I have the following as an exercise.

A measure $\mu$ is doubling if there exists a constant $c_d$ such that $\mu(B(x,2R)) \leq c_d \mu(B(x,R))$ for all balls in the metric space. The claim is that this implies the metric space is doubling, that is there exists $N\in \mathbb{N}$ such that a ball of radius R can be covered with at most $N$ balls of radius $R/2$.

I tried proving this indirectly: Suppose you need countably many balls of radius $R/2$ to cover a ball of radius $R$: $$\bigcup_{i=1}^N B(x_i,R/2) \subset B(x,R) \Rightarrow \mu\left(\bigcup_{i=1}^N B(x_i,R/2)\right) \leq \mu(B(x,R)) $$ for all $N$. Then I would use doubling condition to see that $\mu(B(x,R)) \geq \infty$ which is a contradiction. However, I don't know how to get sum of measures between the inequality. I was also thinking that maybe I can arrange balls $B(x_i,R/2)$ such a way that every ball covers alone some set of positive measure, so that the measure of $B(x,R)$ would again be infinite, but I couldn't figure out that rigorously. (How I ensure that there wouldn't be alot of of non-open sets to cover after some point?)

I also thought that maybe it should be proven directly using $$ c_1\left(\dfrac{r}{R}\right)^{Q_1} \leq \dfrac{\mu(B(x,r))}{\mu(B(y,R))} \leq c_2\left(\dfrac{r}{R}\right)^{Q_2}, $$ but again I didn't figure out how to control overlapping in the covering.

Some hint to get to the right direction would be appreciated.


The doubling property of a metric space is a uniform bound on the cardinality of bounded, uniformly separated subsets. That is, in a doubling space a set $E\subset B(x,R)$ with $d(y,z)>r$ for all distinct $y,z\in E$ can have at most $C$ elements where $C$ depends on $R/r$ only (and this property characterizes doubling spaces).

The doubling measure property prevents large uniformly separated subsets from existing, because the $(r/2)$- neighborhoods of the points in $E$ would be disjoint, contained in $B(a, R+r/2)$, and each of them would have a measure comparable to $B(a,R+r/2)$ itself.

In details: pick a point $x_1\in B(x,R)$. If possible, pick $x_2\in B(x,R)\setminus B(x_1,R/2)$ and continue with $x_{n+1}\in B(x,R)\setminus \bigcup_{k=1}^n B(x_k,R/2)$. When the process stops, we have a cover of $B(x,R)$. And it has to stop soon because the balls $B(x_k,R/4)$ are all disjoint, they are contained in $B(x,5R/4)$, and $$B(x,5R/4) \subset B(x_k, 4R)$$ hence $$\mu(B(x,5R/4)) \le c_d^4\,\mu( B(x_k,R/4))$$ Thus the doubling constant $N$ is at most $c_d^4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.