Editing Vitali set (section)

===Non-measurability===
[[File:diagonal argument.svg|thumb|A possible enumeration of the positive rational numbers]]
A Vitali set is non-measurable. To show this, we assume that <math>V</math> is measurable and we derive a contradiction. Let <math>q_1,q_2,\dots</math> be an enumeration of the rational numbers in <math>[-1,1]</math> (recall that the rational numbers are [[countable]]). From the construction of <math>V</math>, we can show that the translated sets <math>V_k=V+q_k=\{v+q_k : v \in V\}</math>, <math>k=1,2,\dots</math> are pairwise disjoint.  (If not, then there exists distinct <math>v,u \in V</math> and <math>k,\ell \in \mathbb{N}</math> such that <math>v+q_k = u + q_{\ell} \implies v-u = q_{\ell}-q_k \in \mathbb{Q}</math>, a contradiction.) 

Next, note that 
:<math>[0,1]\subseteq\bigcup_k V_k\subseteq[-1,2].</math>
To see the first inclusion, consider any real number <math>r</math> in <math>[0,1]</math> and let <math>v</math> be the representative in <math>V</math> for the equivalence class <math>[r]</math>; then 
<math>r-v=q_i</math> for some rational number <math>q_i</math> in <math>[-1,1]</math> which implies that <math>r</math> is in <math>V_i</math>.

Apply the Lebesgue measure to these inclusions using [[sigma additivity]]:
:<math>1 \leq \sum_{k=1}^\infty \lambda(V_k) \leq 3.</math>

Because the Lebesgue measure is translation invariant, <math>\lambda(V_k) = \lambda(V)</math> and therefore
:<math>1 \leq \sum_{k=1}^\infty \lambda(V) \leq 3.</math>

But this is impossible.  Summing infinitely many copies of the constant <math>\lambda(V)</math> yields either zero or infinity, according to whether the constant is zero or positive.  In neither case is the sum in <math>[1,3]</math>.  So <math>V</math> cannot have been measurable after all, i.e., the Lebesgue measure <math>\lambda</math> must not define any value for <math>\lambda(V)</math>.