Borel regular implies inner regular and outer regular

Question

Let $\mu$ be a measure on $\mathbb{R}^d$ whose domain contains all open sets (and hence all Borel sets). I'm trying to show that if for every $\mu$-measurable set, $A$, there exists a Borel set, $B\supset A$, such that $\mu(A)=\mu(B)$, then $\mu$ is both inner regular and outer regular. That is, for all $\mu$-measurable sets, $A$, I want to show that $$\mu(A)=\inf\{\mu(G):G\supset A\text{ and }G\text{ is open}\}=\sup\{\mu(F):F\subset A\text{ and }F\text{ is compact}\}$$ I was able to show the converse without any trouble, but this direction is troubling me.