$\mu(B)=\sup\{\mu(K): K\subseteq B: \text{$K$ compact}\}$, $B\in \mathscr B(\mathbb R^d)$

Question

Let $\mu$ be a regular borel measure on $\mathbb R^d$. Show: $\mu(B)=\sup\{\mu(K): K\subseteq B: \text{$K$ compact}\}$, $B\in \mathscr B(\mathbb R^d)$. I tried to show $\mu(B)\leq \sup\{\mu(K): K\subseteq B: \text{$K$ compact}\}$ and $\mu(B)\geq \sup\{\mu(K): K\subseteq B: \text{$K$ compact}\}$ but I got no good result. Do you have a hint how to show this?

Answer 1

More generally, any regular measure $\mu$ (according to your definition) that is finite on compact sets, on a $\sigma$-compact metric space $X$ is inner regular. Pick an increasing sequence $(K_n)$ of compact sets that covers $X$ (given by the $\sigma$-compactness assumption) and let $B$ be any Borel set.

Assume for the moment that $B$ has finite measure. Since $(K_n)$ is increasing, we have $\mu(B) = \lim \mu(B\cap K_n)$. Now approximate $\mu(B) \approx \mu(B\cap K_n)$ and $\mu(B) \approx \mu(F)$ where $F \subset B$ is closed, with errors less than a given $\varepsilon>0$. Then consider an appropriate compact set $K\subset B$ such that $\mu(B) \approx \mu(K)$ (hint : closed subsets of a compact set are compact).

Back to the general case, you have $B_n \uparrow B$ where $B_n = B \cap K_n$ and $$ \mu(B_n) = \sup\{\mu(K) : K \subset B_n, K \text{ compact}\} \le \sup\{\mu(K) : K \subset B, K \text{ compact}\} $$ After taking the (increasing) limit, you get inner regularity of $B$.

Here $\mathbb R^d$ fulfills all the requirements (metrizable, $\sigma$-compact and locally compact).