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This comes from an exercise from Real Analysis by Folland.

Let $\mathcal{A}\subset P(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$. Let $\mu_{0}$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

a.) For any $E\subset X$ and $\epsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A) \leq \mu^*(E) + \epsilon$.

b.) If $\mu^*(E) < \infty$, then $E$ is $\mu^*$-measurable if and only if there exists $B\in \mathcal{A}_{\sigma\delta}$ with $E\subset B$ and $\mu^*(B\setminus E) = 0$

proof of a.) was done in an earlier post: Outer measure problem

proof of b.) Using part a.) $\forall \epsilon_i$ there exists $A_i\in \mathcal{A}_\sigma$ such that $E\subset A_i$. Let $B = \bigcap_{i=1}^{n}A_i$ then $B\in\mathcal{A}_{\sigma\delta}$. Now, we need to show that $\mu^*(B\cap E^c) = 0$. From the definition of outer measure we know that: $$\mu^*(B\cap E^c) = \inf\left\{\sum_{1}^{\infty}\mu^*(E_j):E_j\in\mathcal A \ \ \text{and} \ \ (B\cap E^c)\subset \bigcup_{1}^{\infty}E_j\right\}$$ From part a.) we know that $\mu^*(A)\leq \mu^*(E) + \epsilon$. I am not sure if this fact helps and I am stuck where to go from here, any suggestions would be great. Also, any hints on how to prove the converse would be great as well.

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  • $\begingroup$ What exactly is $\mathcal{A}\subset \mathcal{A}$ supposed to mean? $\endgroup$ – zoli Aug 19 '15 at 17:28
  • $\begingroup$ Sorry, that was a mis-type, see edit $\endgroup$ – Wolfy Aug 19 '15 at 17:29
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A hint for the forward case:

Since by assumption $E$ is $\mu^{*}$-measurable we have that for all $F \subset X$ we have:

$\mu^{*}(F) = \mu^{*}(F \cap E) + \mu^{*}(F\cap E^{c})$

Since this holds for all $F \subset X$ it must also hold for $B$ as you constructed above. Now use the fact that $E \subset B$ and part (a) to rearrange the above and show that $0 \leq \mu^{*}(B\setminus E)< \varepsilon$.

A hint for the reverse case:

First if there exists such a set $B$ show that $B$ is $\mu^{*}$ measurable. Then you get that for any $F \subset X$:

$\mu^{*}(F) = \mu^{*}(F \cap B) + \mu^{*}(F\cap B^{c})$

Now try to show that:

$\mu^{*}(F \cap B) \geq \mu^{*}(F\cap B\cap E) + \mu^{*}(F\cap B \cap E^{c})$

and

$\mu^{*}(F \cap B^{c}) = \mu^{*}(F\cap B^{c} \cap E^{c} ) + \mu^{*}(F\cap B^{c} \cap E)$

Finally note that:

$F \cap E = (F\cap B\cap E)\cup(F\cap B^{c} \cap E)$

and

$F\cap E^{c} = (F\cap B\cap E^{c})\cup(F\cap B^{c} \cap E^{c})$

Combining all of this information with subadditivity of $\mu^{*}$ it is possible to show that $\mu^{*}(F) \geq \mu^{*}(F\cap E) + \mu^{*}(F \cap E^{c})$ which is sufficient to prove $E$ is $\mu^{*}$ measurable.

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