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 ?

We assume for the sake of contradiction $V$ is measurable, so $\lambda(V)$ is a real number. By construction, each $V_k$ is disjoint, and is merely a shifted copy of $V$, so each of these sets has the same measure. The sequence $(V_k)_{k=1}^\infty$ is countable.
Sigma additivity (the assumption in the definition of Lebesgue measure) guarantees that the measure of a countable union of disjoint sets will be the sum of the measures of these sets. Applying this principle along with the equal measures of all $V_k$, we find:
$$
\lambda\left(\bigcup_k V_k\right)=\sum_k\lambda\left(V_k\right)=\sum_k\lambda(V).
$$
What can the infinite sum of a constant term be? If the value of the constant term $\lambda(V)$ is $0$, then the sum is zero. On the other hand, if $\lambda(V)>0$, then the sum is unbounded. It follows that $\lambda\left(\bigcup_k V_k\right)$ is either $0$ or $\infty$.
However, by construction, this union is a superset of the interval $[0,1]$, which has measure $1$, and a subset of the interval $[-1, 2]$, which has a measure $3$. The measure must therefore fall between $1$ and $3$.
We cannot have a real number be simultaneously in $\{0, \infty\}$ and $[1,3]$, so we have our contradiction.
