# Atoms of regular Borel measure

Let $X$ be locally compact Hausdorff space. Let $\mu$ be a Borel measure on it which is finite on compact and outer regular with respect to open sets and inner regular with respect to compact sets. Does an atom of such measure have to be a singleton (up to set of zero measure)?

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What if we take the Dirac measure at a point x ? Then each subset A containing x is an atom. –  Ralph Apr 4 '12 at 11:11

Without loss of generality your atom $A$ is compact (by inner regularity). Call a point $a\in A$ negligible if it has a neighborhood with zero measure inside $A$. Clearly the set $B$ of negligible points is open in $A$. If it is $A$ itself, then choose finite subcovering by those neighborhoods to get that measure of $A$ is zero. So, $A\setminus B=C$ is a non-empty compact set in $A$. It has either measure 0 or measure equal to $|A|$ (let $|\cdot|$ denote measure.) If $|C|=0$, then choose a neighborhood $V$ of $C$ with measure at most $|A|/2$ by outer regularity. $V\cap A$ has measure strictly less then $|A|$, hence 0, hence each point of $C$ is negligible. A contradiction. So $|C|=|A|$. Then replace $A$ for $C$ and we get an atom $C$ in each no point is negligible. It may contain only one point, else take two points $u$, $v$ and their disjoint neighborhoods. Both must have positive measure in $C$, a contradiction.
So the answer is affirmative? i.e. all atoms $A$, with respect to Borel measures $\mu$ on locally compact Hausdorff spaces $X$, must be singletons? (sorry, the proof is nice but it takes some unentangling) –  Emilio Pisanty Apr 4 '12 at 13:18