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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.

Is there an example of complete length-metric space which is doubling, but the Besicovitch covering theorem does not hold?

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@Stas: Can't you just repeat the proof of BCT, say, the one given in math.wustl.edu/~sk/books/root.pdf : All you need is the packing lemma 4.2.2, which, I think, follows from the doubling condition? –  Misha Apr 18 '13 at 4:48
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up vote 8 down vote accepted

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 $\{B(j,2^{-j}+2^{-k}) ~:~ j = 1, \dots, k-1 \}$.

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Tapio: Yes, you are right, of course! (I was just not thinking clearly.) However, once one has the packing lemma, which one could regard as a strengthening (rather than weakening!) of the doubling condition, the proof in the Krantz-Parks' book goes through. Packing lemma: There exists a number $p=p(X)$ so that for every $r>0$, the number of pairwise disjoint $r$-balls contained in $B(x, 4r)$, is at most $p$. –  Misha Apr 18 '13 at 16:05
    
Misha: perhaps I am misunderstanding the packing lemma, but it seems to be equivalent with the doubling condition. In particular, the packing lemma would seem to follow directly for example from the existence of a doubling measure on the (completion of the) doubling metric space. –  Tapio Rajala Apr 18 '13 at 19:17
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Heuristically, problems should arise only when one starts to consider balls of different radii. –  Tapio Rajala Apr 18 '13 at 19:18
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