# Probability metric spaces for which $r \leq R$ implies $\frac{Pr(B(x,r))}{Pr(B(x,R))} \geq \frac{r}{R}$

I'm looking for examples of spaces $X$ such that:

1. $X$ is a probability space.
2. $X$ is a metric space.
3. If $x \in X$ and $0 < r \leq R$ then $\frac{Pr(B(x,r))}{Pr(B(x,R))} \geq \frac{r}{R}$.

I already have some examples:

1. A segment $[a,b] \subset \mathbb{R}$ with the natural probability and metric.
2. Generalization of $1$: A finite tree with positive weights on the edges (where the weights sum up to 1, each edge $e$ is equivalent to a segment $[0,weight(e)]$ and the entire space is a quotient of the disjoint union of the edges which identifies the ends of the edges according to the structure of the tree). See Didier's comment.
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Sorry but I fail to see how example 2. works except when the tree is in fact linear. When one crosses the distance $d$ between $x$ and its nearest neighbor(s), for small positive $t$, it seems that $P(B(x,d-t))/P(B(x,d+t))$ becomes too small to exceed $(d-t)/(d+t)$ because of the new branch(es) appearing at distance $d$. – Did Jul 4 '11 at 15:22
@Didier: You are definitely right! Thank you for the correction. – user3533 Jul 4 '11 at 15:40
The first thing that comes to mind is that if you're working in the $d$-dimensional unit cube, then (except near the edges) $Pr(B(x,r))/Pr(B(x,R)) = (r/R)^d$. So in some sense you want spaces which are of dimension one or less. – Michael Lugo Jul 4 '11 at 17:25
For a very slightly different example, let $X$ be the unit circle with uniform probability distribution. Also $d(x,y)$ is the length of the smaller arc between $x$ and $y$. – Srivatsan Aug 1 '11 at 21:38
Here's a fancier version of the above example. (I hope this works; I cannot trust my calculations yet.) Take a unit interval, and a circle of circumference $2$. Identify one endpoint of the interval with one of the points in the circle. The metric is the natural metric. The probability density is uniform over the interval and over the circle individually, and the total probability of the interval (or the circle) is $1/2$. (Note that since the circumference of the circle is twice the length of the interval, the density over the interval is double that over the circle.) – Srivatsan Aug 1 '11 at 21:52