Let $m(B) \leq \mu(B)$ for all closed balls $B$ then $m(U) \leq \mu(U)$ for all open sets

Question

I have a problem.

If $\mu : Bor(\mathbf{R}^n) \to [0,\infty]$ and $m(B) \leq \mu(B)$ for all closed balls $B$ then $m(U) \leq \mu(U)$ for all open sets.

I want to solve this by using Vitali's covering theorem but I don't understand how.

Vitali's covering theorem: Let $E \subset R^n$ and let $V$ be a closed Vitali covering of $E$. Then there exists a countable subfamily $G \subset V$ of pairwise disjoint balls s.t. $m(E\setminus \bigcup_{B_j\in G} B_j)=0$

Answer 1

Let $U$ be open. The collection of all closed cubes which are contained in $U$ is a Vitali cover of $U$. Hence there is a sequence of disjoint closed cubes $B_i$ each contained in $U$ such that $m(U\setminus \cup_i B_i)=0$. Now $\mu (U) \geq \mu (\cup_i B_i)= \sum \mu (B_i) \geq \sum m (B_i)=m(\cup_i B_i) =m(U)$.