Levy prokhorov distance is a metric

By F1, $0\le \pi \le 1$ and symmetry holds. Since $F\subseteq F^{[\epsilon ]}$, every positive $\epsilon$ is admissible for equal measures, giving self-distance zero. Admissibility is upward closed because larger radii enlarge sets and increase the error.

If $\pi$($\mu$,$\nu$)=0, for each positive integer m the upward-closure observation makes 1/m admissible. For nonempty closed F the sets $F^{[1\mathrm{/}m]}$ are closed by F2 and decrease to F. F3 gives $\mu$(F)<=$\nu$(F) and, symmetrically, the reverse. Equality also holds on the empty set. The sets on which the two probabilities agree form a lambda-system; closed sets form a generating $\pi$-system, so F4 yields $\mu$=$\nu$.

If a is admissible between $\mu$ and $\nu$, and b between $\nu$ and $\sigma$, then for every closed F, $\mu (F)\le \nu (F^{[a]})+a\le \sigma ((F^{[a]}{)}^{[b]})+a+b\le \sigma (F^{[a+b]})+a+b$. The last inclusion follows from the metric triangle inequality by approximating each infimum within any positive slack and then taking its infimum; empty F is separate. The reverse inequality interchanges $\mu$ and $\sigma$. Thus a+b is admissible, and taking a and b arbitrarily close above their infima proves the triangle inequality.

If $\epsilon$ is admissible in the all-Borel open convention, it is admissible for closed sets and closed enlargements, since $F^{\epsilon }\subseteq F^{[\epsilon ]}$. Conversely, if a is admissible in the closed convention, any Borel B satisfies $\mu (B)\le \mu (\bar{B})\le \nu ((\bar{B}{)}^{[a]})+a\le \nu (B^{a+\delta })+a+\delta$ for every $\delta$>0, because distance to B and its closure agree. Interchanging the measures gives the other inequality. Infima and arbitrary positive slack prove equality of the two conventions.