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$E_x = \{y \in [0,1]: x-y \in \Bbb{Q}\}$, $ \varepsilon=\{A \subset [0,1]: \exists x \quad A=E_x\} $ .We chose one element from each set of family $\varepsilon$. This is a Vitaly set $V$.

Prove that if $E$ is measurable and $E \subset V$ then $E$ has measure $0$.

$E_q = [0,1] \bigcap \Bbb{Q} $, $q \in \Bbb{Q} $

I don't know how $E$ looks. I know for example that every singleton is measurable and has measure zero. But I don't know how to explain that every measurable set of $V$ has measure zero.

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  • $\begingroup$ I never saw the notation $[0;1]$ for the interval $[0,1]$ before. $\endgroup$ – miracle173 Jan 30 '18 at 21:22
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Consider

$$E_{\mathbb Q} = \bigcup_{\substack{r \in \mathbb Q \\ -1 \le r \le 1}} (E+r) \subseteq [-1,2]$$

This is a countable infinite union of disjoints subsets, each of those having the measure of $E$. If the measure of $E$ would be strictly positive, $E_{\mathbb Q}$ would have an infinite measure, in contradiction with $E_{\mathbb Q} \subseteq [-1,2]$.

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  • $\begingroup$ Why is important that sets are disjoint? $\endgroup$ – Hikicianka Jan 31 '18 at 11:49
  • $\begingroup$ Imagine, extreme case that all those sets would be equal to $E$. Then the measure of the union will be equal to the one of $E$ and the argument of infinite measure will fall apart. $\endgroup$ – mathcounterexamples.net Jan 31 '18 at 12:48

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