I've find this statement in Wikipedia page about support of measures:

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 $supp(\mu)\neq\emptyset$

Unfortunately I couldn't find it stated elsewhere. Can you please me give a proof. I am also interested to know if it's true under different hypothesis, can you please give me some references.

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 $x\in supp(\mu)$


1 Answer 1


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.


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