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 ?
 A: The sets $V_k$ are disjoint and countable, hence the measure of the union is exactly equal to the sum of measures.
A: 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.
A: Notation: $x+S=S+x=\{x+s:s\in S\}$ when $x\in \Bbb R$ and $S\subset \Bbb R.$
Let $q_0$ be the unique member of $V\cap \Bbb Q.$ Let $V_0=V+(-q_0).$ Let $W_0=\cup_{n\in \Bbb Z}(n+V_0).$ For $q\in (0,1)\cap \Bbb Q$ let $W_q=q+W_0.$
Observe that $(-q)+W_q=W_0$ when $q\in \Bbb Q\cap [0,1)$ and that if $q,q'$ are distinct members of $[0,1)\cap \Bbb Q$ then $W_q$ and $W_{q'}$ are disjoint.
Observe that  $\{W_q\cap [0,1):q\in \Bbb Q \cap [0,1)\}$ is a partition of $[0,1).$ 
Observe that if $V$ is Lebesgue-measurable then so is $W_q\cap [0,1)$ for each $q\in \Bbb Q \cap [0,1).$ 
Let $m'(S)$ denote  the Lebesgue outer measure of any  $S\subset \Bbb R.$ 
For any  $q\in \Bbb Q \cap [0,1)$ we have $$m'(W_q\cap [0,1))=m'((\;(-q)+W_q)\;)\cap [-q,1-q))=m'(W_0\cap [-q,1-q))=$$ $$=m'(W_0\cap [-q,0))+m'(W_0\cap [0,1-q))=$$ $$=m'(1+(W_0\cap [-q,0)\;)+m'(W_0\cap [0,1-q))=$$ $$=m'(W_0\cap [1-q,1))+m'(W_0\cap [0,1-q))=$$ $$=m'(W_0\cap [0,1)).$$
Suppose $V$ is Lebesgue-measurable. Then $\{W_q\cap [0,1): q\in \Bbb Q\cap [0,1)\}$ is a partition of $[0,1)$ into a countably infinite family of pair-wise disjoint measurable subsets, each with the same measure $r.$ But if $r=0$ this implies that $[0,1)$ is a measure-$0$ set, and if $r>0$ this implies the measure of $[0,1)$ is $\infty.$
The sets $W_q$ can be described as follows: Consider $\Bbb R$ as a vector space over the field $\Bbb Q.$ Let $B$ be  a Hamel (vector-space ) basis for $\Bbb R$ over $\Bbb Q,$ with $1=b_1\in B.$ Let  $p:\Bbb R\to \Bbb Q$ be the projection of $\Bbb R$ onto the "$b_1$" co-ordinate. Let $W_q=\{x\in \Bbb R: p(x)-q\in \Bbb Z\}.$ The Axiom of Choice is needed to prove that $B$ exists.
