Let $P\subset \mathbb{R}^2$ be a set of positive Lebesgue measure. Is it always true that a suitable rotation and translation of $P$ always contains a set of the form $\{re^{i\theta}:r\in E, \theta\in [0,2\pi)\}$ or $A×B,$ where $E,A,B$ are sets of positive Lebesgue measure in $\mathbb{R}?$


Note: I can show that the two types of positive measure sets, mentioned above, are different (in the sense that no one type contains the other type always).



Property of sets of positive Lebesgue measure in $\mathbb{R}^2$
Firstly, a set $P$ of positive measure need not contain anything of the form $A\times B$, for example consider for some $k\in\mathbb{R}\setminus\{0\}$ the set $P=\{(x,y)\in\mathbb{R}^2;x-ky\not\in\mathbb{Q}\}$. Then the complement of $P$ is a null set, and any set $A\times B$ where $A,B$ have positive measure contains some point $(x,y)$ with $x-ky\in\mathbb{Q}$. This is because the set $A-kB$ [contains an open interval](https://math.stackexchange.com/questions/596273/sum-of-positive-measure-set-contains-an-open-interval). We can substitute $\mathbb{Q}$ by any other set dense in $\mathbb{R}$.


Now let $P=\{(x,y)\in\mathbb{R}^2;x\not\in\mathbb{Q},y\not\in\mathbb{Q},x-y\not\in\mathbb{Q}\}$. Any rotation/translation of $P$ cannot contain a set $A\times B$, with $A,B$ of positive measure, because then $A\times B$ would not intersect some set of parallel lines dense in $\mathbb{R}^2$ and with non zero slope, and we get a contradiction as in the previous case.


The same set $P$ also doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure. To check that, it will be enough to check that every point $p\_0\in\mathbb{R}^2\setminus\{0\}$ with polar coordinates $(r\_0,\theta\_0)$ has a neighborhood $U$ such that $U$ doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure.


This is true because near $p\_0$, any set of parallel lines dense in $\mathbb{R}^2$ can be expressed as $\{r(k\cos(\theta)+l\sin(\theta))\not\in Q\}$, for $k,l$ not both $0$ and $Q$ some dense set in $\mathbb{R}$. We can suppose that $p\_0\not\in\{k\cos(\theta)+l\sin(\theta)=0\}$, so after changing $Q$ by $-Q$ if necessary, we can take logarithms and the equation of the parallel lines becomes $f(r)+g(\theta)\not\in\{\ln(q);q\in Q\cap\mathbb{R}^+\}$, where $f(r)=\ln(r),g(\theta)=\ln(k\cos(\theta)+l\sin(\theta))$. However, if $E,F$ have positive measure, then $f(E)$ and $g(F)$ have positive measure, so $f(E)+g(F)$ contains some open interval, so the same reasoning of before works.


