Vitali set

The set of Vitali ( by Giuseppe Vitali ) implies the existence of a non Lebesgue - measurable quantity. We call each of the through the ( nonconstructive ) proof of the theorem of Vitali incurred amounts, as Vitali set. Their existence is thereby proved with the aid of the axiom of choice, in particular they are not explicitly specified. The Vitali - Amounts are considered as standard examples of non- Lebesgue measurable sets.

The importance of non - measurable quantities

Certain amounts can cause an ' length ' or a ' degree ' are allocated. The interval [ 0, 1] is assigned to the length 1 and is generally an interval [a, b], a ≤ b, the length b - a If we consider such intervals as metal rods, they also have a well-defined mass. When the [ 0, 1] - rod weighs 1 kg weighs the [3, 9] - rod 6 kg. The set [0, 1] ∪ [2, 3] is composed of two intervals of length one composed, and their total length is thus 2, or, if one refers back to the masses, two rods with a mass of 1 to give the total mass 2

Here, naturally, the question arises: If E is an arbitrary subset of the real axis, it then has a " mass " or " length"? For example, we can ask ourselves, what is the measure of rational numbers. These are close to the real axis, and thus it is not immediately clear what level you want to assign reasonably here.

In this situation, ultimately turns out that the meaningful assignment, the degree is 0 - assigns a - in accordance with what the Lebesgue measure provides that the interval [ a, b] the length b. Lots with well-defined measure is called " measurable". In the construction of the Lebesgue measure (for example, on the outer measure ), it is not immediately clear whether the non- measurable quantities exist.

Construction and proof

If x and y are real numbers and x -y is a rational number, then we write x ~ y and say that x and y are equivalent - you can show that ~ is an equivalence relation. There is a subset of [x ] = {y in R: x ~ y} for every x in R, the equivalence class of x. The set of equivalence classes forms a partition of R. The axiom of choice allows us to select a set containing one representative of each equivalence class (for each equivalence class [ x] contains the set V ∩ [ x] only a single element ). We call V then a Vitali set.

The Vitali set is not measurable. To show this, we assume that V would be measurable. From this assumption, we conclude that the infinite sum a a a ... identical numbers a is between 1 and 3 - this is obviously false, and by contradiction, the assumption is refuted.

Now let first q1, q2, ... be a counting of rational numbers in [-1, 1] ( the rational numbers are countable ). The quantities k = 1, 2, ... are by construction of V disjoint pairs, is also

( To see the first inclusion, let us consider a real number x in [0,1] and a representative v of V the equivalence class [ x], then there exists a rational number q from [-1,1] so that x -v = q. , it is q = ql for l, so x is in Vl. )

From the definition of Lebesgue measurable quantities follows that all these quantities have the following two properties:

The first measure is σ - additive, that is, for countably many pairwise disjoint applies

2 The measure is translation invariant, that is, for real numbers x we have

Now we consider the degree of the above association. Μ σ as additive, it is also monotonic, that is to. It follows:

Because σ - additivity implies that the Vk are disjoint:

For translational, for every k = 1, 2, ..., μ ( V k ) = μ ( v). Along with the above, we obtain:

This infinite sum of one and the same real constants is not negative. If the constant 0, and the sum must be 0, and is therefore certainly not greater than or equal to 1, when the term is positive, the sum does not converge, and is particularly not less than or equal to 3

This yields the contradiction, and therefore V is not measurable.