In the answer given by uniquesolution, there is this statement:
A measure has finite support if and only if it is a linear combination of a finite number of measures concentrated at single points
Question: How to prove this? I think it should have something to do with dirac measures but I don't know how to start.
Any hint would be appreciated.
Wikipedia implies that this is not correct even for Hausdorff spaces and Borel measures. However, under the assumptions of $\mu$ being a Radon measure on a Hausdorff space $(X, \mathcal{T})$, we can prove that it is true:
In this case, we have the important property that the complement of the support has measure zero. Let the finite support be given as $$\{x\in X| x\in N_x\in \mathcal{T} \Rightarrow \mu(N_x)>0\}=\{x_1, \dots, x_n\}.$$ Because singletons are measurable, and because $\mu(\{x_1, \dots, x_n\}^c)=0$, the measure $\mu$ is uniquely defined by the values $\mu(\{x_i\})$. We can now construct measures $\mu_i(A):=\mu(A\cap \{x_i\})$ which are "measures concentrated on a single point." Obviously, we have $\mu=\sum_{i=1}^n\mu_i$ which proves the "only if" dircetion.
The "if" direction is quite trivial: Let $\mu_i$ be measures satisfying $\mu_i(\{x_i\}^c)=0$, then any linear combination $\sum_{i=1}^n a_i\mu_i$ satisfies $\mu(\{x_1, \dots, x_n\}^c)=0$. Because $X$ is a Hausdorff space, there exists for each $x\notin \{x_1, \dots, x_n\}$ an open neighborhood $N_x$ such that $N_x \cap \{x_1, \dots, x_n\}=\emptyset$ and thus $x \notin \operatorname{supp}(\mu)$. This implies that the support of $\mu$ is a subset of $\{x_1, \dots , x_n\}$ and, hence, finite.