# This metric in the space of probability generates the weak* topology?

Let $\Omega$ be a compact metric space and $\mathscr{P}(\Omega,\mathcal{F})$ the space of all probability measures definid on $\sigma$-field Borel of $\mathbb{X}$. $UC(\Omega,\mathbb{R})$ stands for the space of all bounded real valued continuous functions on $\Omega$.

A metric to $\mathscr{P}(\Omega,\mathcal{F})$ is

$$d(\mu,\nu)=\sum_{\varphi_n\in\mathcal{A}}\frac{1}{2^n}\left|\int_\Omega \varphi_n \;d\mu -\int_\Omega \varphi_n \;d\nu \right|$$

where $\mathcal{A}$ is countable and dense set of functions in $UC(\Omega,\mathbb{R})$. The symmetry and the triangle inequality can be easily verified. To check $d(\mu,\nu)=0 \Leftrightarrow \mu=\nu$ we use the Riesz-Markov theorem.

Question: How can we prove that this metric generates the weak* topology in $\mathscr{P}(\Omega,\mathcal{F})$?

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How can you define weak-$^*$ topology in $\mathscr{P}(\Omega,\mathcal{F})$ if this is not a linear space. – Norbert Sep 27 '12 at 3:02
weak*-topology in probability gives convergence in distribution, so just read is as convergence in distribution, which do not depend upon linearity. – kjetil b halvorsen Sep 27 '12 at 7:52
In a compact metric space, we hae a characterisation of the topological dual space of continous functions (boundedness is redundant): it's given by the signed measures of bounded variation. Actually, you have to show that this metric is a metric on the unit ball of the dual space of $C(\Omega)$, then show that in a separable normed space, the weak-$*$ topology of the unit ball is metrisable by the analoguous metric. – Davide Giraudo Sep 27 '12 at 11:47
@Norbert The space of signed measures is isomorphic to the space of positive linear functional. This is the statement of the theorem of Riesz-Markov. google.com.br/… – MathOverview Sep 27 '12 at 16:46

There is a quite general fact that might reveal general pattern here.

Let $(X,\tau)$ be a compact topological space. Let $\{f_n:n\in\mathbb{N}\}$ be a bounded sequence in $(C(X),\Vert\cdot\Vert_\infty)$ separating points of $X$. Then $X$ is a metrizable via distance $$d: X\times X\to\mathbb{R}_+:(p,q)\mapsto\sum\limits_{n=1}^\infty2^{-n}|f_n(p)-f_n(q)|$$ In fact topology $\tau_d$ induced by metric $d$ coincide with original topology $\tau$. For the elegant proof see section 3.8 in Rudin's Functional analysis.

Now you can apply this general result to your situation. The role of $(X,\tau)$ is played by $\mathscr{P}(\Omega,\mathcal{F})$ with weak-$^*$ topology. As Davide Giraudo pointed out this a compact topological space. The only thing you had to check is that $\{\varphi_n:n\in\mathbb{N}\}$ separates points in $\mathscr{P}(\Omega,\mathcal{F})$. It is not difficult.

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The definitive result in this direction is the fact that the space of probability metrics on a polish space is also a polish space in the above topology (recall that a polish space is a complete, separable metric space). This can be found in Kechris' book on descriptive topoogy.

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