Let $\mathcal{M}^{+}(\mathbb{R}\_{+})$ be space of non-negative [Radon measures](https://en.wikipedia.org/wiki/Radon_measure) 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 ?



How to prove space of non-negative Radon measures is complete?
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 $\lim\_{n \to \infty} \rho(\mu,\mu\_n) = 0$. 


