MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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 $(\Omega,\Sigma)$ be a measurable space. Denote by $ba(\Sigma)$ the set of all bounded and finitely additive measures on $(\Omega,\Sigma)$ (see for a definition). Is the set of all probability measures $\mathcal{M}_1(\Sigma)\subseteq ba(\Sigma)$ weak*-closed? The weak*-topology on $ba(\Sigma)$ is the weakest topology such that the maps $l_Z:ba(\Sigma)\rightarrow \mathbb{R}$, mapping $\mu\mapsto \int_\Omega Z d\mu$, are continuous for all bounded and measurable maps $Z:\Omega\rightarrow \mathbb{R}$.

share|cite|improve this question
You really need to put these questions on math.stackexchange. – Nik Weaver Jun 13 '12 at 14:26
up vote 1 down vote accepted

No. The most elementary case is $\Omega=\mathbb N$ and $\Sigma$ the power set. A non-fixed ultrafilter is a good example of a finitely-additive but not countably-additive measure: each set has either measure $0$ or $1$. But such a thing is a limit of fixed ultrafilters, that is, countably additive zero-one measures. Indeed, it is the limit along that very ultrafilter.

Nik seems to think this is too elementary. But there is no reason this should be clear to a non-specialist.

share|cite|improve this answer
It's like asking whether the unit ball of $l^1$ is weak* closed in its double dual. – Nik Weaver Jun 13 '12 at 15:02
@Nik: since it is weak* sequentially closed, even that is an interesting question. – Gerald Edgar Jun 13 '12 at 15:46
Nik: yes, it is exactly like that in the sense that neither is clear to a non-specialist. – Vidit Nanda Jun 13 '12 at 15:47
Annoyingly, this is exactly the same as an answer I gave over at math.stackexchange:… I had already suggested to Peter (who may or may not also be Andy) that it was polite to point out when you are cross-posting... – Matthew Daws Jun 13 '12 at 19:25
@Peter: your comment is inappropriate. This is a professional forum and we don't talk that way here. – Nik Weaver Jun 14 '12 at 13:22

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.