Consider the following definitions.
Let $M = (V,T,d)$ be a compact metric space with finite diameter
$$D = D(M) = \max d(x,y), ( x, y \in M)$$
and a finite normalized measure $\mu$$M$(.), ($\mu(.)$ for short).
Let $x \in M$ and $B_x(d)$ = { $y\in M$ | $d(x,y) \leq$ d}, then the measure $\mu$ has the property $\mu(B_x(d))=\mu(d), (0 \leq d \leq D)$ (1). The normalization expresses in $\mu(D) = \mu(M) = 1$ (2).
When $M$ is finite, the measure $\mu$(.) is defined as (3) $$\mu(d)= \frac {1}{|M|} \sum_{i \leq d}k_i $$
where $k_i$ is the number of elements at distance $i$ of an arbitrary fixed element.
I don't understand the definition of $\mu(.)$. What is the domain of the so called measure, in (1) it takes $B_x$ as argument which is defined as a set and at the same $\mu$ takes d which is a real number ($d$ is a metric, defined as $d: V \times V \to \Bbb R$). Also why does the equality (2) holds when $\mu$ is defined as (3)?
Any comments, ideas and suggestions would be very much welcome! As usual, thank you in advance!
I think that answer is revealed in Lebesgue's number lemma. In this case $B_x(d)$ is ball of radius $d$ centered at $x$. $\mu(d)$ behaves as part of elements in $B_x(d)$ and $0 \le \mu(d) \le 1$.