# 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 ?

• Note that the $V_k$ are pairwise disjoint. – Suzet Aug 1 '18 at 0:07

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

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.

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.