Let $K$ be a compact Hausdorff space and $\mu$ be a regular Borel measure on $K$ with infinite support. How to find a decreasing sequence of open sets $\{V_{n}\}$ such that $\mu(V_{n}) > 0$ and $\lim_{n \rightarrow \infty} \mu(V_{n}) = 0$?
If $(X, \tau)$ is a topological Hausdorff space and $\mu$ is a measure on $X$, then definition of support of a measure is $\{x \in X: x \in N_{x} \in \tau \implies \mu(N_{x})>0\}$.
I am sure that the regularity of $\mu$ plays a role in finding the above decreasing sequence, but I don't know how. Please provide hints.
Consider $\{x\in supp (\mu): \mu (\{x\}) >0\}$. If this is infinite then there exist $x_n$'s with $\mu (\{x_n\}) >0$ for all $n$ and $\sum \mu (\{x_n\}<\infty$. Hence, $\mu (\{x_n\}) \to 0$. By going to a subsequence we may suppose $\mu (\{x_n\}) <\frac 1 {2^{n}}$. Now there exist open sets $U_n$ with $\mu(U_n)<\mu (\{x_n\})+\frac 1 {2^{n}}$ and $x_n \in U_n$. Note that $\mu(U_n) >0$ for all $n$ because $x_n$ belongs to the support of $\mu$. Take $V_n=\bigcup_{k\geq n} U_k$.
The case where $\{x\in supp (\mu): \mu (\{x\}) >0\}$ is a finite set is easier. Here there exists at least one point $x$ in the support with $\mu (\{x\})=0$. Can you finish?