Let $M(X)$ be the set of probability measures on a Polish space $X$ with Borel $\sigma$-field. Further consider the properties of $M(X)$ when considered as members of the dual space of $Y:=C_b(X)$ - the supremum-normed space of bounded continuous functions on $X$.
In this context, I struggle to see the reason, wyh the following statement is false: $M(X)$ is weak*-compact.
My reasoning: According to Banach-Alaoglu the following set is weak*-compact: $$K:=\{\Lambda \in Y^* | |\Lambda(V)|\leq 1 \},$$ wherein $V$ denotes any neighborhood of zero. Choose the open unit-ball as $V$ then, clearly, every probability measure $\mu \in M(X)$ lies in $K$.
Hence, the set of probability measures is at least a subset of a weak*-compact set. Further, if I have a sequence of probability measures which converge weakly to another measure $\delta$, then I think one can directly show that $\delta$ itself is also a probability measure (using $\int f\mathrm{d}\mu_n \to \int f\mathrm{d}\delta$ for any $f \in Y$). Therefore, I have almost proven that M(X) is a closed subset of a compact set - and therefore compact. The only way this can be wrong is if
a) someone can point out an error in the above derivation
or
b) there are members of the dual-space of $Y$ which aren't representable as measures $\delta$, in that case I would appreciate it, if someone could give an example of such a member.
In any case: Many thanks for your help!
Let $\delta_n$ be the point measures $\delta_n(f)=f(n)$ on $C_b(\Bbb{R})$. The $\delta_n$ are in the unit ball of $(C_b(\Bbb{R}))^*$ which is weak$^*$ compact, so the sequence $\delta_n$ has a weak$^*$ convergent subnet. Call the limit $\Lambda$. Denote the index set of the subnet as $\Gamma$, and a function $n:\Gamma\to\Bbb{N}$ defines the subnet.
For $N>0$ let $E_N=\{x\in\Bbb{R}:|x|\ge N\}$. Let $f_N\in C_b(\Bbb{R})$ be $$\begin{cases} 0 & |x| < N\\ |x| - N & N\le |x|<N+1\\ 1 & N+1\le |x| \end{cases}$$ By the properties of subnets, there exists some $\gamma_1\in\Gamma$ such that $n(\gamma_1)>N+1$. Then for all $\gamma\in \Gamma$, if $\gamma_1 \le \gamma$ then $N + 1 < n(\gamma_1)\le n(\gamma)$, so $\delta_{n(\gamma)}(f_N)=1$. Since $\Lambda(f_N)$ is the limit of the subnet on $f_N$ it follows that $\Lambda(f_N)=1$.
On the other hand, if $p$ is any probability measure on $\Bbb{R}$, there is some $N>0$ such that $p(E_N)<1/2$ and $$\int_{\Bbb{R}} f_N dp \le p(E_n) < 1/2$$
It follows that $\Lambda$ can't be a probability measure.