Vitali Set - Construction and Proof

Construction and Proof

The rational numbers Q form a subgroup of the real numbers R under addition. The additive quotient group R/Q (the group formed by the cosets of the rational numbers as a subgroup of the real numbers under addition) consists of disjoint "shifted copies" of the rational numbers in the sense that each element of this quotient group is a set of the form Q + r for some r in R.

The elements of the quotient group R/Q partition R. There are uncountably many elements, and each element is dense in R. Each element of R/Q intersects, and the axiom of choice guarantees the existence of a subset of containing exactly one representative out of each element of R/Q. A set formed this way is called a Vitali set, i.e., a Vitali set V is a subset of which, for each real number r ∈ R, contains exactly one number v such that v−r is rational. Every Vitali set V is uncountable, and v−u is irrational for any .

A Vitali set is non-measurable. To show this, we assume that V is measurable and we derive a contradiction. Let q₁, q₂, ... be an enumeration of the rational numbers in (recall that the rational numbers are countable). From the construction of V, note that the translated sets, k = 1, 2, ... are pairwise disjoint, and further note that . (To see the first inclusion, consider any real number r in and let v be the representative in V for the equivalence class ; then r−v = q for some rational number q in .)

Apply the Lebesgue measure to these inclusions using sigma additivity:

Because the Lebesgue measure is translation invariant, and therefore

But this is impossible. Summing infinitely many copies of the constant λ(V) yields either zero or infinity, according to whether the constant is zero or positive. In neither case is the sum in . So V cannot have been measurable after all, i.e., the Lebesgue measure λ must not define any value for λ(V).