If $\operatorname{supp}\mu\subseteq \{x_1,\dots, x_n\}$, then $\mu=\sum_{i=1}^na_i\delta_{x_i}$ for some $a_1,\dots, a_n\geq 0$

Question

First, we begin with a definition:

Let $\mu$ be a Radon measure on a Hausdorff space $X$. Let $\{G_i\}_{i\in I}$ be the family of open $\mu$-null sets, and put $G:=\bigcup_{i\in I}G_i$. Then $G$ is the biggest largest open $\mu$-null set. By the support of $\mu$, we understand it the set $\operatorname{supp}\mu:=X\setminus G$.

The question is, how to show that,

If $\operatorname{supp}\mu\subseteq \{x_1,\dots, x_n\}$, then $\mu=\sum_{i=1}^na_i\delta_{x_i}$ for some $a_1,\dots, a_n\geq 0$, (where $\delta_x$ is the Dirac measure concentrated at $x\in X$).

Okay, we begin with an easy case, we look at $\operatorname{supp}\mu\subseteq \{0\}$. I try to show that it implies $\mu=a\delta_0$ for some $a\geq 0$. But I do not know how to do that directly, unless it is done by contrapositive, I think.

Answer 1

Given supp $\mu \subseteq \{x_1, \ldots, x_n\}$. Define $$a_i = \mu(\{x_i\}) .$$

Now, take any measurable set $A$. If $A \cap \text{supp}~\mu = \emptyset$, then $$\mu(A) = 0.$$ Suppose $A \cap \text{supp}~\mu \neq \emptyset$, let $J = \{j | x_j \in A \cap \text{supp}~\mu\}$ $$\mu(A) = \mu(A \cap \text{supp} ~ \mu) = \sum_{i \in J} \mu(\{x_i\}) = \sum_{i \in J} a_i = \sum_{i=1}^n a_i \delta_{x_i}(A).$$