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Let $(X,d)$ be a compact metric space and $\mu$ be a finite measure on the Borel $\sigma$-algebra $B_X$ on $X$. Then we have for all $A \in B_X$:

$\mu (A) = \inf \{\mu(Q) \ \vert \ A\subseteq Q, Q \ \text{open} \} = \sup \{\mu(K) \ \vert \ K\subseteq A, K \ \text{compact} \} $

So far, I have not had an successful idea how to tackle this problem. Thanks for any inspiration!

Joker123
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  • I already answered essentially the same (slightly more general) question here: http://math.stackexchange.com/questions/1574021/locally-compact-metric-space-regular-borel-measure/1574054#1574054 – PhoemueX Dec 16 '15 at 18:13

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HINT:

Start with a compact subset $K$. $K$ is the intersection of a decreasing sequence of open subsets $$K = \bigcap_{n=1}^{\infty} \{ x \in X \ | \ d(x,K) < \frac{1}{n} \}$$

Hence $\mu(K) = \lim \mu(U_n)$ by general properties of finite measures.

Consider the family $\mathcal{A}$ of subsets $A$ of $X$ with the following property: for every $\epsilon > 0$ there exist $K \subset A \subset U$, $K$ compact, $U$ open so that $$\mu(U\backslash K ) < \epsilon$$ We want to show that all the borel subsets satisfy this property. So we'll show that the family of these subsets if a $\Sigma$ algebra containing the compact subsets.

From the above, the compact subsets are in $\mathcal{A}$. Then if $A$ is in $\mathcal{A}$, so is $X \backslash A$. Indeed if $K \subset A \subset U$ then $X\backslash U \subset X\backslash A \subset X\backslash K$ ...

Now for countable unions. Let $A_1$, $A_2$ $\ldots$ in $\mathcal{A}$. Take $K_i \subset A_i \subset U_i$ and note that $(U_1\cup U_2\cup\ldots ) \backslash (K_1 \cup K_2\cup \ldots ) \subset (U_1\backslash K_1)\cup (U_2\backslash K_2)\cup \ldots$. Take $K_i \subset A_i \subset U_i$ so that $\mu(U_i\backslash K_i) < \epsilon/2^i$, and also note that we can approximate the infinite union of the $K_i$ with a finite union, which is compact.

Hence, the algebra $\mathcal{A}$ contains all the Borel subsets, which means that the measure $\mu$ is regular.

We only needed that compact subsets are countable intersections of open subsets. That is true if $X$ is a (compact) metric space, but may also hold for more general compact spaces ( see this example).

orangeskid
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  • Do you know if it is possible for a non-metric compact space to admit a finite Borel measure which is not regular? – Cronus Feb 09 '18 at 02:25
  • @Cronus: Thanks for the input. I would have thought there are. Is there an example in Rudin- Real and Complex Analysis? – orangeskid Feb 10 '18 at 10:07
  • Good idea, he does give an example: the space is $[1,\omega_1]$ (where $\omega_1$ is the first uncountable ordinal) with the order topology, the $\sigma$-algebra is the collection of all $E$ such that either $E\cup{\omega_1}$ or $E^c\cup{\omega_1}$ contains an uncountable closed set, and the measure gives sets of the former type measure one and of the latter type measure zero. It is finite and Borel but not regular. (Apparently this is also given in Remark 4 before Theorem 29.2 in Measure and Integration Theory by Bauer). – Cronus Feb 10 '18 at 13:17
  • @Cronus: Thank you! I was vaguely recalling something. Will have to study it a bit closer. Wonder what the regularization of this measure is. Also, there is something about Borel measures vs Radon measures. Interesting stuff. – orangeskid Feb 10 '18 at 14:31
  • Yeah, it's pretty interesting! Also very surprising how complicated things get when you don't assume 2nd countability. – Cronus Feb 10 '18 at 15:56
  • And I think the regularisation of this measure is the dirac measure on $\omega_1$. – Cronus Feb 10 '18 at 15:57