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Let $E \subseteq \mathbb{R}$ s.t. $\mu^*(E) < \infty$ for some outer measure $\mu^*$ on $\mathcal{P(\mathbb{R})}$. Must there exist an open set $O$ s.t. $O \supseteq E$ and $\mu(O) < \infty$? What if we let $\mu$ and $\mu^*$ be the Lebesgue Outer Measure and Lebesgue Measure respectively?

I'm reading a proof which seems to assume that, at least the case of the Lebesgue setting, there must exist such an $O$. But I'm thinking that if we let $E = \mathbb{N}$ then $m^*(E) = 0 < \infty$ yet I can't think of an open set $O$ that contains $\mathbb{N}$ with finite outer measure so I'm confused.

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What is your definition of Lebesgue outer measure? $m^*(E)$ is the infimum of the sums of lengths of intervals covering $E$. The intervals can be taken to be open. –  Jonas Meyer Dec 3 '12 at 6:53
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2 Answers

This is true for Lebesgue measure, but not for all outer measures. Let $A=\{1/n :n\in\mathbb N\}$, and let $\mu^*(E)$ be the cardinality of $A\cap E$ if finite, $+\infty$ otherwise. If $E=\{0\}$ then $\mu^*(E)=0$, but for every open set $O$ containing $E$, $\mu^*(O)=+\infty$.

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Since Lebesgue outer measure is regular, then sure such a set exists. In fact, for each $\varepsilon>0$ we find an open set $O_{\varepsilon}\supset E$ so that $\mu^{*}(O_{\varepsilon})\leq \mu^{*}(E)+\varepsilon$. I can give you hints how to prove this depending on your definition of the Lebesgue measure.

For $\mathbb{N}$ choose for each $n\in\mathbb{N}$ $O_{n}=]n-\frac{1}{2^{n+1}},n+\frac{1}{2^{n+1}}[$. Then $\mu^{*}(O_{n})=\frac{1}{2^{n}}$ for all $n\in\mathbb{N}$. Take $O=\bigcup O_{n}$, whence $O$ is an open set such that $\mathbb{N}\subset O$ and by full-additivity (notice that the sets $O_{n}$ are disjoint): \begin{equation*} \mu^{*}(O)=\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1<\infty. \end{equation*}

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