$B_x = \{y \in [0,1]: x-y \in \Bbb{Q}\}$, $ \varepsilon=\{C \subset [0,1]: \exists x \quad C=B_x\} $. By the axiom of choice we can choose exactly one element of each equivalence class $\varepsilon$ and make a set $P$. Let $m$ be the Lebesgue measure

I could prove that if $E \subset P$ is measurable then it must be that $m(E) = 0$. Now I have to prove that if $E \subset [0,1)$ and $m^*(E)>0$ then there is a subset of $E$ that is not Lebesgue measurable.

I am confused with proving that this is not Lebesgue measurable with the outer measure. I was trying to use subadditivity, but I could not find a solution. Any help is appreciated!

  • $\begingroup$ I guess you are not assuming throughout that $E$ is a subset of $P$? $\endgroup$ – Anguepa Feb 22 at 11:48
  • $\begingroup$ The question is actually two steps, but it says that I should use my proof that $m(E)=0$. $\endgroup$ – The Bosco Feb 22 at 11:49
  • $\begingroup$ If $E$ were measurable then $m(E)=m^*(E)>0$. Contradiction. That is if you are assuming $E \subset P$. $\endgroup$ – Anguepa Feb 22 at 11:54
  • $\begingroup$ I know that is the case, but it seems too simple for this question. My guess is the question is just badly worded. What would be the case if then $m(E) \neq 0$? Or how can I prove with countable subadditivity that both measures have to match? $\endgroup$ – The Bosco Feb 22 at 11:55
  • $\begingroup$ Both measures have to match by definition of the Lebesgue measure as a completion of the Borel measure. What I was thinking is that maybe you were not assuming that $E$ is a subset of the Vitali set, and that if $m^*(E)>0$ then you might always be able to construct a subset similar to the Vitali set that is not measurable. I don't know if this is true. $\endgroup$ – Anguepa Feb 22 at 12:36

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