0
$\begingroup$

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?

  • 0
    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.2012-05-08

1 Answers 1