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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?

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It's a long, you would want to look at Rudin's Real and Complex Analysis, page 28, theorem 1.36. What you basically do is you define a complete sigma algebra: adding new measurable sets: if A is measurable, m(B) =0 then $A\bigcup B, A\B$ measurable. and if m(A) =0 then every subset of A is measurable. – Mai May 8 '12 at 17:48

You know $E \subset A$, and since $A \in \overline{M}$, you know there exists a $C \in M$ such that $A \subset C$ and $\mu C = \overline{\mu} A = 0$. So, you know there exists a $C$ such that $E \subset C$, and $\mu C = 0$.

Can you think of a special set (that must be in in every $\sigma$-algebra) that has measure $0$? This set will serve as the '$B$' set above.

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