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A metric space $X$ with metric $d$ is said to be doubling on $\Bbb R^2$ if there is some constant $C > 0$ such that for any $x \in X$ and $r > 0$, the Euclidean ball $B(x, r) = \{y:|x − y| < r\}$ may be contained in a union of no more than $C$ many balls with radius $r/2$.

That is:
$$B(x, 2r) < C \cdot B(x,r).$$

Can we prove that any such measure gives measure zero to a straight line $L$?

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    You're defining "$C$-doubling" as a property of _metrics_, but in the question you suddenly ask for a property of _measures_. Are you deriving the measure you're asking about from the metric in some canonical way?2012-11-04
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    http://en.wikipedia.org/wiki/Doubling_measure2012-11-04
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    This is the wiki-web about C-doubling measure. Intuitively, a C-doubling measure on R^2 should gives measure zero to a straight line. But I cannot prove it strictly. Maybe my intuition is wrong.2012-11-04
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    x @Yuhan: If I read that article correctly, having a doubling _metric_ is not enough to have a doubling _measure_: you need compactness as an additional assumption. And even with compactness you appear to get only that a doubling measure _exists_, not that it is unique. _A priori_ it seems to be conceivable to have a doubling metric space with two different doubling measures, one of which gives measure zero to straight lines and the other doesn't. But .. hmm .. What is a "straight line" in a _general_ metric space anyway? If it were also a manifold it could mean a geodesic, but ...2012-11-04
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    @Henning Makholm Yes u r correct! I am going to modify my measure as "a locally finite Borel measure on R^2 with doubling property."2012-11-04
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    cmnts hr llw 600 chrs; \neq twitr, \neq sms. Is nuf room 2 wrt ordn.y Englsh wds out n fll.2012-11-04
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    @Henning Makholm I have solved this question, thank you for ur help!2012-11-07

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