Set of all probability measure is not compact

Question

I am trying to find an example of the following fact :

If $X$ is not a compact space, then set of all probability measure $\mathcal M(X)$ may not be compact in weak star topology. As my intuition says, I have taken $X=\mathbb R$ with usual topology and then I am trying to show that sequence of all dirac measure $\{\delta_n\}_{n\in \mathbb N}$ does not have a weak star convergent subsequence.

How do I show that there is no probability measure $\mu$ on the real line such that $\int f \ d\delta_n\to \int f\ d\mu$ for all $f\in BC(X)\ ?$.

Thanks in advance.

Answer 1

For any positive integer $N$ there exists a continuous function $f$ with $0\leq f(x) \leq 1$ for all $x$, $f(x)=1$ for $|x| \leq N$ and $f(x)=0$ for $|x|>N+1$. We get $\mu ([-N,N]) \leq \int fd\mu =\lim \int f d\mu_n=0$ since $\int fd\mu_n=0$ for $n >N+1$. But $N$ is arbitrary so $\mu (\mathbb R)=0$, a contradiction.