3
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We know that the Lebesgue measure obtained via the usual Caratheodory extension is complete. As such, the subset of every null set is null.

Is it possible to prove that every non-null measurable subset $A\subseteq \mathbb R$ contains a non-measurable subset $B\subseteq A$?

I suppose it must be true, if it has non-empty interior then it must be true (just use an isometry to take a "copy" of a vitali set inside the interval).

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You can just imitate the usual construction - let $ A $ be a measurable subset of $ \mathbb R $, and choose a bounded interval $ I $ such that $ A \cap I = X $ has nonzero measure. By translating and scaling, we may wlog assume $ I = [0, 1] $. Let $ V $ be a set of representatives in $ I $ for the cosets of $ \mathbb R / \mathbb Q $ intersecting $ X $ (obviously $ V $ may be chosen to be a subset of $ X $, and thus of $ A $), and note that choosing an enumeration $ q_k $ of the rationals in $ [-1, 1] $ and defining $ V_k = V + q_k $ gives that

$$ \mu(X) \leq \sum_{k=1}^{\infty} \mu(V_k) = \sum_{k=1}^{\infty} \mu(V) \leq 3 $$

which is a contradiction, since $ \mu(X) \neq 0 $.

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  • $\begingroup$ $3$?? ${}{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Jan 1 '17 at 15:32
  • $\begingroup$ $ 3 = \mu([-1, 2]) $. $\endgroup$ – Starfall Jan 1 '17 at 15:36
  • $\begingroup$ So what is the contradiction? $\endgroup$ – Jorge Fernández Hidalgo Jan 1 '17 at 15:41
  • $\begingroup$ Summing countably many copies of a real number can only yield $ 0 $ or $ \pm \infty $, it cannot yield a nonzero real number (as it does in this case). $\endgroup$ – Starfall Jan 1 '17 at 15:43
  • $\begingroup$ thank you very much, I get it now. $\endgroup$ – Jorge Fernández Hidalgo Jan 1 '17 at 15:49

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