Let $\displaystyle \mathbb {A} \subset P(X) $ be an algebra,
and let $\displaystyle \mathbb {A} _ \sigma $ be the collection of countable union of sets in $\displaystyle \mathbb {A} $.
Let $\displaystyle \mathbb {A} _ { \sigma \delta } $ be the collection of countable intersection of sets in $\displaystyle \mathbb {A} $.
Let $\displaystyle \mu _0 $ be a premeasure on $\displaystyle \mathbb {A} $.
Let $\displaystyle \mu ^* $ be outer measure with [tex] \mu* \mid _ $\displaystyle \mathbb {A} = \mu _0 $

Prove that if $\displaystyle \mu ^*(E) < \infty $, then $\displaystyle E $ is $\displaystyle \mu ^* $-measurable iff $\displaystyle \exists B \in \mathbb {A} _{ \sigma \delta } $ with $\displaystyle E \subset B $ and $\displaystyle \mu ^* (B \backslash E ) = 0 $

Note: $\displaystyle \mu ^* (E) := inf \{ \sum ^ \infty _{n=1} \mu _0 (F_n) : F_n \in \mathbb {A} , E \subset \bigcup _{n=1} ^ \infty F_n \} $

Proof so far.

First, I assume that $\displaystyle \exists B = \bigcap _{n=1}^k A_n \in \mathbb {A} _{ \sigma \delta } $ with $\displaystyle E \subset B $ and $\displaystyle \mu ^* (B \backslash E ) = 0 $

So I have $\displaystyle \mu ^* ( \bigcap _{n=1}^k A_n \backslash E ) = 0 $

Implies that $\displaystyle \mu ^* ( \bigcap _{n=1}^k A_n) - \mu ^*( E ) = 0 $

Implies that $\displaystyle \mu ^* (E) = \mu * ( \bigcap _{n=1}^k A_n) $

But what I need is $\displaystyle \mu * (E) = \mu ^* (E \cap An ) + \mu ^* (E \cap A_n^c ) \ \ \ \forall n $, then of course, all I have to show is that $\displaystyle \mu ^* (E) \geq \mu ^* (E \cap An ) + \mu ^* (E \cap A_n^c ) \ \ \ \forall n$ since the other inequality follows from the property of outer measure.

Any hints? Thank you.