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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 $\mu^*(A) \leq \mu^*(E) + \epsilon$.

proof: Suppose, $\{E_j\}_{1}^{\infty}\subset P(X)$ and $\epsilon > 0$. For each $j$ there exists $\{A_{j}^{k}\}_{1}^{\infty}\subset \mathcal{A}$ such that $E_j\subset A_\sigma$ and $\sum_{k=1}^{\infty}\mu_{0}(A_{j}^{k})\leq \mu^*(E_j) + \epsilon 2^{-j}$. But then, if $E = \bigcup_{j=1}^{\infty}E_j$, we have $E\subset \bigcup_{j,k=1}^{\infty}A_{j}^{k}$ and $\sum_{j,k}\mu_{0}(A_{j}^{k}) \leq \sum_{j}\mu^*(E_j) + \epsilon$, whence $\mu^*(A) \leq \sum_{j}\mu^*(E_j) + \epsilon$, since $\epsilon$ is arbitrary we are done.

I am not sure if I am right, any suggestions is greatly appreciated.

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  • $\begingroup$ I'm not sure I understand the notation. What does $\mathcal{A}(A_j^k)$ mean? Did you mean $\mu^*(A_j^k)$? $\endgroup$ – bartgol Aug 18 '15 at 21:16
  • $\begingroup$ Be careful, it should be $\{E_{j} \} _{1}^{\infty} \subset \mathcal{A}$ not $\mathcal{P}(X)$. $\endgroup$ – möbius Aug 18 '15 at 21:26
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You are on the right track.

Here is a proof I came up with:

By the definition of the outer measure we know that

$\mu^{*}(E) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} ( A_{j} ) : A_{j} \in \mathcal{A}, E \subset \bigcup_{j=1}^{\infty} A_{j} \right \}$.

Let $A = \bigcup_{j=1}^{\infty} A_{j}$ as above. Then $A \in \mathcal{A}_{\sigma}$ and $E \subset A$.

As you said, for each $A_{j}$ we can construct a sequence $\{ B_{j}^{k} \} _{k=1}^{\infty}$ where we might have trivially that $B_{j}^{k}=A_{j}$ for $k=j$ and $\emptyset$ otherwise. It follows that since:

$\mu^{*}(A_{j}) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} ( B_{j}^{k} ) : B_{j}^{k} \in \mathcal{A}, A_{j} \subset \bigcup_{k=1}^{\infty} B_{j}^{k} \right \}$

we have trivially that:

$\mu^{*}(A_{j}) \leq \mu_{0}(A_{j}) + \varepsilon 2^{-j}, \forall j$ and $\varepsilon>0$.

Thus:

$\mu^{*}(A) \leq \sum_{j=1}^{\infty} \mu^{*} ( A_{j} ) \leq \sum_{j=1}^{\infty} ( \mu_{0}(A_{j}) + \varepsilon 2^{-j}) = \mu^{*}(E) + \varepsilon$, where the result follows since $\varepsilon$ is arbitrary.

I think your strategy was mostly correct just some confusion with notation.

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