# Convergence in distribution of random measures

Let $M$ denote the space of real Radon measures on $\mathbb{R}$ as the topological dual of $C_c(\mathbb{R})$ equipped with the inductive limit topology (for possibly unbounded Radon measures) or uniform norm (for bounded Radon measures). In both cases, $M$ with the vague topology (i.e. the weak-* topology on $C_c'(\mathbb{R})$) is not metrizable. Let $M_+ \subseteq M$ be the subset of positive Radon measures, which is metrizable (and thus we can speak of weak convergence of probability measures on $M_+$ and have the Portmanteau theorem in hand).

Let $\xi_n$ be a sequence of random measures in $M_+$ and $\xi \in M_+$. Kallenberg, "Random Measures", Theorem 4.2 shows that $\xi_n$ converges to $\xi$ in distribution iff $\xi_n f$ converges to $\xi f$ in distribution for all $f \in C_c(\mathbb{R})$ where $\xi f$ is the random variable that takes the values $\omega \mapsto \int f(x) d\xi^\omega(x)$ and $\xi^\omega := \xi(\omega)$.

Q: Does a similar statement holds also for sequences of random measures in $M$? Can it be transfered from the convergence in distribution on $M_+$ by the Jordan decomposition of $\xi_n = \xi_n^+ - \xi_n^-$ where $\xi_n^+(\omega) := (\xi_n(\omega))^+$?

• In the case of bounded measures, the topology you've chosen on $M$ isn't the most common one. For instance, I think in your topology the sequence $n \delta_n$ converges to 0. But the usual definition of the "vague topology" on the bounded measures is the weak-* topology induced by continuous functions vanishing at infinity ($C_0(\mathbb{R})$), and in that topology $n \delta_n$ does not converge. – Nate Eldredge Feb 9 '15 at 19:53
• In fact, I'm considering unbounded Radon measures ala Bourbaki, e.g. the Lebesgue measure on $\mathbb{R}$. Kallenberg also defines vague topology by testing against $C_c$ but on $M_+$. In the inductive limit topology $C_c$ is complete. – yadaddy Feb 9 '15 at 22:30