A metric space is *doubling* if any ball of radius $2R$ can be covered by $N$ balls of radius $R$ and $N$ is fixed once forever.



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> Is there an example of complete length-metric space which is doubling, but the [Besicovitch covering theorem](http://en.wikipedia.org/wiki/Besicovitch_covering_theorem) does not hold?
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Doubling space without Besicovitch covering theorem?
The Besicovitch covering theorem fails for example in the Heisenberg group, see 
[ E. Sawyer and R. L. Wheeden,
*Weighted inequalities for fractional integrals on Euclidean and homogeneous spaces*, 
Amer. J. Math. **114** (1992), no. 4, 813–874. <http://www.jstor.org/stable/2374799> ] 


*The following was more of a comment to Misha's question.*


It is easier to see (without assuming the space to be a length-space) that the Lemma 4.2.2. in <http://www.math.wustl.edu/~sk/books/root.pdf> is not true for general complete doubling metric spaces nor is the Besicovitch covering theorem:


Take for instance a space $X = \mathbb{N} \cup\{0\}$ with the distance
$$d(0,j) = 2^{-j} \text{ for } j \ne 0$$ and $$d(i,j) = 2^{-j}+2^{-i} \text{ for }0 \ne i\ne j \ne 0.$$


To see that this does not satisfy the Lemma nor the Besicovitch covering theorem consider for any $k \in \mathbb{N}$ the collection $\{B(j,2^{-j}+2^{-k}) ~:~ j = 1, \dots, k-1 \}$.


