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Let $A$ be a compact set of positive measure. Let $\bar A$ the set of accumulation points of A. Can $\bar A$ be of zero measure?

(accumulation point=limit point=cluster point)

M.A
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  • If $\mu(\bar A)=0$ then let $A_2=A-\bar A$. Then $\mu(A_2)=\mu(A)>0$. Hence, $A_2$ has positive measure without any accumulation point which can't happen?? – M.A Nov 08 '18 at 20:50
  • But I see your point. let me think. but A is closed. – M.A Nov 08 '18 at 20:54
  • Doesn't A $\subseteq \bar{A}$? – Joel Pereira Nov 08 '18 at 21:02
  • Sorry I deleted my comments, because I was confusing myself. The point I wanted to make is that $A_2$ is the set of isolated points of $A$, which can have limit points. Consider the sequence $\frac{1}{n}$. Every point of this sequence is an isolated point, but it has a limit point. – jgon Nov 08 '18 at 21:02
  • @JoelPereira The OP is using nonstandard notation, $\bar{A}$ is the set of limit points of $A$, rather than the closure of $A$. – jgon Nov 08 '18 at 21:03
  • @JoelPereira No. For instance A can have isolated points – M.A Nov 08 '18 at 21:05

2 Answers2

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Don't use $\bar{A}$ for the set of limit points, that notation is almost universally used to mean the closure of $A$. Instead, one better option is to use $A'$.

Now note that if $A$ is closed, $A$ contains all of its limit points, so $A$ is a disjoint union of the isolated points of $A$ and $A'$. Thus it suffices to show that the isolated points of $A$ have measure 0. To see this, note that that $A$ can only have countably many isolated points. This has a beautiful proof given in the answer here. Thus $\mu(A\setminus A')=0$, so $\mu(A')=\mu(A)>0$.

The result still follows if $A$ is not closed, since then $\mu(\bar{A})\ge \mu(A)$, and we have by the closed case that $\mu(A')=\mu(\bar{A})\ge \mu(A)>0$.

jgon
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Let $S$ be the set of isolated points of $A$. Then the accumulation points are precisely $A\setminus S$. For each $x\in S$, there exists $r_x>0$ such that $B(r_x,x)\cap S=\{x\}$. Then the open balls $B(\frac12r_x,x)$ are pairwise disjoint and contain a rational point. We conclude that $S$ is countable, hence of measure $0$.

Hagen von Eitzen
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