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Vitali set

If you are having difficulty understanding this article, you might want to learn about measure theory and Lebesgue integration.

In measure theory, the Vitali set is a typical, elementary example of a non-measurable set.

Despite the terminology, there are many Vitali sets. They are constructed using the axiom of choice, and for reasons too complex to discuss here, Vitali sets are impossible to describe explicitly.

See also: Banach-Tarski paradox

The importance of non-measurable sets

Certain sets have a definite length or mass. For instance, the interval [0,1] is deemed to have length 1; more generally, an interval [a,b], a≤b, is deemed to have length b-a. If we think of such intervals as metal rods, they likewise have well-defined masses. If the [0,1] rod weighs 1 kilogram, then the [3,9] rod weighs 6 kilograms. The set [0,1]∪[2,3] is composed of two intervals of length one, so we take its total length to be 2. In terms of mass, we'd have two rods of mass 1, so the total mass is 2.

There is a natural question here: if E is an arbitrary subset of the real line, does it have a "mass" or "length?" As an example, we might ask what is the mass of the set of rational numbers. They are very finely spread over all of the real line, so any answer may appear reasonable at first pass.

As it turns out, the physically relevant solution is to use measure theory. In this setting, the Lebesgue measure, which assigns weight b-a to the interval [a,b] will assign weight 0 to the set of rational numbers. Any set which has a well-defined weight is said to be "measurable." The construction of the Lebesgue measure (for instance, using the outer measure) does not make obvious whether there are non-measurable sets.

Technical discussion

Sets which aren't Lebesgue measurable can be detected in a variety of ways. For one, if E is Lebesgue measurable with measure μ(E), then the set E+x, for any real number x, is also Lebesgue measurable, and its measure is also μ(E). This property is called translation invariance. In addition, if E1, E2, ... is a sequence of disjoint Lebesgue measurable sets, then we have that μ(∪Ek)=∑μ(Ek); this is called countable additivity.

Our construction and proof will proceed by contradiction. As a first step, we will construct a set V and assume that it is measurable. Our second step is to show that either translation invariance or countable additivity is violated. Since translation invariance and countable additivity are definetly true for the Lebesgue measure μ, it must be that our initial assumption is incorrect; namely, the set V must not be measurable.

The construction has many parallels to the construction of the paradoxical decompositions in the Banach-Tarski paradox.

Construction of the Vitali set

If x,y are real numbers and x-y is a rational number, then we write x~y and we say that x and y are equivalent; ~ is an equivalence relation. For each x, there is a subset [x]={y∈R;x~y} called the equivalence class of x. The set of equivalent classes partitions R. By the axiom of choice, we are able to choose a set V⊂[0,1] such that for any equivalence class [x], the set V∩[x] is a singleton, that is, a set consisting of exactly one point.

V is the Vitali set. Note that there are in fact several choices of V; the axiom of choice lets you say there is such a V, but there are clearly infinitely many.

The Vitali set is non-measurable

We assume that V is measurable. From this assumption, we carefully work and prove something absurd: namely that a+a+a+... (an infinite sum of identical numbers) is between 1 and 3. Since an absurd conclusion is reached, it must be that the only unproved hypothesis (V is measurable) is at fault.

First we let x1,x2,... be an enumeration of the rational numbers in [-1,1] (Recall that the rational numbers are countable.) From the construction of V, note that the sets Vk=V+xk, k=1,2,... are pairwise disjoint, and further note that [0,1]⊂∪Vk⊂[-1,2]. Because μ is countably additive, it must also have the propriety of being monotone; that is, if A⊂B, then μ(A)≤μ(B). Hence, we know that

1≤μ(∪Vk)≤3 (*)

But now, because of translation invariance, we see that for each k=1,2,..., μ(Vk)=μ(V). Combining with countable additivity, and (*) we obtain


The sum is an infinite sum of a single constant, non-negative term. If the term is zero, the sum is likewise zero, and hence it is certainly not greater than or equal to one. If the term is nonzero then the sum is infinite, and in particular it isn't smaller than or equal to 3.

This conclusion is absurd, and since all we've used is translation invariance and countable additivity, it must be that V is non-measurable.