Let \(V\subseteq[0,1]\) be a Vitali set and \(A=\bigcup_{n}(V+q_n)\) with \(q_n\) an enumeration of \(\mathbb{Q}\cap[-1,1]\). If \(V\) were measurable, what inequality do you get for \(m(V)\) and why is it a contradiction?
Choose one option.
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B. \(m(V)\) would satisfy \(1\le\sum_n m(V)\le 3\), impossible for \(m(V)=0\) or \(m(V)>0\)
If \(m(V)=0\), the union \(A\) would have measure zero, yet it contains \([0,1]\) and so measures at least 1. If \(m(V)>0\), countable additivity gives infinite measure, yet \(A\subseteq[-1,2]\) forces measure at most 3.
The Vitali set is a subset of \([0,1]\) containing exactly one representative of each class of \(\mathbb{R}\) modulo \(\mathbb{Q}\), and it is not Lebesgue measurable.