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 $\mathscr{P}(\Omega,\mathcal{F})$?

 A: 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.
A: 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.
