On the compacity of the space of probability measures

Let $X$ be a complete metric space, and denote by $\mathcal{P}(X)$ the set of probability measures on $X$. I am interested in proving that if $X$ is compact then $\mathcal{P}(X)$ must be compact in the weak-* topology, which is induced by the convergence against $C_b(X)$, that is, the bounded continuous functions.

Since $\mathcal{P}(X) \subset (C_b(X))^*$, and the unit ball of $(C_b(X))^*$ is compact in the weak-* topology (by Banach-Alaoglu), all that is left to do is prove that $\mathcal{P}(X)$ is weak-* closed. This should be easy, but I'm stuck: where does the compacity of $X$ come in? (The result is obviously false for non-compact $X$: consider a sequence of delta measures $\delta_{x_n}$ and $x_n \in \mathbb{R}, x_n \to \infty$.)

• Hm..."compacity"? – Michael Jan 16 '14 at 4:16

Let's begin with the Riesz representation: For a locally compact Hausdorff space $X$, the map taking the finite signed regular Borel measure $\mu$ to the functional $f\mapsto \langle f,\mu\rangle:= \int f\,d\mu$ is an isometric isomorphism of the Banach space $M_r(X)$ onto the dual of the Banach space $C_0(X)$. Here $C_0(X)$ is the space of continuous functions that vanish at infinity, and the norm on $M_r(X)$ is the total variation norm.
Now Banach-Alaoglu guarantees that the unit ball in $C_0(X)^*$ is weak${}^*$ compact, which corresponds to vague compactness for the measures. However, generally, the unit sphere in $C_0(X)^*$ is not weak${}^*$ closed and hence not compact. Note however that the cone $\cal P$ of positive elements in $M_r(X)$ is vaguely closed: $${\cal P}=\bigcap_{f\in C_0(X)_+} \left\{\mu: \langle f,\mu\rangle \geq 0\right\},$$ so its intersection with the unit ball is compact.
So how does the compactness of $X$ enter into the picture? If $X$ is compact, then $C_0(X)=C_b(X)$ and the vague and weak topologies of measures coincide. In particular, the constant function "1" belongs to $C_0(X)$ so the space of probability measures is the compact set $${\cal P}\cap \{\mu: \|\mu\|\leq 1\}\cap \{\mu: \langle 1,\mu\rangle =1\}.$$
Use Riesz representation theorem. Suppose you have a weak-* limit. This is necessarily a positive functional on $C_0(X) = C_b(X)$, with norm $1$. So you're done.