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 ?
 A: 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 $\lim_{n \to \infty} \rho(\mu,\mu_n) = 0$. 
