# Doubling measure implies doubling metric space

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 $c_d^4$.