Given $(\mathbb R^n,\mathcal B,\mu)$, where $\mu$ is a positive finite Radon measure, define the function $$ \mu^*(A) = \inf \left\{ \sum_{i=1}^n\mu(B_i) \,\,|\,\, A\subseteq \cup_i B_i, \,\, n \in \mathbb N \right\} $$ where $B_i$ are balls (both closed and open).
Take now $G:\mathbb R^n \to \mathbb R_+$ a continuous function with compact support $E$ and call $$ E_r = \{x\in \mathbb R^n | G(x)>r \}. $$ Given any $\varepsilon>0$, is it true that $$ \mu^*(E_\epsilon) \le \mu(E_0)? $$
Partial answer.
Here's a proof for $n=1$. We actually show that $E_\epsilon$ is the union of finitely many pairwise disjoint balls. Since $\{x : G(x) > 0\}$ is open, we may write it as $\{G > 0\} = \sqcup_{n=1}^\infty (a_n,b_n)$, a countable union of disjoint intervals (the proof is just to take maximal intervals in $\{G > 0\}$). Now, since $G$ is uniformly continuous (it has compact support), there is some $\delta > 0$ with $|x-y| < \delta \implies |G(x)-G(y)| < \epsilon$. Since $\{G > 0\}$ is bounded, $\infty > m(\{G > 0\}) = \sum_n m((a_n,b_n)) = \sum_n b_n-a_n$, where $m$ is the Lebesgue measure. Therefore, for all except $N < \infty$ intervals, $b_n-a_n < \delta$. If $b_n-a_n < \delta$, then since $G(a_n) = 0$, $G(x) < \epsilon$ for $x \in (a_n,b_n)$. We conclude that $E_\epsilon$ is the union of finitely many pairwise disjoint balls.
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Here's a proof for $n=2,3$ for any measure $\mu$ that is absolutely continuous with respect to the Lebesgue measure. Throughout, $n$ refers to $2$ or $3$ and $|\cdot|$ refers to $\mu$.
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Lemma: let $Q$ be a closed square in $\mathbb{R}^n$. Then, for any $\epsilon > 0$, there is a collection of open balls $\{B_j\}_j$ such that $\sum_{j=1}^\infty |B_j| \le |Q|+\epsilon$.
Proof: Start with $\{B_j'\}$ an Apollonian gasket in $\mathbb{R}^n$. (This is why we need $n \le 3$, I think). The set $E_0 := Q\setminus \cup_j B_j'$ has measure $0$, so let $U$ be an open set containing $E_0$ with $|U| \le \epsilon$. By Vitali's covering Lemma, there is a collection $\{B_j''\}_j$ covering $U$ with $\sum_j |B_j''| \le C_n|U| \le C_n\epsilon$. So, $\{B_j\}_j := \{B_j'\}_j \cup \{B_j''\}_j$ covers $E_0$ and has $\sum_j |B_j| \le |Q|+C_n\epsilon$.
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Main Proof: For $r > 0$, we show $\mu^*(E_r) \le \mu(E_r)$. Take $\epsilon > 0$. Take $0 < r'' < r$ so that $|E_{r''}| < |E_r|+\epsilon$. Take any $r' \in (r'',r)$. Since $E_{r''}$ is open, there exist a collection of closed cubes $\{Q_j\}_j$ that are disjoint except for overlapping boundaries and $E_{r''} = \cup_j Q_j$. By compactness (and looking at the interior of the $Q_j$'s), there is a finite subcollection $Q_1,\dots,Q_N$ whose union covers $E_{r'}$. By the Lemma, for each $1 \le i \le N$, there is $\{B_j^{(i)}\}_j$ with $Q_i \subseteq \cup_j B_j^{(i)}$ and $\sum_{j=1}^\infty |B_j^{(i)}| \le |Q_i|+\frac{\epsilon}{N}$. Let $\{B_j\}_j$ denote $\{B_j^{(i)}\}_{i,j}$. Then, by compactness, there are $B_1,\dots,B_M$ covering $\overline{E_r}$. We therefore have $\mu^*(E_r) \le \mu^*(\overline{E_r}) \le \sum_{j=1}^M |B_j| \le \sum_{i=1}^N \sum_j |B_j^{(i)}| \le \sum_{i=1}^N |Q_i|+\epsilon \le |E_{r''}|+\epsilon \le |E_r|+2\epsilon$.
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Remark: absolute continuity of $\mu$ was used in not caring about the overlapping boundaries.
Yes, it is true, and to prove it I will use the following proposition.
Proposition. Let $\mu$ be any positive outer-regular Borel measure on $\mathbb R^n$, and let $K$ be any compact subset of $\mathbb R^n$. Then $\mu(K)= \inf \sum_{i=1}^k \mu(B_i)$, where the infimum is taken over all finite open covers $\lbrace B_1,\ldots,B_k\rbrace$ of $K$ by open balls $B_i$.
Before presenting the (somewhat lengthy) proof of the proposition, I will first show how it resolves the question. Consider the set $K=G^{-1}\bigl([\epsilon,\infty)\bigr)$, which is closed (by continuity of $G$) and compact (since it is a subset of the support of $G$, which was assumed to be compact). Since Radon measures are outer-regular and Borel, the proposition applies, yielding that $\mu^\star(E_\epsilon)\leq \mu(K)$. Since $K\subseteq E_0$, the bound $\mu^\star(E_\epsilon)\leq \mu(E_0)$ follows.
Proof of the proposition. Obviously $\mu(K)\leq \mu(B_1\cup\cdots\cup B_k)\leq \sum_{i=1}^k \mu(B_i)$ by subbaditivity, so $\mu(K)\leq \inf\sum_{i=1}^k \mu(B_i)$.
To establish the other inequality, we apply the following result:
Besicovitch Covering Theorem. Let $A$ be any subset of $\mathbb R^d$ and let $r\colon A\to\mathbb R_+$ be any function. Then there exists an integer $m\leq 5^d$ and disjoint subsets $A_1,\ldots,A_m$ of $A$ with the following properties:
For all $1\leq i\leq m$, the collection of open balls centered at $x\in A_i$ with radius $f(x)$ is mutually disjoint.
The union of all balls appearing in (1) contains $A$.
Fix $\epsilon>0$. If $\mu(K)=\infty$ there is nothing to prove, so suppose that $\mu(K)<\infty$. We will iteratively construct a cover of $K$ using open balls $B_i$ that are nearly disjoint. To start the iteration, set $K_0=K$. By outer regularity of $\mu$, there exists an open set $U_0$ containing $K_0$ such that $\mu(U_0)\leq(1+\epsilon)\mu(K_0)$. Cover each element of $K$ with a centered open ball contained in $U$. Apply the Besicovitch Covering Theorem to this cover. A disjoint subfamily with maximal $\mu$-measure will have measure at least $5^{-d}\mu(U_0)$. Since $\mu(K)<\infty$, there exists a finite disjoint subfamily - call it $C_0$ - with $\mu(C_0)\geq 6^{-d}\mu(U_0)$.
Given $(K_i,U_i,C_i)_{i=0}^n$ we continue the iteration by setting $K_{n+1}=K_n\setminus C_n$ and applying outer regularity to $K_{n+1}$ to find an open set $U_{n+1}$ containing $K_{n+1}$ with $\mu(U_{n+1}) \leq (1+\epsilon)\mu(K_{n+1})$. In fact, since $K_{n+1}\subseteq U_n$ we can further choose $U_{n+1}$ to be a subset of $U_n$ (by intersecting with $U_n$ if necessary). Cover $K_{n+1}$ with open balls in $U_{n+1}$ and apply Besicovitch to obtain a disjoint finite subcover $C_{n+1}$ with $\mu(C_{n+1})\geq 6^{-d}\mu(U_{n+1})$. These inequalities imply that $$ \mu(K_{n+1})\leq \mu(U_n)-\mu(C_n)\leq (1-6^{-d})\mu(U_n)\leq (1-7^{-d})\mu(K_n), $$ provided that $(1+\epsilon)(1-6^{-d})\leq (1-7^{-d})$ - which we can ensure by taking $\epsilon$ sufficiently small. Thus, we find that $\mu(K_n)$ decreases exponentially in $n$, and in fact $$ \sum_{n=0}^{\infty}\mu(K_n)\leq 7^d\mu(K). $$
Now it is time to bound the overlap between the $C_i$. Observe that $$ C_n\cap (C_0\cup\ldots\cup C_{n-1})\subseteq U_n\setminus K_n, $$ from which it follows by induction that $$ \mu(C_0)+\cdots+\mu(C_n)\leq\mu(C_0\cup \cdots\cup C_n)+\mu(U_1\setminus K_1)+\cdots+\mu(U_n\setminus K_n), $$ and we can bound the right side by $\mu(U_0)+\epsilon\sum_n \mu(K_n)$ (using the monotonicity $U_i\subseteq U_0$ for all $i$) to obtain an upper bound of $$ \mu(C_0)+\cdots+\mu(C_n)\leq (1+\epsilon+\epsilon 7^d)\mu(K).\qquad (\star) $$
We stop the iteration when $n$ becomes large enough so that $(1-7^{-d})^n\leq \epsilon$, in which case $\mu(K_n)\leq \epsilon\mu(K)$ and $\mu(U_n)\leq \epsilon(1+\epsilon)\mu(K)$. Rather than taking $C_{n+1}$ to be a disjoint subfamily as before, we allow overlaps in this final step. First, apply Besicovitch to the balls in $U_{n+1}$ covering $K_{n+1}$ to obtain $m\leq 5^d$ subfamilies whose union still covers $K_{n+1}$. Then apply compactness of $K_{n+1}$ to extract a finite subcover $C_{n+1}'$. Observe that even though this finite subcover is not disjoint, thanks to Besicovitch we still have a weak bound of $$ \sum_{B\in C_{n+1}'}\mu(B)\leq 5^d\mu(U_{n+1}).\qquad (\star\star) $$
Finally, we form the cover $C_0\cup\cdots\cup C_n\cup C_{n+1}'$ of $K$, which we now reindex in terms of the underlying open balls as $B_1\cup\cdots\cup B_k$. Combining $(\star)$ and $(\star\star)$ then yields $$ \sum_{i=1}^k\mu(B_i)\leq (1+\epsilon+\epsilon 7^d+5^d\epsilon(1+\epsilon))\mu(K). $$ Taking $\epsilon$ to $0$ yields the inequality $$ \mu(K)\geq \inf\sum_{i=1}^k \mu(B_i), $$ completing the proof of the proposition. $\square$
