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 $$B\_1^s\times B\_2^t\subset \overline{P\setminus Z}.$$



Problem regarding Lebesgue measure in $\mathbb{R}^2$
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.


