Set of finite outer measure is contained in an open set of finite measure? In a proof I'm reading, the author remarks in the first line of the proof, with no justification whatsoever, that a set of finite outer measure is contained in an open set of finite measure. I have spent some time thinking about this and it is not obvious to me that this is the case. Can someone please explain why this should be obvious? Thank you.
 A: If you mean Lebesgue outer measure on $\mathbb R$, then the outer measure of a set $E$ is the infimum of all sums $\sum_{n=1}^{\infty} |I_n|$ where $\{I_n \mid n \in \mathbb{N}\}$ is a countable collection of open intervals whose union contains $E$. If $E$ has finite outer measure $\lambda^*(E)$, then for any $\epsilon > 0$, there is some collection $\{J_n\}$ with $\sum_{n=1}^{\infty}|J_n| < \lambda^*(E) + \epsilon$, which is certainly finite. Since the union of any collection of open sets is open, the set $U = \bigcup_{n=1}^{\infty}J_n$ meets your requirements.
A: $m_*(E)=\inf \sum_{j=1}^\infty |Q_j|$ where the inf is over all countable coverings of $E$ by closed cubes. If $m_*(E)$ is finite, there exists a covering $E\subset \bigcup_{j=1}^\infty Q_j$ with $\sum_{j=1}^\infty |Q_j|<\infty$. Enlarge the cube $Q_j$ to an open cube $\widetilde{Q}_j$ such that $|\widetilde{Q}_j| \leq |Q_j|+2^{-j}$. Then $\bigcup_{j=1}^\infty \widetilde{Q}_j$ is open, hence measurable, and
$$m\left(\bigcup_{j=1}^\infty \tilde{Q}_j\right) \leq \sum_{j=1}^\infty |\widetilde{Q}_j| \leq \sum_{j=1}^\infty (|Q_j|+2^{-j})= 1+ \sum_{j=1}^\infty |Q_j|<\infty.$$
So $E$ is contained in the open set $\bigcup_{j=1}^\infty \widetilde{Q}_j$, which has finite measure.
Edit: This holds in general for $E \subseteq \mathbb{R}^d$, which is why I said "cubes" and not "intervals."
