# Subset of Lebesgue measurable subset of Vitali set is NOT Lebesgue measurable

$$B_x = \{y \in [0,1]: x-y \in \Bbb{Q}\}$$, $$\varepsilon=\{C \subset [0,1]: \exists x \quad C=B_x\}$$. 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!

• I guess you are not assuming throughout that $E$ is a subset of $P$? – Anguepa Feb 22 at 11:48
• The question is actually two steps, but it says that I should use my proof that $m(E)=0$. – The Bosco Feb 22 at 11:49
• If $E$ were measurable then $m(E)=m^*(E)>0$. Contradiction. That is if you are assuming $E \subset P$. – Anguepa Feb 22 at 11:54
• 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? – The Bosco Feb 22 at 11:55
• 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. – Anguepa Feb 22 at 12:36