Open set U whose outer measure is less than A + epsilon

Question

Let $A$ be any given set with $m^*(A)<\infty$ (outer measure finite).

Then for any $\epsilon>0$, there is an open set $U$ such that $A\subseteq U$ and $m^*(U)\leq m^*(A)+\epsilon$.

How do we prove it rigorously? My lecture notes says the proof is immediate but I don't really get how to prove it.

Intuitively I understand it, take $U$ to be "smallest" such set and the inequality will intuitively hold.

Answer 1

$m^*(A)=\inf \{\sum I_n:A\subset \cup I_n\}$

So for any given $\epsilon>0$ by definition of greatest lower bound of a set we will get a sequence of open intervals such that $A\subset \cup I_n$ and $m^*(A)+\epsilon >\sum I_n$.

Take the open set $O= \cup I_n$ and result follows