Weak* topology is Hausdorff? Problem: Let $S$ be a complete separable metric space, show that the space of probability measures on $S$, denoted as $PM(S)$ is Hausdorff
Attempt: By definition of Hausdorffness, two points $P_1$, $P_2$ in $PM(S)$ can se separated by open disjoint neighborhood: $\exists G_i\in\mathcal{G}(PM(S))$ with $G_1\cap G_2 =\emptyset$ and $P_i\in\mathcal{G_i}$. So I guess I am trying to show this. Also notice that for $\mu_1=\mu_2$ iff $\int fd\mu_1=\int fd\mu_2$ for all continuous function with compact support. in a broader sense, I think this is equivalent to showing that the weak* star topology is Hausdorff..
 A: Here's an argument that in general weak$^*$ topologies are Hausdorff (even $T_{3\frac{1}{2}}$).
Weak$^*$ topology on an arbitrary dual $X^*$ can be seen as pointwise convergence (or Tychonoff) topology (when we consider $X^*$ as a subset of ${\bf R}^X$, or its complex counterpart, whichever you're more interested in), which makes it easy to show that it is Hausdorff just by noting that ${\bf R}$ (as well as ${\bf C}$) is Hausdorff.
In fact, this also shows that they are completely regular (as complete regularity is preserved by products and taking subspaces).
Another, similarly general way to see this is to note that addition and subtraction are weakly$^*$-continuous, so $X^*$ is a topological group. It's easy to see that $\{0\}$ is closed, and that implies Hausdorffness, as the preimage of $\{0\}$ by $(x,y)\mapsto x-y$ is the diagonal.
The same arguments (both) work for weak topologies, as long as the space in question has the property that functionals separate points (as is the case for Banach spaces, and more generally locally convex spaces, as well as dual spaces).
A: The topology is generated by sets of the form 
$$O(\mu,r,f)=\left\{\nu\in PM(S)\mid \left|\int_Sf(x)\mathrm d\mu(x)-\int_Sf(x)\mathrm d\nu(x)\right|\lt r\right\},$$
where $\mu \in PS(M)$, $r>0$ and $f$ is a continuous function on $S$ with compact support. 
If $\mu_1$ and $\mu_2$ are two different probability measures, then by the result recalled in the opening post we have $\int_S f(x)\mathrm d\mu_1(x)\neq \int_S f(x)\mathrm d\mu_2(x)$, and the difference between these two numbers has an absolute value of $3r$. Then define 
$$G_1:=O(\mu_1,r,f)\mbox{ and }G_2:=O(\mu_2,r,f).$$
A: Suppose there exists $f\in C_0(S)$ such that $\int fd\mu_1 \neq \int fd\mu_2$, w.o.l.g. there is an $\alpha \in \mathbb{R}$ such that 
$$\int fd\mu_1 < \alpha < \int fd\mu_2.$$
Define the two disjoint weak$^*$ open sets 
$$\mu_1 \in \left\{\mu\in PM(S): \int fd\mu< \alpha\right\}$$
and
$$\mu_2 \in \left\{\mu\in PM(S): \int fd\mu> \alpha\right\}.$$
