What is Lebesgue measure of the set $\{(a,b) \in \mathbb{R}^2 \mid a-b \in\mathbb{Q}\}$ in $\mathbb{R}^2$ ? I am guessing this is exactly measure of the diagonal subset, but unable to say rigorously.
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Rotate it through $45^\circ$ and it becomes $\sqrt 2(\Bbb Q\times\Bbb R)$, which has measure $0$.
TonyK
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Why $\sqrt 2(\Bbb Q\times\Bbb R)$ has measure zero ? – dragoboy May 27 '20 at 08:54
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The section of this set by $a$ is $E_a=\{b: a-b \in \mathbb Q\}$ which is a contable set. Hence each section has measure $0$ and Fubini's Theorem tells you that the set has measure $0$.
[By Fubini's Theorem $m_2(E)=\int m_1(E_a) dm_1(a)$ where $m_1$ is Lebesgue measure on $\mathbb R$ and $m_2$ is Lebesgue measure on $\mathbb R^{2}$].
Kavi Rama Murthy
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I am not getting how zero measure of $E_a$ implies my required set has zero measure. – dragoboy May 27 '20 at 08:56
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@dragoboy By Fubini's Theorem $m_2(E)=\int m_1(E_a) dm_1(a)$ where $m_1$ is Lebesgue measure on $\mathbb R$ and $m_2$ is Lebesgue measure on $\mathbb R^{2}$ – Kavi Rama Murthy May 27 '20 at 08:58