## Outer measure on collections of unions and intersections

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.