Lebesgue measure · Definition

Vitali set

measure theory and integrationlebesgue measure

Before you start you need to know

The Vitali set shows that not every subset of the real line can be measured consistently with translation invariance. Its existence depends on a choice that cannot be made explicitly.

Definition

The Vitali set is a subset of \([0,1]\) containing exactly one representative of each class of \(\mathbb{R}\) modulo \(\mathbb{Q}\), and it is not Lebesgue measurable.

\[x \sim y \iff x - y \in \mathbb{Q}\]

The relation \(x \sim y\) if \(x-y\in\mathbb{Q}\) is an equivalence relation and splits \([0,1]\) into disjoint classes. By the axiom of choice, there exists a set \(V\subseteq[0,1]\) containing exactly one element of each class. Lebesgue measure is translation invariant and countably additive, which leads to a contradiction if \(V\) were measurable.

Taking \(q_n\) an enumeration of \(\mathbb{Q}\cap[-1,1]\), the translates \(V+q_n\) are pairwise disjoint and their union contains \([0,1]\) and lies inside \([-1,2]\). If \(m(V)=0\) the union would have measure zero; if \(m(V)>0\) it would have infinite measure, both impossible.

A typical mistake is to think every uncountable subset of the line is non-measurable, when the Cantor set, also uncountable, has measure zero. Another error is to believe the Vitali set can be described by explicit formulas; its existence depends essentially on the axiom of choice.

Think about it

What would happen if a translation-invariant, countably additive measure with \(m([0,1])=1\) were defined on all subsets of \(\mathbb{R}\)?

Vitali's argument would give a contradiction, so such a measure cannot exist; that is why Lebesgue measure is restricted to a smaller class of measurable sets.

Exercises 10

References

  • Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd edition, Wiley, 1999.
  • H. L. Royden, P. M. Fitzpatrick, Real Analysis, 4th edition, Pearson, 2010.