Discrete support for Radon measures

Question

I have a sequence of Radon measures (on some set $X$, compact subspace of $\mathbb{R}^d$ so nothing too fancy), say $\mu_n$, which are actually $L^1(X)$ functions. In the limit I want to prove that I obtain a measure supported on a finite set $\{x_1, \ldots, x_N\}$.

One way to prove this seems to be taking a (continuous) function $g$ which is supported outside this set and prove that $\left\langle \mu_n, g \right\rangle$ converges to $0$ as $n$ goes to $+\infty$.

So here are my two questions:

• do you have any reference explaining why this is sufficient? I found tons of definitions of the support of a measure, but always in a (much) too general manner.

• once I have proved this result, it seems to be also asserted in several publications that the limit must be a sum of Dirac masses located on the set. Do you have any reference for the fact that Radon measures supported on a finite set must be sums of Dirac masses on these points?

Answer 1

Let $\mu$ be your limiting measure and $F$ be your finite set. Let $B$ be any closed ball disjoint from $F$, and let $f_B$ be a positive continuous function which equals 1 on $B$ and is supported in $F^c$. (For instance, you can take $f_B(x) = \max(1 - n d(x,B), 0)$ for sufficiently large $n$.) By assumption, $\int f\,d\mu = 0$ so we have $\mu(B) = 0$. Now you can write $F^c$ as a countable union of closed balls disjoint from $F$, so $\mu(F^c) = 0$ and the support of $\mu$ is thus contained in $F$.

(Support of the measure is perhaps most easily defined as follows: let $V$ be the union of all open sets which have measure zero; the support of $\mu$ is then $V^c$. In a separable metric space you may easily verify that $\mu(V) = 0$.)

The fact that such $\mu$ is a sum of Dirac masses is more or less obvious: given a Borel set $A$, use additivity to write $$\mu(A) = \mu(A \cap\{x_1\}) + \dots + \mu(A \cap \{x_N\}) + \mu(A \cap F^c)$$ where the last term is zero. Since $A \cap \{x_i\}$ is either $\{x_i\}$ or $\emptyset$, we have $\mu(A) = \sum_{x_i \in A} \mu(\{x_i\})$ which is the same as what you get for the measure $\mu' = \sum_{i} \mu(\{x_i\}) \delta_{x_i}$. So $\mu(A) = \mu'(A)$ for all $A$.