Can someone prove me the following measure theory Lesbegue outer measure property. I have been looking for its proof and intuition but I have not found it yet. $\forall E \in\mathscr{P}(\mathbb{R}), \epsilon > 0, \exists O\:\:\:\text{open set in}\:\mathbb{R}:O\supset E\:\:\:\text{and}\:\mu^*_l(E)\leqslant\mu^*_l(O)\leqslant\mu^*_l(E)+\epsilon$
This property implies that $\mu^*_l(E)=\mu^*_l(O)$ but how? If $O\supset E$. Is this just a property to help prove equalities? Which are the examples of this on the real line? I am asking this because this property has been used to solve the following problem whose statement seems to be paradoxical, I guess.
Prove that for every $E \subset \mathbb{R}$ , there exists a there exists a $G\subset\mathbb{R}$ such that:
$G\supset E$ and $\mu^*_l(E)=\mu^*_l(G)$
How can $G\supset E$ and $\mu^*_l(E)=\mu^*_l(G)$? It seems a contradiction.
This property says that
i.e. for every $E\subseteq\Bbb R$ and $\varepsilon>0$ there is an open $O\supseteq E$ with $\mu^*(O)\le\mu^*(E)+\varepsilon$.
And it doesn't follow anything more than that, in particular, we don't necessarily have an open set $O$ with $\mu^*(E)=\mu^*(O)$.
But, we can always intersect (countably) infinitely many of them to get a $G_\delta$-set $G$ which already satisfies the sharper $\mu^*(E)=\mu^*(G)$, using the above approximations for $\varepsilon=1/n$ values.
For a lightweight illustration, take $E$ to be a cheese with tiny-tiny holes (or more mathematically, e.g. $E:=(0,1)\setminus\Bbb Q$) and just fill those holes (taking $O:=(0,1)$).
Finally, even in this example we have infinitely many points in $O\setminus E$, but still $\mu^*(E)=\mu^*(O)$.