MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer
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
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. – Ramiro de la Vega Apr 4 '12 at 14:17
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 set-theoretic equality, then of course no, as is pointed out by Ralph. – Fedor Petrov Apr 4 '12 at 23:02
So, like real word atoms, a measure-theoretic atom of a compact space is a nucleus surrounded by a no-weight cloud :) – Pietro Majer Apr 4 '12 at 23:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.