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

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

Note:  \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  \exists B = \bigcap _{n=1}^k A_n \in \mathbb {A} _{ \sigma \delta } with  E \subset B and  \mu ^* (B \backslash E ) = 0

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

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

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

But what I need is  \mu * (E) = \mu ^* (E \cap An ) + \mu ^* (E \cap A_n^c ) \ \ \ \forall n , then of course, all I have to show is that  \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.