# How to prove space of non-negative Radon measures is complete?

Let $$\mathcal{M}^{+}(\mathbb{R}_{+})$$ be space of non-negative Radon measures on $$\mathbb{R}_{+}$$ with bounded total variation and define the metric $$\rho$$ on $$\mathcal{M}^{+} (\mathbb{R}_{+})$$ as $$\rho(\mu,\nu)= \sup \left \{ \int_{\mathbb{R}_{+}} \psi d (\mu - \nu) ~|~ \psi \in C^{1}(\mathbb{R}_{+}), \|\psi \|_{\infty} \le 1 , \|\partial_{x} \psi \|_{\infty} \le 1 \right \} .$$ How to prove $$\mathcal{M}^{+}(\mathbb{R}_{+})$$ is complete w.r.t. $$\rho$$. I know that $$\lim_{n \to \infty} \rho(\mu_{n},\mu) = 0 \iff \mu_{n} \to \mu~ \text{narrowly for}~ n \to \infty.$$ But how above equivalence can help us to prove the completeness ?

• I think your "metric" is ill defined. Since each $\mu \in M$ may be unbounded, in the integral you may have the difference $\infty-\infty$. You need some condition which implies that this does not happen. – Dieter Kadelka Feb 15 at 9:24
• @DieterKadelka You are right, I edited my question. – Manoj Kumar Feb 15 at 9:37

Assuming that the above equivalence is valid, all you have to show that $$\lim_{m,n \to \infty} \rho(\mu_m,\mu_n) = 0$$ implies uniform tightness of the sequence $$(\mu_n)$$. So let $$\epsilon > 0$$ be arbitrary. Then there is $$n_0 \in \mathbb{N}$$ with $$\rho(\mu_m,\mu_n) \leq \epsilon$$ for $$m,n \geq n_0$$, in particular $$\mu_n(\mathbb{R}_+) - \mu_m(\mathbb{R}_+) < \epsilon$$. Since $$\mu_{n_0}$$ is tight, there is $$t_0 \in \mathbb{R}_+$$ with $$\mu_{n_0}((t_0,\infty)) \leq \epsilon$$. Now let $$\psi \geq 0$$ be any function with the properties above and with $$\psi|{[0,t_0]} \equiv 0$$, $$\psi|{[t_0+2,\infty]} \equiv 1$$ and $$\psi$$ increasing. Such function exists. Then $$\mu_n((t_0+2,\infty)) \leq \int \psi~d\mu_n \leq \int \psi~d\mu_{n_0} + \epsilon \leq 2\epsilon.$$ It follows that $$(\mu_n)$$ is uniformly tight, hence there are limit points $$\mu$$ with respect to the narrow topology. But then $$\mu_{n_k} \to \mu$$ weakly for some subsequence $$(\mu_{n_k})$$, hence $$\rho(\mu_{n_k},\mu) \to 0$$ by the above equivalence. Since $$(\mu_n)$$ is Cauchy neceesarily $$\mu$$ is unique and $$\lim_{n \to \infty} \rho(\mu,\mu_n) = 0$$.
• It is well known that $\mathcal{M}^+$ is Polish. Proving that a concrete metric is complete almost always is done as in my answer. – Dieter Kadelka Feb 15 at 10:50
• Thanks for the answer, I just want to know what do you mean by- $\psi$ with the properties above in your answer? Also, will the existence of $\psi$ be shown through the standard argument of mollifiers? – Manoj Kumar Feb 15 at 13:52
• The properties of $\psi$ in your definition of $\rho$, such a $\psi \in C^1$ etc. – Dieter Kadelka Feb 15 at 14:29
• Concerning mollifier: You can do it by mollifier and also directly. Note that the interval $[t_0,t_0+2]$ has length $2$. – Dieter Kadelka Feb 15 at 14:57