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Let $S = \mathbb{R}^2 \setminus \{ (x,y): x+y \in \mathbb{Q} \}.$ Show that there are no Lebesgue measurable sets $A, B \subset \mathbb{R}$ of positive Lebesgue measure for which $A \times B \subset S$.

I don't really know what I'm given and what to work with. It's obvious that $A, B$ cannot have any rational numbers. But I'm not sure how to start. Could someone help me with this?

Airdish
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  • Why is it obvious that $A$ and $B$ have no rational numbers? – Julian Mejia May 28 '19 at 18:47
  • Show that $S$ has measure zero and $A \times B$ has positive measure. – copper.hat May 28 '19 at 18:51
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    @copper.hat I do not think that $S$ has measure $0$. Because the complement of $S$ is the set of all lines with gradient $-1$ and rational $y$-intercepts. Since each line has measure $0$ and it is a countable collection, the complement must have measure $0$. – Airdish May 28 '19 at 21:16
  • My apologies, I thought it was bad MathJax. You are correct. – copper.hat May 28 '19 at 21:17
  • @Airdish: Perhaps you can use a result like https://math.stackexchange.com/q/1079464/27978? – copper.hat May 28 '19 at 21:18
  • @copper.hat not really sure how I could use that. Do you have any suggestions? – Airdish May 28 '19 at 21:23
  • Here is a more appropriate result https://math.stackexchange.com/a/1276487/27978. Here is a possible approach: (i) Note that $S$ cannot contain an line segment perpendicular to the planes in $S$. (ii) Show that if $A,B$ have positive measure, then $A-B$ contains an open interval (of the real line). – copper.hat May 28 '19 at 22:57

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The main result here is that if $A,B$ have positive measure then $A+B$ contains a non trivial interval (see https://math.stackexchange.com/a/1276487/27978 for a proof using convolution).

Consider the map $T((x,y)) = x+y$ and note that $(x,y) \in S$ iff $T((x,y)) \notin \mathbb{Q}$.

The above result shows that $T(A \times B) = A+B$ contains a non trivial interval, in particular, there is some $(a,b) \in A \times B$ such that $T((a,b)) \in \mathbb{Q}$ and so $(a,b) \notin S$.

copper.hat
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