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Let $(X,d)$ be a compact metric space and $\{\mu_{n}\}$ be a sequence of Borel probability measure on $X$ which converges in the weak* topology to a Borel measure $\mu$. Show that if the diameter of the support of $\mu_{n}$ tends to zero as $n \rightarrow \infty$, then $\mu$ is a point mass.

I try to find some $x\in X$ such that $\mu$ is point mass at this point by using the diameter of $\mu_{n}$ tends to zero. I think it is easy to think geometrically. But I fail to write it down.

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  • $\begingroup$ What does weak$^*$ topology mean on a metric space? $\endgroup$ – copper.hat May 1 '19 at 4:27
  • $\begingroup$ @copper.hat I guess it's the topology induced by the Prokhorov metric $\endgroup$ – Sudheesh Surendranath May 1 '19 at 4:53
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Let $x_n \in K_n$ for all $n$ where $K_n$ is the support of $\mu_n$. There is a subsequence $x_{n_i}$ converging to some point $x$. Let $f$ be a bounded continuous function on $X$. Consider $\int_{K_{n_i}} (f(y)-f(x_{n_i}))d\mu_{n_i}$. By uniform continuity of $f$ it follows that $|f(y)-f(x)| <\epsilon$ for $i$ sufficiently large and hence $\int_{K_{n_i}} (f(y)-f(x_{n_i}))d\mu_{n_i} \to 0$. Now$\int (f(y)-f(x))d\mu_{n_i}=\int_{K_{n_i}} (f(y)-f(x_{n_i}))d\mu_{n_i}+f(x_{n_i})-f(x) \to 0$. Hence $\int_{K_{n_i}} f(y)d\mu_{n_i} \to f(x)$. This proves that the limiting measure is $\delta_x$.

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  • $\begingroup$ Thank you! It looks correct! $\endgroup$ – user387147 May 1 '19 at 18:03

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