$\mu(L)=0$ for every line $L\subset\mathbb{R}^2$ when $\mu$ is translational invariant Borel-measure

Question

I have next kind of problem: if $\mu$ is Borel measure in $\mathbb{R}^2$ which is translational invariant for every Borel set and $\mu(K)<\infty$ for every compact set $K\subset\mathbb{R}^2$, then $\mu(L)=0$ for every line $L$. I have literally no idea how to start, so every hint is welcome.

Answer 1

It suffices to show that the measure of segments is zero, since lines are countable unions of segments.

Let k be the finite measure of $(0,1]^2$. Note that every translation of a given square has the same measure. Now by additivity and translation invariance you can show that $(0,2^{-j}]^2=4^{-j}k$.

Then you can cover any given segment by a large number of small squares, and the total measure of these squares can be made small.

I let you fill the details.

Answer 2

To simplify, I consider the line $D=\{y=0\}$. Consider $I=\{(x,y),0\leq x\leq 1, 0\leq y\leq 1\}$. You decompose this as $\bigcup I^p_n$ such that $I_n=\{(x,y), 0\leq x\leq 1, p/n\leq y\leq {{p+1}\over n}\}$, $\mu(I_n^i)=\mu(I_n^j)$ since $I_n^i$ is a translation of $I_n^j$.

Write This implies that $\mu(I_n^i)=\mu(I)/n$. You can divides $D$ as the union of $U_1=\{[n,n+1], n\geq 0\}, U_2=\{[n,n+1] n<0\}$. For every $m>0$, there exists a translation of $I_{ m^{n+1}}^0$ which contains $[n,n+1], n>0$. This implies that $\mu(U_1)\leq \sum_{n>0}{\mu(I)\over m^{n+1}}={\mu(I)\over m}{1\over{1-{1\over m}}}.$ This implies that $\mu(U_1)=\mu(U_2)=\mu(D)=0$.

Answer 3

Suppose that it is not true and $L$ denotes a line with $\mu(L)>0$.

WLOG we may assume that $L$ contains the origin.

Then if $a$ is another element of $L$ we can write:$$L=\bigcup_{n\in\mathbb Z}(na+D)$$where $D$ denotes the bounded set $\{\lambda a\mid\lambda\in[0,1)\}\subset L$.

Then: $$0<\lambda(L)=\sum_{n\in\mathbb Z}\mu(na+D)=\sum_{n\in\mathbb Z}\mu(D)$$and consequently $\mu(D)>0$.

However it is easy to create a compact set that contains infinitely many disjoint translations of $D$, so that such a set must have measure $+\infty$.