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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!

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  • $\begingroup$ Maybe this helps: the last inequality can be rewritten: $$\frac{1}{2} >\frac{a_2 - b_2}{a_1 - b_1}\text{,}$$ therefore it suffices to find a line $p$ with the equation $y = kx + n$, such that $0 < k < 1/2$ and $|p\cap S|\geq 2$. Then you take the ball $B$ (maybe squares are more appropriate) and find the two points in the ball. $\endgroup$ – Antoine Aug 26 '15 at 23:27
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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).

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  • $\begingroup$ Thank you very much, zhw! This really helps. $\endgroup$ – MerylStreep Aug 27 '15 at 21:14
  • $\begingroup$ Dear @zhw. , I started to go through your argument in detail and got stuck at getting the inequality above for the contradiction. Please could you clarify the steps to obtain m((a1,a2)+Dr)/m(Dr)≤(2π−arcsin(1/2))/2π? $\endgroup$ – MerylStreep Aug 31 '15 at 14:39
  • $\begingroup$ Thank you for the question, there was a typo. On the left we want to intersect with $E.$ I edited it. $\endgroup$ – zhw. Aug 31 '15 at 19:19
  • $\begingroup$ This is what I thought. Thanks! Would upvote you if I could. $\endgroup$ – MerylStreep Sep 1 '15 at 1:05

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