Set of finite outer measure is contained in an open set of finite measure?

Question

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.

Answer 1

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.

Answer 2

$m_*(E)=\inf \sum_{j=1}^\infty |Q_j|$ 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."