Let $E$ be a subset of $\mathbb R$ with $m^*(E)>0$. Prove that for each > $\alpha\in]0,1[$, there exist an open interval $I$ so that $$m^*(E\cap I)\geq \alpha m^*(I).$$
Notation: Here $m^*$ is the exterior mesure of Lebesgue.
In the correction it's written:
$E=\bigcup_{n\in\mathbb Z} E\cap [n-1,n]$, therefore $$m^*(E)\leq\sum_{n\in\mathbb Z}m^*\left(E\cap [n-1,n]\right).$$
Moreover, we can suppose $E$ bounded, therefore $m^*(E)$ is finite.
Question 1: Why can we suppose $E$ bounded ? If it's not bounded it's obvious ?
If $\alpha\in]0,1[$, there exist an open set $O\supset E$ such that $$m^*(E)\geq m(O)-(1-\alpha)m^*(E)\geq \alpha m(O).$$
Question 2: Here is my problem, if $E$ would be mesurable I would agree, but here $E$ is not supposed mesurable, that's why I don't understand why there exist ans open $O\supset E$ such that $m^*(E)+\varepsilon\geq m(O)$. Here the $\varepsilon=(1-\alpha)m^*(E)$.