# Is the space of finite Radon measures, equipped with weak topology, locally compact?

The space of Radon measures on a complete separable metric space $E$, endowed with the Borel σ-algebra, is denoted by $\mathcal M(E)$, while $\mathcal M_F(E)$ is the subspace of finite measures in $\mathcal M(E)$. The space $\mathcal M_F (E)$ is equipped with the weak topology. Well known fact is that the space $\mathcal M_F (E)$ is a Polish space as well.

Suppose $E$ is a half open interval $[0,H)$ for some $H\in (0,\infty]$. Let $\mathcal M_D[0,H)$ be the subset of measures in $\mathcal M_F ([0,H))$ that can be reprsented as the sum of a finite number of unit Dirac measures in $[0, H)$, that is, measures that take the form $\sum_{i=1}^k \delta_{x_i}$ for some $k\geq1$ and $x_i \in [0,H)$, $i = 1,\ldots,k$. The space $\mathcal M_D [0,H)$ is endowed with the topology of weak convergence.

Question: Is $\mathcal M_D [0,H)$ locally compact?

• Are you using weak topology in the sense of funcitional analysis? Meaning that $\mu_n \to \mu$ iff $\langle \mu_n, \psi \rangle \to \langle \mu, \psi \rangle$ for $\psi \in M_F(E)^\ast = C_0(E)^{\ast \ast}$. Or you mean evaluating against $\psi \in C_0(E)$? I think the last meaning is more common. – Adrián González-Pérez Mar 27 '18 at 9:45
• If your meaning is the second you have that $M_D[0,H)$ is dense inside $M_F[0,H)$, so the answer will be no. – Adrián González-Pérez Mar 27 '18 at 9:50
• Thanks much for your comments! By the weak topology, I meant $\mu_n\rightarrow \mu$ iff for every $\psi \in C_b(E)$ (that is, the space of bounded and continuous functions), $\int_E\psi(x)\mu_n(dx)\rightarrow \int_E \psi(x)\mu(dx)$ as $n\rightarrow \infty$. I am particularly interested in whether the space of finite point measures on the interval $[0,H)$ with such topology is locally compact or not. Thanks again for your time and consideration! – chlee Mar 27 '18 at 11:53