$\nu$ is a Dirac delta

Question

Let $X$ be an locally compact Hausdorff space and $m$ a positive regular Borel probability measure where $m(Y)$ is 0 or 1 for any Borel set of $X$. Does it necessarily follow that $m$ is a Dirac delta?

Answer 1

I assume that "regular" means "inner regular and outer regular", i.e. for any Borel $A \subset X$ and any $\epsilon > 0$ there exist $K$ compact and $U$ open with $K \subset A \subset U$ and $m(U \setminus K) < \epsilon$.

Then yes, $m$ is a Dirac delta. For suppose not; then every singleton $\{x\}$ has measure zero. By outer regularity, there is an open set $U_x$ containing $x$ with $m(U_x) = 0$.

Now suppose $K$ is compact. Then $\{U_x : x \in K\}$ is an open cover of $K$ so it has a finite subcover $\{U_{x_1}, \dots, U_{x_n}\}$. By subadditivity $m(K) \le m(U_{x_1}) + \dots + m(U_{x_n}) = 0$.

So every compact set has measure zero. This contradicts inner regularity.