we can always find an open set with outer measure slightly more

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Theorem

for every $A \subset R$ and $ϵ > 0$ , there exist an open set $O$ such that $A \subset O$ and

m^{*} (A) \leq m^{*} (O) \leq m^{*} (A) + ϵ

Proof

The result is obvious if $m^{*} (A) = \infty$ (just take $O = R$ ), so we assume $m^{*} (A) < \infty$ .

Let ${I_{n}}$ be a collection of open intervals that cover $A$ and have total length at most $m^{*} (A) + ϵ$ . Then $O = ⋃_{n} I_{n}$ is open and obviously $A \subset O$ . Thus by subadditivity ( theorem from Lecture 6 )

\begin{aligned} m^{*} (O) & = m^{*} (⋃_{n} I_{n}) \\ \leq \sum_{n} m^{*} (I_{n}) \\ \leq \sum_{n} ℓ (I_{n}) \\ \leq m^{*} (A) + ϵ \end{aligned}

we can always find an open set with outer measure slightly more

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