= Solution
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 $\mathcal L^*(G)=\mathcal L^*(A)$. <Lebesgue outer measure> has even a Borel hull. If its value on $A$ is finite, choose open-box covers of total cost below $\mathcal L^*(A)+1/j$, and let $U_j$ be their open unions. Then $G=\bigcap_jU_j$ is Borel and
$$
\mathcal L^*(A)\leq\mathcal L^*(G)\leq\mathcal L^*(U_j)\leq\mathcal L^*(A)+1/j.
$$
For infinite <outer measure> the whole space is a suitable hull. Thus \b[<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 $\mathcal L(B)=\mathcal L^*(A)$. Since $B$ is measurable, its splitting of $M\setminus A$ gives
$$
\mathcal L^*(M\setminus A)=\mathcal L(M\setminus B)+\mathcal L^*(B\setminus A).
$$
The assumed equality and $\mathcal L(M)=\mathcal L(B)+\mathcal L(M\setminus B)$ imply $\mathcal L^*(B\setminus A)=0$. 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 \b[the equivalence].
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