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.)

  • $\begingroup$ Would be nice if you gave some context or evidence of some effort; otherwise just looks like a classic case of copypasting your homework $\endgroup$
    – T_M
    Aug 5 at 17:49
  • $\begingroup$ @T_M I added a context of my question. This is not a homework problem. $\endgroup$
    – user483450
    Aug 6 at 3:00
  • $\begingroup$ Is symbol $\cup$ missing in the formula $\ L\ =\ \ldots\ $ (the second line from the top)? $\endgroup$
    – Wlod AA
    Aug 9 at 3:08

1 Answer 1


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 $(a,b)\in (A\times B)\cap L$.

  • $\begingroup$ Thanks. I missed the fact the lines are $y=qx, q\in \mathbb{Q}^+$. Appreciate your efforts. $\endgroup$
    – user483450
    Aug 11 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.