# Support of a measure on Compact Hausdorff space

Let $$(X,\mathcal{B})$$ a compact Hausdorff topological space equipped with Borel $$\sigma$$-algebra. If $$\mu$$ is a measure on $$\mathcal{B}$$ s.t. $$\mu(X)>0$$ then $$supp(\mu)\neq\emptyset$$
I think I can prove it if X is a metric space. In fact I can take a finite cover of X with closed(hence compact since $$X$$ is compact) balls of radius 1. By additivity of measure I can find a ball $$B_1$$ s.t. $$\mu(B_1)>0$$. Then I can repeat the process covering $$B_1$$ with balls of radius $$1/2$$ and so on. In this way I have constructed a sequence $$\{B_n\}_n$$ of compact balls with radius going to zero. Hence $$\exists!x\in\cap_n B_n$$ and it is easy to show that such $$x\in supp(\mu)$$
Suppose $$supp (\mu)=\emptyset$$. If $$x \in X$$ then $$x \notin supp (\mu)$$ so there is an open set $$U_x$$ conatining $$x$$ such that $$\mu (U_x)=0$$, $$(U_x)_x$$ is an open cover of $$X$$. Let $$U_{x_i}, 1\leq i \le n$$ be a finite subcover. Then $$\mu(X) \leq \sum_i \mu (U_{x_i})=0$$, a contradiction.