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Nov 13, 2010 at 6:48 vote accept gondolier
Nov 12, 2010 at 0:19 comment added gondolier Great. The same proof shows that it is impossible to have $\mu(B(x,r)) \sim r^{\alpha}$ for $\alpha > d+1$. I think the tight result should be $\alpha > d$.
Nov 11, 2010 at 17:51 comment added R W Yes - it's OK now - except for one detail. In the last line you, of course, wanted to refer to summability of the series $\sum n^d e^{-n/k}$ (which is the easy direction of the Borel-Cantelli Lemma).
Nov 11, 2010 at 16:58 comment added rpotrie Sorry, there were some holes in the proof. I hope now it is correct.
Nov 11, 2010 at 16:57 history edited rpotrie CC BY-SA 2.5
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Nov 11, 2010 at 16:32 comment added R W Why is it that $A_k=\bigcup B_n$, and even it were so, why is it enough to show that $\mu(B_n)\to 0$?
Nov 11, 2010 at 16:10 history edited rpotrie CC BY-SA 2.5
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Nov 11, 2010 at 12:27 history answered rpotrie CC BY-SA 2.5