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

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  • $\begingroup$ 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 $r > 0$? $\endgroup$ – mathworker21 Sep 11 at 9:27
  • $\begingroup$ @mathworker21 in the definition of $\mu^*$ the sum is over a finite number of balls that cover $E_r$ $\endgroup$ – Exodd Sep 11 at 9:31
  • $\begingroup$ 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". $\endgroup$ – mathworker21 Sep 11 at 9:33
  • $\begingroup$ @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.. $\endgroup$ – Exodd Sep 11 at 9:38
  • $\begingroup$ Sorry, you're right, once again. Thanks a lot! $\endgroup$ – mathworker21 Sep 11 at 9:51
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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.

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  • $\begingroup$ I just realized that our argument in chat actually works for the original problem with a little modification.. $\endgroup$ – Exodd Sep 12 at 17:53
  • $\begingroup$ @Exodd I thought this was our argument in chat $\endgroup$ – mathworker21 Sep 13 at 0:07
  • $\begingroup$ I'm saying I can make it work also for all finite measures $\endgroup$ – Exodd Sep 13 at 7:16
  • $\begingroup$ @Exodd post it as an answer $\endgroup$ – mathworker21 Sep 14 at 1:23
  • $\begingroup$ wait. It is false that open $\implies$ union of disjoint closed balls.. $\endgroup$ – Exodd Sep 14 at 7:37

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