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.)



Problem regarding set of positive Lebesgue measure in $\mathbb{R}^2$
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 $(a,b)\in (A\times B)\cap L$.


