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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? – Henning Makholm Nov 4 '12 at 3:02
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. – Yuhan Liu Nov 4 '12 at 3:14
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 ... – Henning Makholm Nov 4 '12 at 3:17
@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." – Yuhan Liu Nov 4 '12 at 3:20

Although the question was solved by the OP, there is no solution here for others to read. Let's fix that.

Definition. A set $E\subset \mathbb R^n$ is porous if there exists $c>0$ such that for any ball $B\subset\mathbb R^n$ the set $B\setminus E$ contains a ball of diameter $c\,\mathrm{diam}\, (B)$.

Observe that a line in $\mathbb R^2$ is porous ($c=1/2$ works).

Definition. A Radon measure $\mu$ on $\mathbb R^n$ is doubling if there exists $C$ such that $\mu(2B)\le C\,\mu (B)$ for any ball $B$. An equivalent form of the definition involves squares: two squares with a common side get about the same measure, up to a multiplicative constant.

Claim. If $\mu$ is a doubling measure and $E$ is a porous set, then $\mu(E)=0$.

Proof. It suffices to show that $\mu(E\cap Q)=0$ for every square $Q$. To begin with, note that $\mu(Q)$ is finite. Divide $Q$ into $N^2$ equal subsquares, where $N$ is something like $100/c$; this will make sure that at least one of the subsquares lies in the complement of $E$. The doubling property implies that $\mu$ is distributed somewhat fairly between subsquares: namely, each subsquare gets at least $\epsilon \,\mu(Q)$ where $\mu$ depends only on $N$ and on the doubling constant of the measure. Removing a subsquare that lies outside of $E$ from consideration, we are left with $N^2-1$ subsquares with total measure at most $(1-\epsilon)\,\mu(Q)$. Repeating this divide-and-remove process, we obtain $\mu(E\cap Q)\le (1-\epsilon)^k\,\mu(Q)$ for every $k$. Thus $\mu(E\cap Q)=0$. $\quad \Box$

This just in! A doubling measure on $\mathbb{R}^d$ can charge a rectifiable curve.

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