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?

  • $\begingroup$ What do you understand by the term "Radon measure"? Is it a sort of linear functional on a space of functions? (Bourbaki) Or is it a set function defined on the Borel sets? (Halmos) $\endgroup$ – Gerald Edgar Jul 24 '17 at 13:55
  • $\begingroup$ I meant the dual of C(X), so in the Bourbaki sense. $\endgroup$ – Camille Pouchol Jul 25 '17 at 8:58

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 $A$.

| cite | improve this answer | |
  • $\begingroup$ Thanks a lot. I think it does answer my question if by Radon measure I mean a measure defined on a topological space endowed with its Borel sigma algebra. In my case I meant as a element of the dual of $C(X)$, but from the Riesz representation theorem it does the trick also in my case. $\endgroup$ – Camille Pouchol Jul 25 '17 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.