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).
Why is it so?
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).
Why is it so?
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.
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.