# Outer measure preserves elementary measure.

Let $E$ be the collection of elementary subsets of $\mathbb{R}$. Define a measure $m : E \rightarrow [0, \infty)$ by $m(\emptyset) = 0$ and if $\bigcup_{n=1}^N I_n \in E$, a disjoint union of intervals, then $m\left(\bigcup_{n=1}^N I_n\right) = \sum_{n=1}^N \ell(I_n)$ where $\ell(I_n)$ is the length of $I_n$.

If $A \in E$, prove that $m(A) = m^*(A)$ where $m^*(A) = \inf\left\{\sum_{n=1}^\infty \ell(I_n) : I_n \ \text{is an open interval and} \ A \subseteq \bigcup_{n=1}^\infty I_n\right\}$ is the outer measure of $A$. I am lost on where to start this. Any help will be appreciated! I was told that I need to use two facts:

1.) If $A \in E$, then for each $\varepsilon > 0$ there is an open set $O \in E$ with $A \subseteq O$ and $m(O) \leq m(A) + \varepsilon$.

2.) Let $A \in E$ and let $\{A_n\}_{n=1}^\infty$ be a collection of sets in $E$. If $A \subseteq \bigcup_{n=1}^\infty A_n$, then $$m(A) \leq \sum_{n=1}^\infty m(A_n).$$