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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 $Z:\Omega\rightarrow \mathbb{R}$.

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I think you need only look at the constant function $Z:\Omega\rightarrow\mathbb{R},\omega\mapsto 1$, then the probabilty measures are those measures that are mapped to $1$ under $l_Z$. – Olivier Bégassat Jun 13 '12 at 14:46
The same question, with very much the same answer, has appeared over at MathOverflow:… – Matthew Daws Jun 13 '12 at 19:45
up vote 5 down vote accepted

No. Take $\Omega=\mathbb N$ and $\Sigma$ the power set. As wikipedia says, then $ba(\Sigma)=ba=(\ell^\infty)^*$. However, the collection of probability measures is just the collection of $(x_n)\in\ell^1$ (as a measures are countably additive) with $x_n\geq 0$ for all $n$, and $\sum_n x_n=1$. This is not weak$^*$-closed in $(\ell^\infty)^*$. For example, any limit point of the set $\{\delta_n:n\in\mathbb N\}$, where $\delta_n\in\ell^1$ is the point mass at $n$, is a member of $ba \setminus \ell^1$.

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This of course assumes I interpreted "probability measure" correctly. If you just meant the finitely additive measures which are positive and with total mass 1, then Olivier's comment says that the answer is "yes". – Matthew Daws Jun 13 '12 at 14:52
Yes, by probability measure I mean countably additive measures and not only finitely additive measures. – Andy Teich Jun 13 '12 at 15:44

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