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)?

2$\begingroup$ What if we take the Dirac measure at a point x ? Then each subset A containing x is an atom. $\endgroup$ – 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 nonempty 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.

$\begingroup$ 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) $\endgroup$ – Emilio Pisanty Apr 4 '12 at 13:18

2$\begingroup$ Emilio: The answer is no, as pointed out by Ralph in his comment. But Fedor has shown that any atom contains a singleton with the same measure (which is probably what the OP wanted). You can't usually expect the singletons to be the only atoms since you can always add a null set to an atom to get another atom. $\endgroup$ – Ramiro de la Vega Apr 4 '12 at 14:17

1$\begingroup$ It depends on what you understand by equality of measurable sets. If equality means that symmetric difference has zero measure (what is most appropriate and usual sense on my opinion), then yes. If you mean settheoretic equality, then of course no, as is pointed out by Ralph. $\endgroup$ – Fedor Petrov Apr 4 '12 at 23:02

4$\begingroup$ So, like real word atoms, a measuretheoretic atom of a compact space is a nucleus surrounded by a noweight cloud :) $\endgroup$ – Pietro Majer Apr 4 '12 at 23:18