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Given a Polish space $X$, I note $C_b(X)$ the set of the continuous bounded functions with the norm of the uniform convergence, and $(C_b(X))^\star$ its topological dual with the $*-$weak convergence $\sigma((C_b(X))^\star, C_b(X))$.

To a Borel probability $P$ on $X$ we can associate a $l(P) \in (C_b(X))^\star$ by $$E_P[\phi]=\langle l(P) , \phi \rangle$$for every $\phi \in C_b(X)$, where $\langle,\rangle$ denotes the duality bracket.

I further consider a sequence $(P_n)_{n\in \mathbb{N}}$ of Borel probabilities on $X$ which is such that $(l(P_n))$ converges in the $*-$weak topology of $(C_b(X))^\star$ to a $m \in (C_b(X))^\star$. Obviously $$\langle m,G_1 \rangle=1$$ (where $G_1$ is the function $G_1(x)=1$ for any $x\in X$), and for any positive $\phi$ $$\langle m,\phi \rangle\geq 0$$ (the positive cone is closed).

However, is it true that one can find a unique probability $P^\star$ on $X$ such that $$m=l(P^\star) \, ?$$ The problem could come from the sigma additivity.

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up vote 1 down vote accepted

I believe that the answer is affirmative. Weak$^{\ast}$ topology on probability measures can be metrized (e.g. via Lévy–Prokhorov metric). When we deal with a Polish space $X$, then this metric is complete. Using this, I would like to conclude that the set of probability measures is sequentially closed; indeed, weak$^{\ast}$ convergent sequence of probability measures satisfies Cauchy condition, so it has to converge to some probability measure. Nevertheless, this space is definitely not weak$^{\ast}$ closed, as the example of $(\delta_{n}) \subset \ell_{\infty}^{\ast}$ shows.

I hope that it is correct.

EDIT The above argument is completely wrong: of course completely metrizable subset isn't necessarily closed, e.g. $(0,1) \subset \mathbb{R}$. Nevertheless, I found a paper by Dimitris Gatzouras, On weak convergence of probability measures in metric spaces, in which he claims that the set of separably supported Borel probability measures is sequentially closed. I haven't read it, so I can't tell if his argument is correct.

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This is a comment, not an answer, but the software won't accept it as such from me. Some of the confusion which is apparent here arises from the fact that $C^b(X)$ has two natural structures and so two duals. As a Banach space, its dual is the space of Radon measures on the Stone-Ćech compactification or, slternatively, the space of bounded, FINITELY additive Borel measures. However, as noted above, it is clear from the context that the query is about the space of bounded TIGHT (equivalently, in this situation, i.e., that of a polish space, $\sigma$-additive measures). This, as noted above is the dual with respect to the strict topology on $C^b(X)$ (Buck, Fremlin, Garling, Haydon and many others). Then the result quoted above applies---we have again a polish space under the weak $\ast$ topology.

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Is it a comment on the question or on one of the answers? If you tell me, I can move it (although it looks a little long for it to appear as a single comment; I can break it in two but my name would be attached to part two, although obviously I'd attribute it to you). – Todd Trimble Mar 10 '14 at 23:36

Yes, the problem is with sigma-additivity. If the space $X$ is not compact, then the dual of $C_b(X)$ consists of finitely additive measures. Think about the case when your $X$ is just a discrete countable space, so that $C_b(X)\cong \ell^\infty$.

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Yes, but i was wondering whether the fact that $m$ was the limit of elements $(l(P_n))$ associated to probabilities (and not any normed positive element of $(C_b(X))^\star$) implies that $P$ was also a probability. Naively i hope that the convergence of the sequence $(l(P_n))$ in the $\star-$weak topology could imply the tightness of $(P_n)$. Do you know whether this last fact is wrong ? – Theluze Mar 17 '13 at 13:48
Sigma-additive probability measures are dense in finitely additive probability measures with respect to the weak topology, whereas tightness is precisely the condition which guarantees that the limit of sigma-additive measures is also a sigma-additive measure. – R W Mar 17 '13 at 14:42
But: also he is asking about limit of a sequence so simply being dense won't produce a counterexample, you would need sequentially dense. – Gerald Edgar Mar 17 '13 at 16:51
"Sigma-additive probability measures are dense in finitely additive probability measures with respect to the weak topology" Do you mean with respect to the $\star-$weak topology ? Is it trivial ? – Theluze Mar 19 '13 at 11:06

It is a standard result of descriptive topology that the space of Borel probability measures on a polish space $S$ is itself a polish space with the weak star topology as the dual of $C^b(S)$ (Nota bene, not as a Banach space but with the so-called strict topology, i.e., the finest locally convex topology which agrees with compact convergence on the unit ball). This result is to be found, e.g. in the text on descriptive topology by Kechris.

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