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Let $\mu$ be a locally finite Borel measure on $\mathbb{R}^2$, and for every $r\in \mathbb{R}^+$ , $\mu(B(x,2r))<C\mu(B(x,r))$ for some $C\in \mathbb{R}$,where $B(x,r)$ is Euclidean open ball at $x$ with diameter $r$.

how to show that every straight line in $\mathbb{R}^2$ is $\mu$-null set.

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May $C$ depend on $r$? On $x$? –  martini Nov 8 '12 at 12:48
    
@martini $C$ is constant. –  Leitingok Nov 8 '12 at 13:52
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up vote 3 down vote accepted

This is a nice little problem of geometric measure theory, solvable by elementary manipulations.

One may start by upgrading the hypothesis to $\mu(B(x,3r))\leqslant C\mu(B(x,r))$ for every $x$ and $r$, for some positive finite $C$.

Next, consider the points $x=(0,1)$ and $y=(0,-1)$. The ball $B(x,1)$ is a subset of the ball $B(y,3)$ hence $\mu(B(x,1))\leqslant\mu(B(y,3))\leqslant C\mu(B(y,1))$.

More generally, any two points $x\ne y$ are such that $\mu(B(x,r))\leqslant C\mu(B(y,r))$ for $r=\frac12\|x-y\|$.

Consider now $I=(0,1)\times\{0\}$ and $R(n)=(0,1)\times(\frac1{2n},\frac3{2n})$, for every positive integer $n$. Thus $I$ is a unit horizontal interval and each $R(n)$ a rectangle of size $1\times\frac1n$. Then $I$ is included in the disjoint union of the $n$ balls $B_k^n=B(x_k^n,\frac1{2n})$, where $x_k^n=(\frac{2k-1}{2n},0)$ and $R(n)$ contains the disjoint union of the $n$ balls $\bar B_k^n=B(\bar x_k^n,\frac1{2n})$, where $\bar x_k^n=(\frac{2k-1}{2n},\frac1n)$, for $1\leqslant k\leqslant n$.

For every $k$, the balls $B_k^n$ and $\bar B_k^n$ are centered at points at distance twice their radius. Hence, summing their contributions yields $C\mu(R(n))\geqslant C\sum\limits_{k=1}^n\mu(\bar B_k^n)\geqslant\sum\limits_{k=1}^n\mu(B_k^n)\geqslant\mu(I)$.

Let us choose some sequence of integers $(n_i)_i$ such that the rectangles $R(n_i)$ are disjoint, for example $n_i=3^i$ for every $i\geqslant1$. Since every $R(n_i)$ is included in the rectangle $Q=(0,1)\times(0,\frac12)$, this yields $\mu(Q)\geqslant\sum\limits_{i\geqslant1}\mu(R(n_i))\geqslant C\sum\limits_{i\geqslant1}\mu(I)$. The only way $\mu(Q)$ can be finite is if $\mu(I)=0$.

Likewise, the measure of every interval is zero and, by countable union, so is the measure of every line.

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Well, you can cover any finite section of the line with one ball. Then you can cover it again using 2 balls of half the radius and so on. That is when you use the inequality you have to show that actually the finite section has measure 0.

Then use that a countable union of null sets is null to get the result for the whole line.

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Could you provide a more formal argument? –  Davide Giraudo Nov 8 '12 at 16:01
    
Not sure this works if $C\geqslant2$. –  Did Nov 8 '12 at 16:08
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