Lebesgue outer measure with open balls

Question

I'm not sure if this has been asked before; if so please redirect me to the appropriate question.

The Lebesgue outer measure of $A \subseteq \mathbb{R}^n$ is defined as $$\mu_*(A) = \inf\left\{ \sum_{i} |R_{i}|: A \subseteq \bigcup_{i \in I} |R_i|, I \,\,\mbox{countable} \right\}$$ where the infimum is taken over open boxes $R_i$.

Now suppose we define a new measure in this exact same manner, except we take the infimum over open balls, defining their volume using the usual formula for the volume of an $n$-sphere. Why is this equivalent to the above definition?

It seems somewhat related to the Vitali covering lemma.

Answer 1

What we need from the Vitali Covering Lemma:

Given a box $R$, there exist disjoint balls $B_1, B_2, \dots$ and a Lebesgue-null set $N$ so that $$ \bigcup_{n=1}^\infty B_n \subseteq R \subseteq N \cup \bigcup_{n=1}^\infty B_n . $$

This will let you prove one inequality between the two outer measures. For the other inequality, you will need some similar inclusions, where you approximate a ball by disjoint cubes, but that one is easier.