TITLE: The construction of a Vitali set
QUESTION [27 upvotes]: Can anyone explain the concept of a Vitali set? I am not able to understand the construction of the set.

REPLY [10 votes]: $\newcommand{\R}{\mathbb{R}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\N}{\mathbb{N}}\newcommand{\nm}{\mathcal{N}}$
I have seen at least two constructions that lead to sets called Vitali sets.
In the following consider the Lebesgue measure.
Let $S\subseteq\R^d$. Define an equivalence relation on $S$ by $$x\sim y\iff x-y\in\Q^d.$$ Let $S/\sim$ be the set of equivalence classes determined by $\sim$. Since each equivalence class is not empty, by the Axiom of Choice, we can choose one element of each equivalence class. The Axiom of Choice says that there exist a function $$f:(S/\sim)\to S=\bigcup_{A\in S/\sim}A,\quad \text{ so that }\quad f(A)\in A.$$ Define $$V_d(S):=\{f(A):A\in S/\sim\},$$
i.e. $V_d(S)$ is the set formed by picking exactly one element of each equivalence class and put them all together.
The first construction I have seen is by setting $S=\R^d$ (actually, it was with $d=1$ and $S=\R$). So, let $V=V_d(\R)$. Let's see why $V$ is not measurable.
Lemma. Let $E\subseteq\R$. If $E$ is measurable and $m(E)\gt 0$, then the set $$E-E:=\{x-y:x,y\in E\}$$ contains an open ball centered at the origin.
The above Lemma is the theorem stated and proved here.
Notice that the set $V-V$ does not contain any points of $\Q^d$, since $\Q^d$ is dense in $\R^d$, this says that there can't be open balls contained in $V-V$. By the lemma, there are only two possibilities: $m(V)=0$ or $V$ is not measurable. If $m(V)=0$, let $\{q_k\}_{k\in \N}$ be an enumeration of $\Q^d$. You can see that the sets $$V+q_k:=\{x+q_k:x\in V\}$$
are disjoint, measurable, and $m(V+q_k)=m(V)=0$. You can also see that $\R^d$ is the disjoint union $$\R^d=\bigcup_{k\in\N} V+q_k$$ and then $$\infty=m(\R^d)=m\left(\bigcup_{k\in\N} V+q_k\right)=\sum_{k\in\N} m(V+q_k)=0.$$ This is an absurd. Therefore $V$ is not measurable and that's our $d-$dimensional Vitali set.
The second one is by setting $d=1$ and $S=[0,1]$. So, let $\nm=S_1([0,1])$. Let $\{r_k\}_{k\in\N}$ be an enumeration of $\Q\cap [-1,1]$. Consider, again, the sets $$\nm_k=\nm+r_k,$$
they are disjoint and you can see that $$[0,1]\subset\bigcup_{k\in\N} \nm_k\subset[-1,2].$$
Once that we assume measurability of $\nm$ we get measurability of the $\nm_k$, moreover, $m(\nm_k)=m(\nm)$, for all $k$. Thus
$$\begin{align*}
m([0,1])\leq & \sum_{k\in\N} m(\nm_k)\leq m([-1,2])\\
1\leq &\sum_{k\in\N } m(\nm)\leq 3.
\end{align*}$$
If $m(\nm)=0$ we get $1\leq 0$, and if $m(\nm)\gt 0$ we get $\infty\leq 3$. Therefore $\nm$ is not measurable.