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 $U \subset \mathbb{R}^{n}$.

Question

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.

Answer 1

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*}