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Given a Lebesgue measurable set $E$ of finite measure, how can one construct a nonmeasurable set $N$ with $N \subset E$?

I asked my professor this, and she said that one needs to perform the construction of the Vitali set on elements of $E$. Her argument was along the following lines:

Suppose $E \subset \mathbb{R}$. Given that $m(E) < \infty$, there exists an interval $[a, b]$ such that $E \subset [a, b]$. Let $\{r_k\}$ be an enumeration of the set $\mathbb{Q} \cap [a, b]$. Then construct a set $N$ whose elements are the representatives of the equivalence classes of the following equivalence relation: $$ x \sim y \iff x - y = r \in \mathbb{Q}\cap[a, b]. $$ Then one can show that $N$ is nonmeasurable by contradicting translation invariance.

However, none of this sits right with me. My professor didn't really give any info on how we get $[a, b]$. Sure such an interval exists, but which do we pick? I can't imagine it works with every such interval. Also, what's even weirder, is that $N$ isn't a subset of $E$.

I asked a similar question before on MSE with no response, so if someone can tell me what my professor was trying to say or if this is just not possible, I would really appreciate it.

trujello
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  • take a look here, or here. Or searching in the web for "proof vitali set is not measurable" you get a bunch of explanations and proofs – Masacroso Oct 31 '19 at 01:15
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    but it is true that the statement "given some $E\subset \Bbb R $ such that $m(E)<\infty $ exists some interval $[a,b]$ that $E\subset [a,b]$". This is clearly false because $\Bbb Q \cup [0,1]$ is not contained in any closed interval in $\Bbb R $ and $m(\Bbb Q\cup [0,1] )=1$ – Masacroso Oct 31 '19 at 01:21
  • @Masacroso Thanks for your response! I understand how to construct the Vitali set. But my professor claims that it can be done inside any set of finite measure, although based on her explanation, which appears to be flawed, I'm not any closer in learning how one would go about doing this. I'm sure she wasn't saying nonsense on purpose, and I'm inclined to believe that there is a way to do what she said is possible. – trujello Oct 31 '19 at 07:12
  • Possible duplicate of Non measurable subset of a positive measure set – AGF Oct 31 '19 at 07:35
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    Are you certain she said there exists $[a,b]$ with $E\subset[a,b]$? That's obviously nonsense. A non-nonsensical thing she might have said: There exists $[a,b]$ such that $E\cap[a,b]$ has positive measure, hence we may assume that $E\subset[a,b]$. – David C. Ullrich Oct 31 '19 at 15:47

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