Prove: For all $E\subseteq \mathbb{R} $ and for all $\epsilon>0$ there is an open set $O$, with $E\subseteq O$ such that: $$m^*(O)<m^*(e)+\epsilon$$
I am acquaint with the following:
For all $E$ and for all $\epsilon>0$ there is a covering $O$ of $E$ by closed set such that:
$$\sum l(O_i)\leq m^*(E)+\epsilon$$
and because the outer measure is defined as the infimum of $\sum l(O_i)$ then $$\sum l(O_i)\leq m^*(E)+\epsilon$$ as it can not be an infimum, does the same hold if the cover is an open set? it seems so, as $l(a,b)=l([a,b])$ so the countable sum will be the same too just that the proof ask $<$ whereas "my claim" is for $\leq$
If $E\subseteq\mathbb{R}$, then by definition
$$m^{*}(E) = \inf\{m^*(O): O \ \text{is open and } \ E\subseteq O\}$$
Let $x$ be some element that is in the infimum. Also for $\epsilon > 0$, let $\mathcal{C}$ be a cover set of $E$ such that
$$\sum_{I\in\mathcal{C}}l(I)\leq m^*(E) + \epsilon$$
Then $O = \bigcup \mathcal{C}$ is an open set that contains $E$, so
$$x\leq m^*(O)\leq \sum_{I\in\mathcal{C}}m(I) = \sum_{I\in\mathcal{C}}l(I)\leq m^*(E) + \epsilon$$