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!
Best Answer
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).