Show that if $m(\bar{B}^{n}(x,r)) \leq \mu(\bar{B}^{n}(x,r))$ then $m(U) \leq \mu(U)$ holds for all open sets $U \subset \mathbb{R}^{n}$. Studying real analysis and came upon this problem in a chapter about Vitali covering lemma.

Let $\mu: Bor\mathbb{R}^{n} \rightarrow [0, + \infty]$ be a measure
  s.t
$m(\bar{B}^{n}(x,r)) \leq \mu(\bar{B}^{n}(x,r))$ (here m is a lebesgue
  measure)
for all closed balls $\bar{B}^{n}(x,r)$ of $\mathbb{R}^{n}$. Show that
$m(U) \leq \mu(U)$
for every open set  $U \subset \mathbb{R}^{n}$.

I have no idea where to even start with this.
 A: In one dimension:
For an open ball $B(x,r)$, we have $B(x,r)=\displaystyle\bigcup_{n=1}^{\infty}\overline{B}(x,r-1/n)$, and $\{\overline{B}(x,r-1/n)\}_{n=1}^{\infty}$ is an increasing sequence of closed balls, so 
\begin{align*}
\mu(B(x,r))&=\lim_{n\rightarrow\infty}\mu(\overline{B}(x,r-1/n))\\
&\geq\lim_{n\rightarrow\infty}m(\overline{B}(x,r-1/n))\\
&=m(B(x,r)).
\end{align*}
Now express each open set $U$ as the countable union of a class of disjoint open balls $\{B(x_{N},r_{N}): N=1,2,...\}$, then
\begin{align*}
m(U)=\displaystyle\sum_{N=1}^{\infty}m(B(x_{N},r_{N}))\leq\sum_{N=1}^{\infty}\mu(B(x_{N},r_{N}))=\mu(U).
\end{align*}
In higher dimension:
Use the following fact which is adopted from Geometric Integration Theory, Steven G. Krantz/Harold R. Parks.

Let $A\subseteq{\bf{R}}^{N}$ and let $\mathcal{B}$ be a family of open balls. Suppose that each point of $A$ is contained in arbitrarily small balls belonging to $\mathcal{B}$. Then there exist pairwise disjoint balls $B_{j}\in\mathcal{B}$ such that 
  \begin{align*}
\mathcal{L}^{N}\left(A-\bigcup_{j}B_{j}\right)=0.
\end{align*} 
  Furthermore, for any $\epsilon>0$, we may choose the balls $B_{j}$ in such a way that
  \begin{align*}
\sum_{j}\mathcal{L}^{N}(B_{j})\leq\mathcal{L}^{N}(A)+\epsilon.
\end{align*}

Back to the question, we now let $\mathcal{B}$ be the set of all open balls contained in $U$. Choose such $B_{j}$ as in the above fact, then
\begin{align*}
m(U)&=m\left(\bigcup_{j}B_{j}\right)+m\left(U-\bigcup_{j}B_{j}\right)\\
&=m\left(\bigcup_{j}B_{j}\right)\\
&=\sum_{j}m(B_{j})\\
&\leq\sum_{j}\mu(B_{j})\\
&=\mu\left(\bigcup_{j}B_{j}\right)\\
&\leq\mu(U).
\end{align*}
