I am trying to see whether it is true that in any set of a positive Lebesgue measure in $R^2$ we can always find two points $(a_1,a_2)$ and $(b_1,b_2)$ such that the following hold:
$a_1>b_1$
$a_2>b_2$
$a_1-2a_2>b_1-2b_2$.
I think the starting point can be the fact that for any set $S$ of a positive Lebesgue measure in $R^2$ we can find a ball $B$ in $R^2$ such that $S\cap B$ has a positive Lebesgue measure in $R^2$ but not sure how to proceed with the proof from here.
Any advice would be greatly appreciated!
Define $A=\{(x,y): x<0,y<0, x-2y <0\}.$ This is the open region in the third quadrant between the $x$-axis and the line $y=x/2.$ Note that if $D_r$ is the open disc centered at $(0,0)$ of radius $r,$ then
$$\frac{m(A\cap D_r)}{m(D_r)} = \frac{\arcsin (1/2)}{2\pi}$$
for all $r>0.$ Here $m$ is Lebesgue measure on $\mathbb {R}^2.$
Assume $E\subset \mathbb {R}^2, m(E)>0.$ All we seek is an $(a_1,a_2) \in E$ such that $[(a_1,a_2) + A]\cap E$ is nonempty. In fact any $(a_1,a_2) \in E$ that is a point of (symmetric) density of $E$ will work. To see this, note that if $(a_1,a_2)$ is a point of density of $E$ and the above set is empty, then
$$\frac{m([(a_1,a_2) + D_r]\cap E)}{m(D_r)} \le \frac{2\pi - \arcsin (1/2)}{2\pi}$$
for all $r.$ Thus $(a_1,a_2)$ can't be a point of density, contradiction. Thus for all points of density $(a_1,a_2)$ in $E,$ we can find $(b_1,b_2)\in E$ to give us the conclusion. By Lebesgue of course, a.e. $(a_1,a_2) \in E$ is a point of density, so we not only have one instance, but an embarrassment of riches (as often happens in the Lebesgue setting).