# Does a given positive measure set in $\mathbb{R}^2$ contains product of two 1 dimensional positive measure set?

In $$\mathbb{R}^2$$ consider the 2 dimensional positive Lebesgue measure set, $$D=\{re^{it}:t\in[0,2\pi),r\in[0,1]\setminus\mathbb{Q}\} .$$ Does there exist sets $$A,B\subset\mathbb{R}$$ of 1 dimensional positive Lebesgue measure such that $$A\times B\subset D?$$

Note: The above is not true if $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin\mathbb{Q}\}.$$

No, it there were two such sets $$A$$, $$B$$, then with the map $$(x,y) \mapsto (x^2, y^2)$$ ($$t \mapsto t^2$$ bi-Lipschitz on compacts inside $$(0, \infty)$$) , it would transform into $$A_1$$, $$B_1$$ of measure non-zero inside $$\{(x,y)\ | \ x+y \not \in \mathbb{Q}^2\}$$, which is not possible.
• $r\notin \mathbb{Q}$ doesn't imply $r^2\notin \mathbb{Q}$. Then how we conclude that $D$ maps inside $\{(x,y):x+y\notin \mathbb{Q}\}?$ Dec 9 '20 at 14:25
• I guess of I defined $D$ to be $\{re^{it}:r^2\notin \mathbb{Q}\}$ then your argument works perfectly. Am I right? Dec 9 '20 at 14:30
• @Oh, I see, it transforms into $x + y \not \in \mathbb{Q}^2$, and again get a contradiction ( only use $\mathbb{Q}^2$ is dense, and nothing else) Dec 9 '20 at 21:40
• I am not getting why you need to take $A,B$ away from 0? Dec 10 '20 at 2:09
• @Prof.Hijibiji: You want to the image not to be of measure $0$. You guarantee that if the inverse $t\mapsto\sqrt{t}$ is Lipschitz. Dec 10 '20 at 2:14
Let $$A$$ and $$B$$ be defined by \begin{align*} A & =\big\{a\in\mathbb{R}: a^2=\sum_{i=1}^\infty a_i10^{-2i}, a_i\in\{1,2\,\dots,9\}, a^2\not\in \mathbb{Q}\big\} \\ B & =\big\{b\in\mathbb{R}: b^2=\sum_{i=1}^\infty b_i10^{-2i+1}, b_i\in\{1,2,\dots,9\}, b^2<\frac12, b^2\not\in \mathbb{Q}\big\}. \end{align*} Note that if $$r^2=a^2+b^2\in \mathbb{Q}$$ then the decimal expansion of $$r^2$$ will eventually have a finite length repeating sequence. Any such sequence would have to come from a unique repeating sequence of even place decimals from $$a^2$$ and repeating odd place decimals from $$b^2$$. Since $$a^2,b^2\not\in \mathbb{Q}$$, such repeating sequences do not exist, thus $$r^2\not\in\mathbb{Q}$$. Not entirely sure about the measure of $$A$$ and $$B$$, but for their squares, I think the measures are positive.
• Actually you need to show that $A\times B\cap D$ is positive Lebesgue measure in $\mathbb{R}^2$ which could not be true as we can see from the other answer. Thank you for your participation. Appreciation Dec 11 '20 at 9:16