu(F),$$ so there is a $N$ so big that $\mu(F_{1/N})-\mu(F) < \varepsilon$. Then $F$ complies with the definition of $\mathcal{A}$ given in equation (1) with $K=F$ and $U=F_{1/N}$: since $\varepsilon$ was arbitrary this shows that $F\in \mathcal{A}$.

Later you show that $\mathcal{A}$ is a $\sigma$-algebra and since it contains all closed sets it contains all Borel sets too.


(*)Which is sometimes called "Minkowski sausage", a name which I find funny.

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Any Borel probability measure on a metric space is regular, and the condition you describe is precisely regularity of mu since your space is compact.

See: http://en.wikipedia.org/wiki/Regular_measure

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I think one needs more than just a metric space (that is, I do not trust wikipedia). If open sets are $\sigma$-compact or if the space is polish (complete and separable) then every Borel measure is regular. –  Jochen Dec 18 '12 at 9:54
    
@Jochen: The passage on Wikipedia is correct in that they are talking about inner regularity with respect to closed sets as opposed to inner regularity with respect to compact sets. The former holds for every semi-finite Borel measure on a metric space while the latter needs additional assumptions such as the ones you note. –  t.b. Dec 19 '12 at 7:48

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