# Problem regarding set of positive Lebesgue measure in $\mathbb{R}^2$

Let $$T_{p,q}$$ be line joining $$(0,0)$$ and $$(p,q).$$ Now let us define the set $$L= \bigcup_{p\in[0,1]\cap \mathbb{Q}}T_{p,1} \bigcup_{q\in[0,1]\cap \mathbb{Q}}T_{1,q}$$ and consider $$P=[0,1]\times[0,1]\setminus L.$$ $$P$$ should be a set of positive Lebesgue measure.

Question: Does there exist set of positive Lebesgue measure, $$A,B \subset \mathbb{R}$$ such that $$A\times B\subset P?$$

Addition after 1st comment:

The above property also not holds for the set $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin \mathbb{Q}\}.$$ This is classic example and can be proved using Steinhaus theorem. But a suitable translation and rotation of $$D$$ contains $$A\times B,$$ where $$A,B$$ are sets of positive Lebesgue measure. This leads me to ask the above question since if $$P$$ doesn't have the above property, then any translation and rotation of $$P$$ would not also have this property.

(I believe that $$P$$ wouldn't contain $$A\times B,$$ where $$A,B$$ are sets of positive Lebesgue measure but I can't prove it as it doesn't have a nice definition, unlike $$D$$, to apply Steinhaus theorem.)

• Would be nice if you gave some context or evidence of some effort; otherwise just looks like a classic case of copypasting your homework
– T_M
Aug 5 at 17:49
• @T_M I added a context of my question. This is not a homework problem. Aug 6 at 3:00
• Is symbol $\cup$ missing in the formula $\ L\ =\ \ldots\$ (the second line from the top)? Aug 9 at 3:08

## 1 Answer

There are no such $$A,B$$. If $$A\subset[0,1]$$ is a Lebesgue measurable set of positive measure, the set $$\mathbb Q_+ A:=\{qa: q\in\mathbb Q_+,\, a\in A\}$$ has full measure in $$\mathbb R_+$$, for it has at least one point of density $$1$$ and it is invariant by multiplication by positive rationals. Therefore if $$B\subset[0,1]$$ is a Lebesgue measurable set of positive measure, $$\mathbb Q_+ A\cap B\neq\emptyset$$, that is $$qa=b$$ for some $$q\in\mathbb Q_+, \ a\in A ,\ b\in B$$, that is $$(a,b)\in (A\times B)\cap L$$.

• Thanks. I missed the fact the lines are $y=qx, q\in \mathbb{Q}^+$. Appreciate your efforts. Aug 11 at 3:57