upper bound of outer measure on compact support of continuous function

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 $$\mu^*(E_\epsilon) \le \mu(E_0)?$$

• Isn't $E_r$ an open set and thus a countable union of balls, implying $\mu^*(E_r) = \mu(E_r)$, which is at most $\mu(E_0)$ for any $r > 0$? – mathworker21 Sep 11 at 9:27
• @mathworker21 in the definition of $\mu^*$ the sum is over a finite number of balls that cover $E_r$ – Exodd Sep 11 at 9:31
• Yes, but you have continuity. That is, if $E_r = \cup_{n=1}^\infty B_n$, then $\sum_{i=1}^n \mu(B_i) = \mu(\cup_{i=1}^n B_i) \to \mu(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty \mu(B_i)$. I should have said in my first comment "countable union of pairwise disjoint closed balls". – mathworker21 Sep 11 at 9:33
• @mathworker21 first of all, is it true that an open set is the union of countable disjoint closed ball? But even if it is true, $\mu^*$ is not countable addictive or subaddictive.. – Exodd Sep 11 at 9:38
• Sorry, you're right, once again. Thanks a lot! – mathworker21 Sep 11 at 9:51

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.
For $$n \ge 2$$, I'm pretty sure the result is false. I don't think one can cover, for example, a square or an annulus by balls with areas summing arbitrarily close to the area of the given square or annulus.
• wait. It is false that open $\implies$ union of disjoint closed balls.. – Exodd Sep 14 at 7:37