Vitali set: Definition from Answers.com

In mathematics, a Vitali set is an elementary example of a set of real numbers that is not Lebesgue measurable, found by Giuseppe Vitali (1905). The Vitali theorem is the existence theorem that there are such sets. There are uncountably many Vitali sets, and their existence is proven on the assumption of the axiom of choice.

1 Measurable sets
2 Construction and proof
3 See also
4 References
5 External links

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 with uniform density, they likewise have well-defined masses. 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 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 'total length'? As an example, we might ask what is the mass of the set of rational numbers, given that the mass of the interval [0, 1] is 1. The rationals are dense in the reals, so any non negative value may appear reasonable.

However the closest generalization to mass is sigma additivity, which gives rise to the Lebesgue measure. It assigns a measure of b − a to the interval [a, b], but will assign a measure of 0 to the set of rational numbers because it is countable. Any set which has a well-defined Lebesgue measure is said to be "measurable", but the construction of the Lebesgue measure (for instance using Carathéodory's extension theorem) does not make it obvious whether there exist non-measurable sets. The answer to that question involves the axiom of choice.

Construction and proof

A Vitali set V is a subset of [0,1] which, for each real number r, contains exactly one number v such that v-r is rational. (This implies that V is uncountable, and also that v-u is irrational for any $u,v \in V, u \neq v$ .) Such sets can be shown to exist given the axiom of choice.

To construct a Vitali set V, consider the additive quotient group $\mathbb{R} / \mathbb{Q}$ . Each element of this group is a "shifted copy" of the rational numbers: a set of the form $\mathbb{Q}+r$ for some $r \in \mathbb{R}$ (actually, for countably many such $r$ ). Thus, the elements of this group are subsets of R and partition R. There are uncountably many elements. Since each element intersects [0,1], we can use the axiom of choice to choose a set $V \subset [0, 1]$ containing exactly one representative out of each element of $\mathbb{R} / \mathbb{Q}$ .

A Vitali set is non-measurable. To show this, we argue by contradiction and assume that V is measurable. 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 translated sets $V_k=V+q_k=\{v+q_k : v \in V\}$ , k = 1, 2, ... are pairwise disjoint, and further note that $[0,1]\subseteq\biguplus_k V_k\subseteq[-1,2]$ . (To see the first inclusion, consider any real number r in [0,1] and let v be the representative in V for the equivalence class [r]; then r −v = q for some rational number q in [-1,1].)

Apply the Lebesgue measure to these inclusions using sigma additivity :

$1 \leq \sum_{k=1}^\infty \lambda(V_k) \leq 3.$

Because the Lebesgue measure is translation invariant, $λ(V k) = λ(V)$ and

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

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 [1,3]. So V cannot have been measurable after all, i.e., the Lebesgue measure λ must not define any value for $λ(V)$ .

References

Herrlich, Horst: Axiom of Choice, page 120. Springer, 2006.
Vitali, Guiseppe (1905), "Sul problema della misura dei gruppi di punti di una retta", Bologna, Tip. Gamberini e Parmeggiani

External links

Vitali sets at Finance wiki Retrieved 03-10-2009

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

Donate to Wikimedia

	Did vitali klitchko beat lennox lewis? Read answer...
	Is there a vitaly scherbo school of gymnastics located in texas? Read answer...
	Where is IT set? Read answer...

	Can you say your names are orji obiora vitalis or your name is obiora orji vitalis which one is correct?
	What is vitali klitchkos arm length?
	Is alex vitali boss?

		Sci-Tech Dictionary. McGraw-Hill Dictionary of Scientific and Technical Terms. Copyright © 2003, 1994, 1989, 1984, 1978, 1976, 1974 by McGraw-Hill Companies, Inc. All rights reserved. Read more
		Wikipedia. This article is licensed under the Creative Commons Attribution/Share-Alike License. It uses material from the Wikipedia article "Vitali set". Read more

Vitali set

Contents

Measurable sets

Construction and proof

See also

References

External links