# Borel regular implies inner regular and outer regular

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.

• According to Wikipedia the domain of a Borel measure IS the set of Borel sets. What is your definition? – DanielWainfleet Jun 29 at 7:24
• @DanielWainfleet My apologies. I meant for it to be some measure whose domain contains the Borel sets. I'll make the edit. – Anonymous Jun 30 at 17:35
• I think you left something else out too. E.g. that $B\subset A$ or that $A\subset B$. – DanielWainfleet Jun 30 at 23:48
• Hint: (1). The hypotheses imply there also exists Borel $C\subset A$ with $\mu(C)=\mu(A)$. (2). Borel sets $B,C$ have the inner and outer regular properties. – DanielWainfleet Jul 3 at 10:15
• @DanielWainfleet Thanks. If the hints are true, then it's clear how to complete the proof. I think I figured out how to prove the second hint using transfinite induction and the fact that the Borel sets can be generated by iterating countable intersection and union to the first uncountable ordinal. I do not, however, see how to prove the first hint. – Anonymous Jul 4 at 22:19