On compactness 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 $x_n \in \mathbb{R}, x_n \to \infty$.)
 A: 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.
A: The other answer is correct, but a little brief so I decided to add a few words.
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. 
The intuition here is that the positive part of the unit sphere is tiny, so 
 could well be compact even though the whole sphere itself is not. 
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
$$ {\cal P}\cap \{\mu: \|\mu\|\leq 1\}\cap \{\mu: \langle 1,\mu\rangle =1\}.$$      
A: $\{\mu:⟨1,\mu⟩=1\}$
$P=\cap _{f\in C_{0}(X)} \{\mu:⟨f,\mu⟩≥0\}$
Those μ should not only be considered from $C_{0}(X)^*$
They are in the product topology space of $C_{0}(X)$ copies of R: the set of all real valued functions defined on $C_{0}(X)$. Then we can say the valuation at any $C_{0}(X)$ is continuous so the preimage of 1 will be closed.
Then by the Banach-Alaoglu theorem, the {$\mu: \mu \in C_{0}(X)^*$ and ∥$\mu$∥≤1} is compact.
By the Rietz reprentation theorem, the norm for this signed measures space(functionals for continuous functions) is the total variation norm.
And $\{\mu: \mu \in C_{0}(X)^*$, ∥$\mu$∥≤1}$\cap\{\mu|\mu$ a real function on $C_{0}(X):⟨1,\mu⟩=1$} is already enough to tell the set of Probability Borel measures is weak-* closed/compact
If a signed measure is with full measure 1 and total variation full measure no more than 1, using Hahn-decomposition, it must be a positive measure with full measure 1 thus probability measure.
