Exterior measure using open balls

Question

I'm trying to show that given $A\subseteq\mathbb{R^n}$ the exterior measure of Lebesgue, $\mu^*$, can also be defined as $$\mu^{*,B}(A) = \inf{\sum_{j}\mu(B_J)},$$ where the infimum is taken over the countable collections of open balls such that $$\bigcup_jB_j\supseteq A$$ and $\mu$ is the Lebesgue's measure of $B_j$ (or just the ball's volume). It's easy to show that $\mu^{*,B}(A)\geq\mu^*(A)$, my problem is the $\leq$ inequality. I've read somewhere that it's an application of Vitali's covering lemma, however I'm not able no figure out a way to do that.

Can someone help me? Thank you!

Answer 1

Currently I'm also working on this problem. Since you've already gotten the proof of $m_{*} (E) \leq m_{*}^B (E)$, I'll only provide the proof of the reverse inequality. My idea proceeds as follows.

1. First we consider $E$ as a L-measurable set. Thus, for any given $\epsilon >0$, there exists an open set $O\supset E$ s.t. $m(O-E) <\epsilon /2$. Let $\frac{m(B_0)}{m(C_0)}=C$, where $B_0$ is the unit ball in $\mathbb{R}^d$ and $C_0$ is its built in cube (we will use this constant later). We consider the following family of open balls: $\mathcal{B}=\bigcup_{x\in E}\bigcup_{n>N(x)} B_{1/n} (x)$, where $N(x)$ is selected to be big enough such that $B_{1/N(x)} (x)\subset O$. Clearly, $\mathcal{B}$ is a Vitali covering of $E$. According to Vitali covering theorem (Corollary 3.10, Ch3, Real Analysis by Stein), there exists $B_1 , B_2 ,\cdots, B_N \in \mathcal{B}$ which are disjoint and satisfy $m(E-\bigcup_{i=1}^{N}B_i )<\epsilon/4C$. Then, consider $E-\bigcup_{i=1}^{N}B_i$, since it's measurable, we can find a collection of cubes $\{C_i \}_{i=N+1}^{\infty}$ covering $E-\bigcup_{i=1}^{N}B_i$ such that $\sum_{i=N+1}^{\infty} m(C_i) <m(E-\bigcup_{i=1}^{N}B_i) + \epsilon/4C$. Thus, if we select $B_i$ as the circumscribed ball of $C_i$ ($i>N$), we have $\bigcup_{i=1}^{\infty} B_i \supset E$ and $$ \begin{aligned} \sum_{i=1}^{\infty} m(B_i)&=\sum_{i=1}^{N} m(B_i)+\sum_{i=N+1}^{\infty} m(B_i) \\ &\leq m(O)+C\sum_{i=N+1}^{\infty} m(C_i) \\ &< m(E)+\epsilon /2 + C(m(E-\bigcup_{i=1}^{N}B_i) + \epsilon/4C)\\ &< m(E)+\epsilon \end{aligned} $$ Thus we conclude that for any $\epsilon>0$, we can find a covering of $E$ by balls, i.e.$E\subset \bigcup_{i=1}^{\infty} m(B_i)$ such that $\sum_{i=1}^{\infty} m(B_i)<m(E)+\epsilon$. Since $E$ is measurable, we have $m_{*}^{B}(E)\leq m_{*}(E)$ (otherwise, there would be a contradiction);
2. When the situation is not limited to a mesurable set $E$, our method is rather similar. According to the definition of exterior measure, for any given $\epsilon >0$, there exists a covering of $E$ by cubes $E\subset \bigcup_i C_i$ such that $\sum_{i=1}^{\infty} m(C_i) <m_* (E) + \epsilon /2$. Since every cube is measurable, according to the analysis in (1), for every $i\geq 1$, there exists a covering of $C_i$ by balls, namely $C_i \subset \bigcup_{j=1}^{\infty} B^{(i)}_j$, such that $\sum_{j=1}^{\infty} m(B^{(i)}_j) < m(C_i) +\epsilon /2^{i+1}$. Then we have $E\subset \bigcup_{i,j} B^{(i)}_j$ and $$ \sum_{i,j} m(B^{(i)}_j) <\sum_{i=1}^{\infty} m(C_i) + \sum_{i=1}^{\infty}\epsilon /2^{i+1} <m_* (E) +\epsilon $$ Hence we conclude the proof.