Let $E\subset \mathbb{R}^2$ such that $\overline{E}$ has a positive measure in $\mathbb{R}^2$. Can we always find $A,B \subset\mathbb{R}$ such that $\overline{A},\overline{B}$ is of positive measure in $\mathbb{R}$ and $A×B\subset E$ ?
I tried this but not sure. Can someone confirm?
Proof: Let $$ M=\{(x,y) \in [0,1]\times[0,1] \mid x-y \not\in\mathbb{Q}\}.$$ Then $|M|_2 = 1 > 0$. So by inner regularity there exists a compact set $C \subseteq M$ such that $|C|_2 > 0$. Now let $E = C$. Then $\overline{E} = C$) (since $C$ is closed), which implies that $|\overline{E}|_2 > 0$.
If possible let $\exists A,B \subseteq \mathbb{R}$ such that $A\times B \subseteq E$ and $|\overline{A}|_1, |\overline{B}|_1 > 0$. Then $$\overline{A}\times\overline{B} = \overline{A\times B} \subseteq \overline{E} = C \subseteq M. $$ So $M$ contains a positive measured square, which is a contradiction. Hence the answer to the question is no.
The answer to the title question is that no, a Borel measurable subset of $\mathbb{R}^2$ need not contain a measurable rectangle of positive measure. See the first answer to the following: Measurable rectangles inside a non-null set.