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Let $A \subset P(X)$ be an algebra, $A_\sigma $ the collection of countable unions of sets in $A$, and $A_{\sigma \delta}$ the collection of countable intersections of sets in $A_\sigma $. Let $\mu_0 $ be a premeasure on $A$ and $\mu ^*$ the induced outer measure. Show that if $\mu^*(E)<\infty$, then

$E$ is $\mu^*$ measurable iff there exists $B\in A_{\sigma \delta}$ with $E\subset B$ and $\mu^*(B\cap E^c)=0$.

I can prove the forward implication. Stuck in proving reverse implication. Can anyone help me?

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Let $C$ be an arbitrary set in $X$. By Caratheodory's theorem the $\mu^{*}$-measurable sets form a $\sigma$-algebra, and so $B$ is $\mu^*$-measurable.

Therefore $\mu^*(C) = \mu^*(C \cap B) + \mu^*(C \cap B^{c}).$ Since $E \subseteq B$ we have that $\mu^*(C \cap E) \leq \mu^*(C \cap B)$. And since $\mu^*(B \setminus E) = 0$ we have $\mu^*(C \cap (B \setminus E)) = 0$. Finally, we use the fact that $E^{c} = B^{c} \cup (B \setminus E)$, which tells us that $\mu^*(C \cap E^{c}) \leq \mu^*(C \cap B^{c}) + \mu^*(C \cap (B \setminus E)) = \mu^*(C \cap B^{c})$.

Putting this all together \begin{equation} \mu^*(C) = \mu^*(C \cap B) + \mu^*(C \cap B^{c}) \geq \mu^*(C \cap E) + \mu^*(C \cap E^{c}), \end{equation} implying that $E$ is $\mu^*$-measurable.

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  • $\begingroup$ How is the statement "$B$ is $\mu^*$-measurable" justified? Just because the $\mu^*$ measurable sets form a $\sigma$-algebra, that doesn't mean that $B$ must be in that $\sigma$-algebra. $\endgroup$ – Emily Feb 8 '15 at 3:03
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    $\begingroup$ Folland Proposition 1.13 b: If $\mu_{0}$ is a premeasure on $A$ and $\mu^{*}$ is the induced outer measure, then every set in $A$ is $\mu^{*}$-measurable. This means that if we let $M$ denote the $\sigma$-algebra of $\mu^{*}$-measurable sets, then $A \subseteq M$, and hence $A_{\sigma \delta} \subseteq M$. $\endgroup$ – unknownymous Feb 8 '15 at 6:54

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