Vitali set

In mathematics, the Vitali set is an elementary example of a set of real numbers that is not Lebesgue measurable. The Vitali theorem is the existence theorem that there are such sets. It is a non-constructive result. The naming is for Giuseppe Vitali.

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

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

We give two different constructions of a Vitali set. The first is the original construction first considered by Vitali; the second is a slight modification of this which lends itself more easily to generalisation.

Vitali's original construction

If x and 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 in 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. To show this, 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 V_k = V + x_k, k = 1, 2, ... are pairwise disjoint, and further note that [0, 1]⊂∪_kV_k⊂[−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(∪V_k)≤3 (*)

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

1≤∑_{k = 1}^∞μ(V)≤3

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.

Another construction

We will need to know the following facts about Lebesgue measure:

If E is Lebesgue measurable with measure λ(E), then for any real number x, the set E + x is also Lebesgue measurable, and

\lambda (E+x)=\lambda (E)

This property is called translation invariance.

If E₁, E₂, ... is a sequence of disjoint Lebesgue measurable sets, then

\lambda \left(\bigcup _{n=1}^{\infty }E_{n}\right)=\sum _{n=1}^{\infty }\lambda (E_{n})

This property is called countable additivity.

If E is Lebesgue measurable, then

\lambda (E)=\sup \,\{\lambda (K):K\subseteq E,\,K{\mbox{ compact}}\}

This property is called inner regularity.

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 if V is measurable, then every compact set contained in V has measure zero, so V must have measure zero. The final step is to show that the real line R is a countable union of translates of V, showing that R itself has measure zero. Since R clearly has nonzero measure, the set V must not be measurable.

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

We define an equivalence relation ≡ on R as follows: if x and y are in R, then x ≡ y iff x − y is in Q, the set of rational numbers. For each x, there is a subset [x] = {y in R : x ≡ y} called the equivalence class of x. The set of all equivalence classes forms a partition of R. By the axiom of choice, we are able to choose a set V, such that V contains exactly one number from each equivalence class; i.e. for each x in R, there is a unique v in V such that x ≡ v. (Note: The reader familiar with group theory may recognize that V is a set of distinct coset representatives of Q in R.)

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

The Vitali set is non-measurable. To show this, suppose V is measurable. We will show that its Lebesgue measure λ(V) = 0. Since λ(V) is equal to the supremum of λ(K), where K ranges over all compact subsets K of V, it suffices to show that for every compact subset K of V, λ(K) = 0.

So, let K be a compact subset of V. The family of sets

$\{K+q:q\in \mathbb {Q} ,0\leq q\leq 1\}$

is a pairwise disjoint family. (If k₁ + q₁ = k₂ + q₂, then k₁ − k₂ is in Q, so that k₁ ≡ k₂; but k₁ and k₂ are both is V, so k₁ = k₂ and q₁ = q₂.) Let H be the (disjoint) union of this family:

$H:=\bigcup _{\begin{matrix}q\in \mathbb {Q} \\0\leq q\leq 1\end{matrix}}^{\bullet }\left(K+q\right)$

then by countable additivity, translation invariance, and the countability of Q, we have

$\lambda (H)=\sum _{\begin{matrix}q\in \mathbb {Q} \\0\leq q\leq 1\end{matrix}}\lambda (K+q)=\lambda (K)+\lambda (K)+\cdots$

If λ(K) > 0, then λ(H) = ∞, which contradicts the fact that H is clearly bounded. Thus, λ(K) = 0. Since K was an arbitrary compact subset of V, by inner regularity, we have λ(V) = 0.

But by a similar argument to the one above, we see that

$\{V+q:q\in \mathbb {Q} \}$

is a pairwise disjoint family of sets, and every real number lies in one of these sets. (If x is in R, then x − v is in Q for some v in V, hence x = v + (x - v) is in V + q, where q = x − v.) Again, by countable additivity, translation invariance, and the countability of Q,

$\lambda (\mathbb {R} )=\sum _{q\in \mathbb {Q} }\lambda (V+q)=\lambda (V)+\lambda (V)+\cdots =0$

and this contradiction completes the proof. Q.E.D.