Let $X\neq\emptyset$ and let $\mu:P(X)\to[0,\infty]$ be an outer measure. Recall that, a set $A\subseteq X$ is $\mu$-measurable if $$ \mu(B)=\mu(A\cap B)+\mu(B\setminus A), \text{ for all }B\subseteq X. $$ Let $M(X,\mu):=\{A\subseteq X:A\text { is }\mu-\text{measurable}\}$ be the set of all $\mu$-measurable sets. We know that $M(X,\mu)$ is a $\sigma$-algebra on $X$.
QUESTION:
My question is about the converse. Given any $\sigma$-algebra on $X$, does there exist an outer measure $\mu$ on $X$ such that $$ \boxed{\sigma=M(X,\mu)?} $$ If not, is there any necessary and sufficient condition on $\sigma$ that ensures the existence of an outer measure $\mu$ on $X$ such that $ \sigma=M(X,\mu)? $
