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If $\mu$ is a Radon measure on $\mathbb{R}^n$ and $B_r$ is a closed ball of radius $r$. Why is $\mu(\partial B_r) = 0$? Or how can I prove that there is at least one $r_0 > 0$ such that $\mu(\partial B_{r_0}) = 0$?

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    $\begingroup$ The first fact is false: just take a Dirac measure and $r=0$. $\endgroup$ Apr 13, 2014 at 12:44
  • $\begingroup$ It is true for all but countably many r. $\endgroup$
    – user55449
    Apr 13, 2014 at 16:57

2 Answers 2

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The following seems to work, unless I overlooked something:

Assume $\mu$ is a real measure and $\mu\ge 0$, otherwise you need to write it as the difference of two positive measures (Jordan decomposition) for the following to work: a radon measure is locally finite and inner regular. With the assumption about the sign, $r\mapsto \mu(\overline{B_r})$ is increasing, hence has at most countably many points of discontinuity. Assume it is continuous in $r_0$. Then $$0 \le \mu(\partial B_{r_0}) = \mu(\overline B_{r_0})-\mu(B_{r_0})=\mu(\overline B_{r_0})-\lim_{r\rightarrow r_0}\mu(\overline{B_{r}}) =0 $$

where the limit is taken with the assumption $r<r_0$ (here inner regularity is used).

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Assume that $\mu(\partial B_r)\neq0$ for all $r\geq 0$. From this we will find for every $n > 0$ a ball $U_{\frac1{n}}(x_n)$ with infinite measure, sequence $x_n$ will be bounded therefore you can pick converging sub-sequence and at its limit point the measure can't be locally finite.

So how to find ball $U_\epsilon$ with infinite measure?

Fix $\epsilon > 0, r>0, \delta>0$. Now you can cover $B_{r+\delta} \setminus B_{r-\delta}$ by finitely many balls $U_\epsilon^k$ of radius $\epsilon$. $$ B_{r+\delta} \setminus B_{r-\delta} \subset \bigcup_{k=1}^n U_\epsilon^k $$ Since $\mu(\partial B_r)\neq0$ for all $r>0$ than for every $s \in [r-\delta,r+\delta]$ there is $k_s$ that $\mu( \partial B_s \cap U_\epsilon^{k_s}) \neq 0$.

Because $k$ can range only from $1$ to $n$ than there is uncountable set $I\subset [r-\delta,r+\delta]$ that for every $u,v\in I$ is $k_u=k_v = \overline{k}$ and $\mu( \partial B_u \cap U_\epsilon^{\overline{k}}) \neq 0$.

From this we can conclude $$ \mu( U^\overline{k}_\epsilon ) \geq \mu\left( \bigcup_{u\in I} \partial B_u \cap U_\epsilon^{\overline{k}} \right) = \sum_{u\in I} \mu \left( \partial B_u \cap U_\epsilon^{\overline{k}} \right) \geq \infty $$ because sum of uncountable many positive numbers is infinite.

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