I saw an example of a sequence of measures for which no subsequence can have a weak limit, but don't think it is correct.
Let $(\delta_{x_n})_{n\in\mathbb{N}} $ be a sequence of Dirac measures and let $ (x_n)_{n\in\mathbb{N}}\rightarrow \infty$
Note that it was not further specified what the probability space of the dirac measures really is, so lets assume $ (\overline{\mathbb{R}}, \mathcal{B}(\overline{\mathbb{R}}) )$ which is the extended Borelspace. Let $f$ be a real bounded continuous function, then
$$\lim_{n\to\infty} \int_{\overline{\mathbb{R}}}f\,\mathrm{d}\delta_{x_n}=\lim_{n\to\infty}f(x_n)=f(\infty)= \int_{\overline{\mathbb{R}}}f\,\mathrm{d}\delta_{\infty}$$
i.e. it converges weakly. What is the problem, why should it not work?
Best Answer
The underlying space is intended to be $\mathbb R$, not its compactification. Thus, bounded and continuous functions may fail to have a limit at infinity. For example, assuming that $x_n=(n+\tfrac12)\pi$ (thank you Nate), consider the bounded and continuous function $$ f(x)=\sin(x).$$ If the sequence $\delta_{x_n}$ had a limit, say $\mu$, then it should be possible to integrate all bounded and continuous functions against it; however, $$ \int f(x)\, d\mu=\lim_{n\to \infty} f(x_n). $$ Now, this last limit does not exists for any subsequence of $x_n$. Thus $\mu$ is not well-defined.