TITLE: Measure of a set on $\mathbb R$
QUESTION [0 upvotes]: Prove that $$\forall A \subset \mathbb R, m^*(A)=\inf\{m^*(O):O \text{ open },A\subset O\}$$

I'm trying to prove both sides of the inequality.
I've got that $m^*(A)\leq \inf\{m^*(O):O \text{ open },A\subset O\}$ because $A\subset O$.
How should I proceed to get the other side of the inequality?
The definition that I can see may be useful is that of an outer measure:
$$\mu^*(A)= \inf\left\{\sum^\infty_{j=1}|I_j| \text{ for any } \{I_j\}^\infty_{j=1} \text{ open cover of }A \text{ by intervals}\right\}$$

REPLY [1 votes]: Hint: An open cover of $A$ by open intervals is exactly the same as an open set containing $A$ because any open set in $\mathbb R$ can be written as a countable disjoint union of open intervals.

REPLY [1 votes]: Choose a sequence of open intervals $(I_n)$ covering $A$ such that $\sum_{n=1}^\infty \mu^*(I_n) \leq \mu^*(A) + \varepsilon$. But $G = \bigcup_{n=1}^\infty I_n$ is an open set containing $A$ and $\mu^*(G) \leq \sum_{n=1}^\infty \mu^*(I_n) \leq \mu^*(A) + \varepsilon$.
Since $\mu^*(A) \leq \mu^*(G)$ whenever $A \subset G$, the assertion follows. (Notice that the only property that is being used is countable subadditivity)
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