M is a $\sigma$-algebra.
$\bar M$ : $A \in \bar M \Longleftrightarrow$ there exist $B,C \in M$ such that
$B \subset A \subset C$ and $\mu (C \sim B) = 0$
hereby, $\mu (B)=\mu (C)$
so let, $\bar{\mu}(A)=\mu(B)=\mu(C)$
In above Notation,
I want to prove it. $E \subset A \in \bar M$ and $\bar \mu (A) =0$, then $E \in \bar M$.
Is it necessarily fact for $\bar M$ is a $\sigma$ algebra and $\mu$ is well defined? If then, why is it?