Let $\mu$ be a locally finite Borel measure on $\mathbb{R}^2$, and for every $r\in \mathbb{R}^+$ , $\mu(B(x,2r))
how to show that every straight line in $\mathbb{R}^2$ is $\mu$-null set.
Let $\mu$ be a locally finite Borel measure on $\mathbb{R}^2$, and for every $r\in \mathbb{R}^+$ , $\mu(B(x,2r))
how to show that every straight line in $\mathbb{R}^2$ is $\mu$-null set.
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.
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.