Proof that Vitali set is non-measurable. A Vitali set is a subset $V$ of $[0,1]$ such that for every $r\in \mathbb R$ there exists one and only one $v\in V$ for which $v-r \in \mathbb Q$. Equivalently, $V$ contains a single representative of every element of $\mathbb R / \mathbb Q$.
The proof I read is in this short article on Wikipedia: https://en.wikipedia.org/wiki/Vitali_set
Under "proof", the second to last inequality $1 \leq \sum \lambda (V_k) \leq 3$ is claimed to result from the previous inequality $[0,1] \subset \bigcup V_k \subset [-1,2]$ simply using sigma-additivity. There must be some missing argument to claim that the sum of the measures, although greater than the measure of the union, is still less than the measure of $[-1,2]$.
What is the missing argument ?

The sets $V_k$ are disjoint and countable, hence the measure of the union is exactly equal to the sum of measures.
