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Context.

Let's denote by $\lambda$ Lebesgue's measure on $\mathbb R^4$. I have a subset of $\mathbb R^4$, let's call it $E$ of full Lebesgue measure. I have another set $F$, and I would like $E\cap F$ to be non-empty. The issue is that I don't have much information about $E$, so if by chance $\lambda(F)>0$, this would solve my problem.

The question.

Let's denote by $F$ the following set:

$$F:=\{(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb R^4,\ \xi_1\xi_4-\xi_2\xi_3\in\mathbb Q\}.$$

Do we have $\lambda(F)>0$ where $\lambda$ stands for Lebesgue's measure?

Remarks.

It seems to me that it could be possible that $\lambda(F)>0$, even if $F$ is not so big, since at least $F$ isn't countable. If I had to guess, I would say that $\lambda(F)=0$, but I am not able to prove it, which gives me hope that the opposite is true.

E. Joseph
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2 Answers2

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For each $r\in \mathbb{Q}$ the set $\{ (\xi_1,...,\xi_4) \in \mathbb{R}^4: \xi_1\xi_4 - \xi_2\xi_3 - r = 0 \}$ is a $0$-set of a polynomial, which is known to have Lebesgue measure $0$ (see here for instance). Now we get a countable union of measure $0$-sets.

Hayk
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It is almost immediate from Fubini's Theorem that $\{(\xi_1,\xi_2,\xi_3,\xi_4): \xi_1\xi_2-\xi_1\xi_3=q\}$ has measure zero for each rational number $q$: just integrate w.r.t $\xi_1$ first (you will get integral over a singleton). Taking union over all $q$ we get $\lambda (F)=0$

Kavi Rama Murthy
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