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 $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin\mathbb{Q}\}.$$
 A: 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.
A: 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.
