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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.

A regular outer measure has, for every $E\subseteq X$, a measurable hull $H\supseteq E$ with $\mu (H)=\mu (E)$.

Suppose $\mathcal{C}$ is an algebra containing ${\mathfrak{M}}_{\mu }^{\ast }$ and $\mu$ is finitely additive on $\mathcal{C}$. Fix $A\in \mathcal{C}$ and an arbitrary $E$ of finite outer measure. Take a measurable hull $H$ of $E$. Both $H\cap A$ and $H\setminus A$ belong to $\mathcal{C}$, soOuter subadditivity gives equality. If $\mu (E)=\infty$, outer subadditivity already forces the sum on the right to be infinite, again giving equality. Therefore $A$ satisfies the Caratheodory criterion. We have proved $\mathcal{C}\subseteq {\mathfrak{M}}_{\mu }^{\ast }$, soThis proves the maximal additivity algebra of a regular outer measure without any sigma-finiteness assumption.

A regular outer measure has a measurable hull of equal outer measure for every set: for each $A$ there is measurable $G\supseteq A$ with ${\mathcal{L}}^{\ast }(G)={\mathcal{L}}^{\ast }(A)$. Lebesgue outer measure has even a Borel hull. If its value on $A$ is finite, choose open-box covers of total cost below ${\mathcal{L}}^{\ast }(A)+1/j$, and let $U_j$ be their open unions. Then $G={\bigcap }_j{U}_j$ is Borel andFor infinite outer measure the whole space is a suitable hull. Thus Lebesgue outer measure is regular.

If $A$ is measurable, additivity immediately gives the requested equality. Conversely suppose the equality holds. Choose a measurable hull $G$ of $A$ and put $B=G\cap M$. Then $A\subseteq B\subseteq M$ and $\mathcal{L}(B)={\mathcal{L}}^{\ast }(A)$. Since $B$ is measurable, its splitting of $M\setminus A$ givesThe assumed equality and $\mathcal{L}(M)=\mathcal{L}(B)+\mathcal{L}(M\setminus B)$ imply ${\mathcal{L}}^{\ast }(B\setminus A)=0$. These subtractions are legitimate because $\mathcal{L}(M)<\infty$. Outer-measure-null sets are measurable by the Caratheodory criterion. Hence $A=B\setminus (B\setminus A)$ is measurable, proving the equivalence.