upper bound of outer measure on compact support of continuous function

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.