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

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 $$B_1^s\times B_2^t\subset \overline{P\setminus Z}.$$

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.