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This is Exercise 1.16 from the Book: Mathematics++: Selected topics beyond the basic courses by Kantor, Matousek and Samal.

Prove that if $E \subseteq [0,1]$ is measurable with $\lambda(E) > 0$, then $E\cap V$ is not measurable, where $V$ is the Vitali set.

I assume one has to show that $$ \lambda(A) \not = \lambda(A \cap E\cap V) + \lambda(A\cap(E\cap V)^c) $$ for an arbitrary subset $A \subset \mathbb R$ but fail to do so.

How does one prove the statement?

Flo
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  • The statement should be like this. If $X$ is a measurable set, then for every measurable set $A$ $$\lambda (A) = \lambda (A\cap X) + \lambda (A\cap X^c) $$ So if you find a measurable set $A$ for which this equality fails, that automatically forces $X$ to be non-measurable. – AlvinL Jul 08 '21 at 14:44
  • The text says its for every set $A\subset \mathbb R$? I was thinking of using $E$ as $A$ but did not see the contradiction. – Flo Jul 08 '21 at 14:49
  • I don't see it happening, either. $\lambda$ can measure measurable sets. – AlvinL Jul 08 '21 at 15:34

1 Answers1

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Hint: $V$ is the disjoint union of its rational translates mod $1$. A measurable set of positive measure can't be disjoint from all its rational translates.

Robert Israel
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