# Prokhorov metric can be extended to all finite Borel measures?

Let $$(X, d)$$ be a metric space and $$\mathcal{P} :=\mathcal{P}(X)$$ the set all Borel probability measures on $$X$$. Let d_{P}(\mu, \nu) := \inf \left\{ \alpha>0 \,\middle\vert\, \begin{align*} \mu(A) \leq \nu\left(A_{\alpha}\right)+\alpha \\ \nu(A) \leq \mu \left(A_{\alpha}\right)+\alpha \end{align*} \quad \forall A \in \mathcal{B}(X) \right\} \quad \forall \mu, \nu \in \mathcal{P}, where $$A_{\alpha} := \{x \mid d(x, A)<\alpha\}$$ and $$d(x, A)=\inf \{d(x, a) \mid a \in A\}$$.

Below is the proof that $$d_P$$ is indeed a metric.

• Any $$\alpha \geq 1$$ is in the set of the defining formula of $$d_{P}$$, so the infimum is well defined.

• Clearly, $$d_{P}(\mu, \nu) \geq 0$$ and $$d_{P}(\mu, \nu)=d_{P}(\nu, \mu)$$ for all $$\mu, \nu \in \mathcal{P}$$.

• Let $$\mu \in \mathcal{P}$$. For every $$A \in \mathcal{B}(X)$$ and $$\alpha>0$$, $$A \subseteq A_{\alpha}$$, so $$\mu(A) \leq \mu\left(A_{\alpha}\right)+\alpha$$. Hence $$d_{P}(\mu, \mu) \leq \alpha$$ and thus $$d_{P}(\mu, \mu)=0$$.

• If $$d_{P}(\mu, \nu)=0$$, then there is a sequence $$\alpha_{n} \downarrow 0$$ such that $$\mu(A) \leq \nu\left(A_{\alpha_{n}}\right)+\alpha_{n}$$ and $$\nu(A) \leq \mu\left(A_{\alpha_{n}}\right)+\alpha_{n}$$ for all $$n$$. As $$\bar{A}=\bigcap_{n} A_{\alpha_{n}}$$, it follows that $$\mu(A) \leq \nu(\bar{A})$$ and $$\nu(A) \leq \mu(\bar{A})$$. In particular, $$\mu(A)=\nu(A)$$ for all closed sets $$A$$ and therefore $$\mu=\nu$$ by inner regularity.

• The triangle inequality is proved here.

It seems to me this metric can be extended to all finite non-negative Borel measures. For the defining formula of $$d_P$$ to be well-defined, we just replace $$\alpha \ge 1$$ by $$\alpha = \max \{\mu(X), \nu(X)\}$$.

Could you confirm if my observation is correct?