Vitali set

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In mathematics, a 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.

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[edit] 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.

[edit] Construction and proof

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} of R called the equivalence class of x. The set of these equivalence classes partitions R. By the axiom of choice, we are able to choose a set V \subset [0, 1] containing exactly one representative out of each equivalence class (for any equivalence class [x], the set V ∩ [x] is a singleton). We say that V is a Vitali set.

A 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 q1, q2, ... 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+q_k : v \in V\}, k = 1, 2, ... are pairwise disjoint, and further note that [0,1]\subseteq\bigcup_k V_k\subseteq[-1,2]. (To see the first inclusion, consider any real number x in [0,1] and let v be the representative in V for the equivalence class [x]; then x −v = q for some rational number in [-1,1] (say q = ql) and so x is in Vl.)

From the definition of Lebesgue measureable sets, it can be shown that all such sets have the following two properties:

1. The measure is countably additive, that is if Ai is a set of at most a countable number of pairwise-disjoint sets, then \mu \left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mu(A_i).

2. The measure is translation invariant, that is, for any real number x, μ(A) = μ(A + x).

Consider now the measure μ of the union given above. Because μ is countably additive, it must also have the property of being monotone; that is, if A⊂B, then μ(A)≤μ(B). Hence, we know that

1 \leq \mu\left(\bigcup_k V_k\right) \leq 3.

By countable additivity, one has

\mu\left(\bigcup_k V_k\right) = \sum_{k=1}^\infty \mu(V_k)

with equality following because the Vk are disjoint. Because of translation invariance, we see that for each k = 1, 2, ..., μ(Vk) = μ(V). Combining this with the above, one obtains

1 \leq \sum_{k=1}^\infty \mu(V) \leq 3.

The sum is an infinite sum of a single real-valued 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 true that V is non-measurable.

[edit] See also

[edit] References

  • Herrlich, Horst: Axiom of Choice, page 120. Springer, 2006.