Can anyone explain the concept of a Vitali set? I am not able to understand the construction of the set.

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    $\begingroup$ What is it exactly in the construction that is puzzling you? $\endgroup$ Apr 28, 2012 at 8:27

2 Answers 2


You have the real numbers, $\mathbb R$. You have the rational numbers inside them, $\mathbb Q$.

We can define an equivalence relation on the real numbers, $x\sim y\iff x-y\in\mathbb Q$. For example $\pi\sim\pi+3$ and $\sqrt 2\nsim\pi$.

Using the axiom of choice we can define $A$ to be a set of representatives for this equivalence class. This means that we choose $A$ such that every real number is equivalent to exactly one element of $A$.

If we do that in the interval $[0,1]$ instead of all $\mathbb R$, this set $A$ is called Vitali set and we can show that it cannot be measurable. Of course any set of representatives is not measurable, it is just easier to show that in the unit interval.

Why is the set not measurable? Well, note that if we take the translations $A_q=\{a+q\pmod 1\mid a\in A\}$ for every $q\in\mathbb Q\cap[0,1)$ then we have a countably family of pairwise disjoint sets whose union is exactly $[0,1]$. Since the Lebesgue measure is translation invariant all sets must have the same measure.

We have that $m([0,1])=\sum m(A_q)=\sum_{i=1}^\infty m(A)$. However the sum of a constant positive number can either be $0$ or $\infty$, neither of which is the measure of $[0,1]$. This shows that the set $A$ cannot be measurable .

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    $\begingroup$ Just a small remark: Your answer is obviously correct, but might mislead someone into thinking the choice set is measurable when one does that with all of $\mathbb{R}$. $\endgroup$ Apr 28, 2012 at 13:57
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    $\begingroup$ @Artus: You don't "add a rational" and suddenly things become countable. Here's a possibly clearer way to define a Vitali set. You define the following equivalence relation on $[0,1]$, $a\sim b\iff a-b\in\Bbb Q$. It's not hard to see that every equivalence class is countable, fix $a$, then $b\sim a$ means $a-b\in\Bbb Q$ and there are only countably many rationals so only a countable of $b$ with this property. So each class is countable, and clearly the union of these classes has to be $[0,1]$ because every element in $[0,1]$ is in some equivalence class. $\endgroup$
    – Asaf Karagila
    Feb 10, 2013 at 0:51
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    $\begingroup$ @Artus: Yes, this is the point, the numbers are non-negative. So either they are all zero or the sum is infinite. $\endgroup$
    – Asaf Karagila
    Feb 10, 2013 at 1:09
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    $\begingroup$ @Artus: You are missing the point. The point is that if $A$ is a system of representatives from the $\sim$ defined above, and $q$ is a fixed rational then $A+q$ is a completely disjoint system of representatives, and by varying on all the rational numbers in $[-1,1]$ we certainly cover $[0,1]$ (show that every $x\in[0,1]$ is equivalent to a unique $y\in A$ and $x-y\in[-1,1]$). This is why this is a countable sum. Because the measure is translation invariant $A+q$ has the same measure as $A$, if it's zero, then $[0,1]$ has measure zero; if its positive then $[-1,2]$ has infinite measure. [cont] $\endgroup$
    – Asaf Karagila
    Feb 10, 2013 at 1:27
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    $\begingroup$ [...] Either way we arrive at a contradiction. Therefore $A$ itself cannot be measurable. $\endgroup$
    – Asaf Karagila
    Feb 10, 2013 at 1:27

$\newcommand{\R}{\mathbb{R}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\N}{\mathbb{N}}\newcommand{\nm}{\mathcal{N}}$ 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.


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