Functional Analysis – Extending Prokhorov Metric to All Finite Borel Measures

Question

I'm reading about Prokhorov metric, i.e.,

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.

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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?

Answer 1

I just check all the proofs, which only use finiteness for finite many measures.

So the answer is yes, you're right.