Measurable sets in $[0,1]$

Question

Suppose $E$ is a Lebesgue measurable sets in $[0,1]$ such that $m(E)$ is positive. Show that there exist two points $x,y$ in $E$ such that $x-y$ is irrational.

I tried to prove that $f(x)=m(E\cap [0,x])$ is continuous, but I don't know how to show that E has two points whose difference is irrational. Can someone give me some hint? Thanks

Have you tried to prove it by contradiction? – Aug 08 '17 at 01:37 — , Aug 08 '17 at 01:37
This is a good idea! Thank you! – user335468 Aug 08 '17 at 01:38 — user335468, Aug 08 '17 at 01:38
Also related to https://math.stackexchange.com/q/38902/9464 – Aug 08 '17 at 01:39 — , Aug 08 '17 at 01:39

score 3 · Accepted Answer · answered Aug 08 '17 at 01:52

3

If $x-y$ is rational for all $x, y\in E$, then take any $x\in E$ and note that $E\subseteq S := \{q+x : q\in \mathbb{Q}\cap [-1, 1]\}$. Therefore, as $m(S) = m(\mathbb{Q}\cap [-1, 1]) = 0$, we have $m(E) = 0$. The contrapositive of this is that if $m(E) > 0$, then $x-y$ cannot be rational for all $x, y\in E$.

answered Aug 08 '17 at 01:52

Michael L.

5,521

Seems to me you don't need measure for this, just that $E$ is uncountable. – bof Aug 08 '17 at 01:58
@bof That's correct. If all differences in $E$ are rational, then $E$ must be countable. – Michael L. Aug 08 '17 at 02:35

Measurable sets in $[0,1]$

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