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


1 Answer 1


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.

  • $\begingroup$ Moreover you are saying that we won't lose any measure in obtaining a rectangle. That is great. $\endgroup$ Jul 28, 2020 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.