1
$\begingroup$

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!

  • 0
    ... and PlanetMath says we need to have a 1 there, as their $e_0$ is $\chi_{(0,\infty)}$.2012-04-23

0 Answers 0