Regular outer measure ($\forall A\exists G\supseteq A:G\text{ measurable},{\mu }^{\ast }(G)={\mu }^{\ast }(A)$)

An outer measure is regular in this sense when every set admits a measurable hull of equal outer measure. This is different from merely specifying inner or outer approximation of already measurable sets by compact or open sets. Lebesgue outer measure has a hull that is a Borel set: for finite outer measure choose open covers of total volumes within $1/j$ of its value and intersect their unions. Monotonicity gives equality of the hull's outer measure with that of the original set. For infinite outer measure the whole Euclidean space is a hull. This regularity is useful for turning numerical equality of inner and outer sizes into measurability.

For a regular outer measure, its Caratheodory measurability criterion defines a maximal algebra of sets on which that outer measure is finitely additive. Indeed, suppose an additive algebra contains all measurable sets and also $A$. Given $E$ of finite outer measure, take a measurable hull $H\supseteq E$ of equal mass. Finite additivity splits $H$ by $A$, and monotonicity gives ${\mu }^{\ast }(E)\ge {\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\setminus A)$. Outer subadditivity gives equality. For infinite mass that equality already follows from outer subadditivity. Thus $A$ was measurable. No sigma-finite measure assumption is necessary.

For a regular outer measure and measurable $M$, the forward implication is countable additivity. For the reverse, choose a measurable hull of $A$ and intersect it with $M$ to obtain $A\subseteq B\subseteq M$ and $\mu (B)={\mu }^{\ast }(A)$. Measurability of $B$ splits $M\setminus A$ into $M\setminus B$ and $B\setminus A$, soThe assumed equality and finiteness force ${\mu }^{\ast }(B\setminus A)=0$. Every outer-measure-null set is measurable by Caratheodory measurability criterion, so $A=B\setminus (B\setminus A)$ is measurable. Finiteness prevents invalid subtraction of infinite masses.