This question has been asked before but the solutions use a version of Vitali Covering Lemma which is not clear to me how it relates to the statement I am familiar with (below). Any help is appreciated.
The problem:
For every Lebesgue measurable set $E$ we have that $$m(E)= \inf\left\{\sum_{i=1}^\infty m(B_i): B_i\mbox{ are balls, } E\subset \cup_{i=1}^\infty B_i\right\}.$$
For the proof, we use Vitali Covering Thm:
Let $E\subset \mathbb R^n$ be Lebesgue measurable with $m(E)<\infty$. Let $\mathcal{C}$ be a covering of $E$ such that for any $x\in E$ and any $\delta<0$ there exists a ball centered at $x$ and of diameter $\delta$ in $\mathcal C$. Then, for any $\varepsilon>0$, there is a finite disjoint sub-collection $B_1,\dots, B_N$ of $\mathcal C$ such that $m(E\setminus\bigcup_{i=1}^N B_i)<\varepsilon$.
Fix $E\subset \mathbb{R}^n$. If $m(E)<\infty$, we apply Vitali Covering Thm: Let $\mathcal{C}$ be a collection of balls centered at any $x\in E$ and of any positive diameter $\delta>0$. Then, take a sequence $\varepsilon_k>0$ which approaches $0$ as $k\to \infty$. Then, by Vitali covering theorem, for each $\varepsilon_k$ there exists a finite disjoint subcollection $B^k_1,\dots,B^k_{N_k}$ of $\mathcal C$ such that $$m(E\setminus\bigcup_{i=1}^{N_k}B_i^k)<\varepsilon_k.$$ Call $F_k:= E\setminus\bigcup_{i=1}^{N_k}B_i^k.$ Since $E\subset F_k\cup \bigcup_{i=1}^{N_k}B_i^k,$ by monotonicity and subadditivity of measures, we get that
$$m(E)\leq m(F_k)+ m(\bigcup_{i=1}^{N_k}B_i^k).$$
Then, as $k\to\infty$, we get that $m(E)\leq m(\cap_{k=1}^\infty\bigcup_{i=1}^{N_k}B_i^k)$ and $m(F_k)\to0$.
From here:
- I don't know how to make an open cover of $E$ using balls $B_i^k$ and their union and intersection; as in the problem.
- I also don't know how to deal with $m(E)= \infty$.
- Lastly, I don't know how to do the other direction. Perhaps a regularity result about Lebesgue measure would help (Theorem 2.40 from Folland) but I don't know how.
There are certainly collections of balls with $E$ as a subset of their union. $B_n = \{x: \|x - 0\| < n\}$, for example. So the set is not empty. Further, since $A \subseteq C \implies m(A) \le m(C)$, we know that $m(E) \le m\left(\bigcup_n B_n\right) \le \sum_n m(B_n)$ for any such collection $\{B_n\}_n$. So the set is bounded below by $m(E)$. Thus the infimum exists and is $\ge m(E)$.
Well, that is easy enough. $m(E)$ is a lower bound of values in the set, and $\infty$ is an upper bound of any set of non-negative extended-real numbers. So if the two are the same...
The purpose of the Vitali Covering Lemma proof is to show that there are collections of balls whose total measure exceeds that of $E$ by arbitrarily small amounts, meaning the infimum cannot be larger than $m(E)$. However, there are parts missing from the argument as you've reproduced it. "$m(E)\leq m(\cap_{k=1}^\infty\bigcup_{i=1}^{N_k}B_i^k)$", heads off in a different direction than you need. Intersections of unions of balls are not unions of balls.
Personally, I think I would try to prove it directly from the definition of outer measure. But that depends on what definition you are using, so I cannot say how to do it in your case. I would note that $n$-cubes are contained in balls the ratio of whose measure to that of the $n$-cube is some fixed factor $M$. For each $\epsilon > 0$, cover $E$ with such $n$-cubes whose total measure is $< m(E) + \epsilon / M$, then replace each of those cubes with its circumferencing ball to get a collection of balls covering $E$ whose measure $< m(E) + \epsilon$.