Does the set of accumelation points of a positive measure set has a positive measure?

Question

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)

Answer 1

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 $0$.

Answer 2

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 $\mu(A')=\mu(\bar{A})\ge \mu(A)>0$.