The Vitali Nonmeasurability Contradiction

The Contradiction Behind a Vitali Set

A Vitali representative set is formed by selecting one point from each class of real numbers in \([0,1]\) that differ by a rational number. The selection gives a set with a special pattern: its rational translates cover the unit interval, are pairwise disjoint when indexed by distinct rationals, and all lie in one bounded interval. If the set were Lebesgue measurable, translation invariance and countable additivity would force an impossible measure value. The contradiction has two cases: the measure cannot be zero, and it cannot be positive.

Work in \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. Define \(x\sim y\) for \(x,y\in[0,1]\) when \(x-y\in\mathbb{Q}\). This is an equivalence relation: \(x-x=0\in\mathbb{Q}\), rational differences remain rational when their signs are reversed, and sums of rational differences are rational. A Vitali representative set \(V\subseteq[0,1]\) contains exactly one point from each equivalence class. The axiom of choice gives such a selection; no measurability is assumed.

Why the Translates Cover and Do Not Overlap

For coverage, take any \(x\in[0,1]\). Let \(v\in V\) be the representative of the equivalence class of \(x\). By the definition of the relation, \(q=x-v\) is rational. Since both \(x\) and \(v\) belong to \([0,1]\), \(-1\le x-v\le1\), so \(q\in D\). Thus \(x=v+q\in V+q\subseteq S\). Conversely, if \(y\in V+q\) for some \(q\in D\), then \(y=v+q\) for some \(v\in[0,1]\). Therefore \(-1\le y\le2\), proving \(S\subseteq[-1,2]\).

To check disjointness, suppose a point belongs to both \(V+q_1\) and \(V+q_2\). Then there are \(v_1,v_2\in V\) such that \(v_1+q_1=v_2+q_2\). Rearranging gives \(v_1-v_2=q_2-q_1\in\mathbb{Q}\). Hence \(v_1\sim v_2\). Since \(V\) contains exactly one representative of each class, \(v_1=v_2\); substituting into the equality gives \(q_1=q_2\). Distinct translates therefore have empty intersection.

The Nonmeasurability Theorem

Proof. Suppose, for contradiction, that \(V\in\mathcal{L}\). For each \(q\in D\), the translate \(V+q\) is Lebesgue measurable. We use the standard translation invariance of Lebesgue measure: for every Lebesgue measurable \(E\subseteq\mathbb{R}\) and every \(q\in\mathbb{R}\), \(E+q\) is measurable and \(m(E+q)=m(E)\). In particular, every translate \(V+q\) has measure \(m(V)\). Also \(V\subseteq[0,1]\), so monotonicity gives \(0\le m(V)\le m([0,1])=1\). Thus \(m(V)\) is finite, and exactly one of the cases \(m(V)=0\) and \(m(V)>0\) holds.

First suppose \(m(V)=0\). The family \((V+q)_{q\in D}\) is a countable family of measurable sets, each of measure zero. By countable subadditivity, which follows from countable additivity, \(m(S)\le\sum_{q\in D}m(V+q)=0\). Hence \(m(S)=0\). But the coverage already proved gives \([0,1]\subseteq S\), so monotonicity gives \(1=m([0,1])\le m(S)=0\), a contradiction.

Now suppose \(m(V)=c>0\). The set \(S\) is measurable as a countable union of measurable sets. Its translates are pairwise disjoint, so countable additivity gives \(m(S)=\sum_{q\in D}m(V+q)=\sum_{q\in D}c=+\infty\). Indeed, the sum of its first \(N\) terms is \(Nc\), which exceeds any fixed finite bound for sufficiently large \(N\). On the other hand, \(S\subseteq[-1,2]\), and \(m([-1,2])=3\), so monotonicity gives \(m(S)\le3\). This is a contradiction as well. Neither possible value of \(m(V)\) is consistent with measurability. Therefore \(V\notin\mathcal{L}\). \(\square\)

Where the Measure Properties Enter

The proof depends on two distinct uses of the measure structure. Translation invariance gives the same measure to each translate as to \(V\), but it can be applied here only under the supposition that \(V\) is Lebesgue measurable. Countable additivity gives the measure of a countable disjoint union as the sum of the measures of its pieces. In the zero case, countable subadditivity suffices to show that a countable union of null sets is null. In the positive case, the disjointness and countable additivity show that the union has infinite measure. Its bounded location then supplies the conflicting finite upper bound.

A common mistake is to say only that “there are infinitely many translates, so their union has infinite measure.” Infinitely many sets need not have a union of infinite measure: they might overlap, or each might have measure zero. The corrected argument checks both missing points. Distinct translates are disjoint, and if \(m(V)=c>0\), each has the same strictly positive measure \(c\). Countable additivity then gives an infinite sum. In the other case, when the translates have measure zero, countable subadditivity gives a null union, which conflicts with coverage. The bounded containment \([-1,2]\) is what rules out the infinite-measure conclusion.

The theorem shows that the selected set is nonmeasurable; it does not assign it an “intermediate” measure or claim that its outer measure is a particular number. The two-case contradiction is conditional on measurability, so translation invariance and countable additivity apply to measurable sets throughout. The location and disjointness of the translates come from the representative construction itself, independently of whether \(V\) is measurable.