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From Wikipedia

Let $D+$ be the set of all probability distribution functions $F$ such that $F(0) = 0$: $F$ is a nondecreasing, left continuous mapping from the real numbers $\mathbb{R}$ into $[0, 1]$ such that $$ \sup_{x \in\mathbb{R}} F(x) = 1 $$

The ordered pair $(S,d)$ is said to be a probabilistic metric space if $S$ is a nonempty set and $$ d: S×S →D+ $$ In the following, $d(p, q)$ is denoted by $d_{p,q}$ and is a distribution function dp,q(x). The distance-distribution function satisfies the following conditions: $$ d_{u,v}(x) = 0 \text{ for all }x > 0 \Leftrightarrow u = v (u, v ∈ S). $$ $$ ... $$

I wonder if $d_{u,v}(x) = 0 \text{ for all }x > 0$, whether $d_{u,v}$ is still a probability distribution function, since $\sup_{x \in \mathbb{R}} d_{u,v}(x) =0$ not $1$?

Thanks and regards!

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I think there is some error in the Wikipedia article. The line should read $d_{u,v}(x) = 1$, all $x > 0$ $\iff u = v$. –  martini Apr 23 '12 at 20:34
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Wikipedia articles that have no references or sources should be considered suspect. At least Planet Math has some references ... planetmath.org/ProbabilisticMetricSpace.html –  GEdgar Apr 23 '12 at 20:57
    
... and PlanetMath says we need to have a 1 there, as their $e_0$ is $\chi_{(0,\infty)}$. –  martini Apr 23 '12 at 21:08

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