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Let $P=A_1\times A_2,$ where $A_1,A_2\subset \mathbb{R}$ are set of positive Lebesgue measure, and $Z\subset \mathbb{R}^2,$ be a set of zero Lebesgue measure. Can we always find positive Lebesgue measure sets $B_1,B_2\subset \mathbb{R}$ such that $$B_1\times B_2 \subset \overline{P\setminus Z}?$$ What extra conditions ensure that the above is true?(I can show that the above is true if $P\setminus \overline{Z}$ is of positive measure then the above is true)

In this question https://math.stackexchange.com/q/3767758/641816, it was shown that the result is true if $A_1=A_2=[0,1]$.

This is my attempt: Since $A_1,A_2$ are positive Lebesgue measure set we can find $a_1\in A_1, a_2\in A_2$ such that for any $r>0$ we have $B(a_1,r)\cap A_1, B(a_2,r)\cap A_2$ are sets of positive measure(in fact this phenomenon is true for almost every $a_1\in A_1,a_2\in A_2$). Consider $$B_1^r=\overline{B(a_1,r)\cap A_1},\quad B_2^r=\overline{B(a_2,r)\cap A_2}$$ Then I think somehow one can show that there exits some $s,t>0$ such that $$B_1^s\times B_2^t\subset \overline{P\setminus Z}.$$

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For $i=1,2$, let $$Q:=C_1\times C_2,$$ where $$C_i:=\{x\in A_i\colon\forall r>0\ |B(x,r)\cap A_i|>0\},$$ $B(x,r):=(x-r,x+r)$, and $|\cdot|$ denotes the Lebesgue measure in $\mathbb R^d$, for any $d\ge1$. Then $|C_i|=|A_i|>0$, by (say) the Lebesgue density theorem.

For all $(x_1,x_2)\in Q$, all real $r>0$, and all $i\in\{1,2\}$ $$|(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)\setminus Z| =|(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)| =|(B(x_1,r)\cap A_1)|\ |(B(x_2,r)\cap A_2)|>0.$$ So, for each $(x_1,x_2)\in Q$ and each $r>0$ there is some $$(y_1,y_2)\in(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)\setminus Z \\ =(B(x_1,r)\times B(x_2,r))\cap P\setminus Z.$$ Thus, $$C_1\times C_2 =Q\subset \overline{P\setminus Z}$$ and $|C_i|=|A_i|>0$ for $i=1,2$, as desired.

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  • $\begingroup$ Moreover you are saying that we won't lose any measure in obtaining a rectangle. That is great. $\endgroup$ – Prof.Hijibiji Jul 28 '20 at 4:10

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