Let $A \subset [-1, 1]$ Lebesgue measurable with $\lambda (A) > 1$. Why does it hold true that $1 \in A-A$, where $ A -A =$ {$x-y: x, y \in A$} ?
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Better think of it as a pigeonhole principle for reals mod 1. – user120527 Nov 07 '19 at 09:58
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Consider $A\cap[0,1]-1$ and $A\cap[-1,0]$. By $\lambda(A)>1$, they must have non-empty intersection.
Arthur
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Thank you for this answer. But why does follow from $\lambda(A) > 1$ that the intersection of the two guys is not empty? – StMan Nov 07 '19 at 10:29
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Because $\lambda(A)= \lambda(A\cap[0,1]-1)+\lambda(A\cap[-1,0])$. And the two latter sets are subsets of $[-1,0]$, which has measure $1$. Two subsets of $[-1,0]$ with total measure greater than $1$ must have non-empty intersection. – Arthur Nov 07 '19 at 10:33
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Thank you for this comment. Now it's clear. But why does then follow from this that $1 \in A-A$? – StMan Nov 07 '19 at 10:47
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Consider what an overlap means. It means that there is some $y\in A\cap [-1,0]$ such that $y+1\in A\cap [0,1]$. Now subtract these two numbers from one another. – Arthur Nov 07 '19 at 11:07
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