# Borel Probability Measure on a compact metric space

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.

• What does weak$^*$ topology mean on a metric space? – copper.hat May 1 '19 at 4:27
• @copper.hat I guess it's the topology induced by the Prokhorov metric – Sudheesh Surendranath May 1 '19 at 4:53

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 $$\delta_x$$.