Vitali's theorem (measure theory)

The set of Vitali (after Giuseppe Vitali ) is a mathematical theorem from the measure theory . It says that there is a subset of the real numbers that is not Lebesgue measurable . Each of the sets resulting from the proof of Vitali's theorem is also called a Vitali set . Their existence is proven with the help of the axiom of choice , in particular they are not explicitly stated. The Vitali sets are standard examples of non-Lebesgue measurable sets.

The importance of non-measurable quantities

Certain quantities can be assigned a length or a dimension. The interval is the length assigned to and generally an interval , the length . If we think of such intervals as metal bars, they also have a well-defined mass. If the spear weighs, weighs the spear . The set is composed of two intervals of length one, and its total length is accordingly , or, if we refer to masses again, two bars with mass make up the total mass . ${\displaystyle [0,1]}$ ${\displaystyle 1}$ ${\displaystyle [a,b]}$ ${\displaystyle a<b}$ ${\displaystyle b-a}$ ${\displaystyle [0,1]}$ ${\displaystyle 1\,\mathrm {kg} }$ ${\displaystyle [3,9]}$ ${\displaystyle 6\,\mathrm {kg} }$ ${\displaystyle [0,1]\cup [2,3]}$ ${\displaystyle 2}$ ${\displaystyle 1}$ ${\displaystyle 2}$

The question naturally arises: If any subset is the real axis, does it have a mass or length? For example, we may wonder what the measure of rational numbers is. These lie close to the real axis, and so it is not initially clear which measure is sensible to assign here. ${\displaystyle E}$

In this situation it turns out that the meaningful assignment is the measure - in accordance with what the Lebesgue measure provides, which assigns the length to the interval . Any quantity with a well-defined measure is called measurable. When constructing the Lebesgue measure (for example using the external measure ) it is initially not clear whether non-measurable quantities exist. ${\displaystyle 0}$ ${\displaystyle [a,b]}$ ${\displaystyle b-a}$

Construction and proof

If and is real numbers and a rational number, then we write and say that and are equivalent - one can show that is an equivalence relation . For each there is a subset , the equivalence class of . The set of equivalence classes forms a partition of . The axiom of choice allows us to choose a set that contains a representative of each equivalence class (for each equivalence class , the set contains only a single element). We then call a Vitali set. ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle x-y}$ ${\displaystyle x\sim y}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle \sim }$ ${\displaystyle x}$ ${\displaystyle [x]=\{y\in \mathbb {R} \mid x\sim y\}\subseteq \mathbb {R} }$ ${\displaystyle x}$ ${\displaystyle \mathbb {R} }$ ${\displaystyle V\subset [0,1]}$ ${\displaystyle [x]}$ ${\displaystyle V\cap [x]}$ ${\displaystyle V}$

The Vitali quantities cannot be measured. To show this, we assume would be measurable. From this assumption we conclude in the following that the infinite sum always lies between and - this is obviously wrong and the assumption is refuted by the contradiction. ${\displaystyle V}$ ${\displaystyle a+a+\dotsb }$ ${\displaystyle a}$ ${\displaystyle 1}$ ${\displaystyle 3}$

Now let us first be a count of the rational numbers in (the rational numbers are countable). The amounts , are by construction of disjoint in pairs, is also ${\displaystyle q_{1},q_{2},\dots }$ ${\displaystyle [-1,1]}$ ${\displaystyle V_{k}=\{v+q_{k}\mid v\in V\}}$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle V}$

${\displaystyle [0,1]\subseteq \bigcup _{k=1}^{\infty }V_{k}\subseteq [-1,2].}$

(To see the first inclusion, consider a real number from and a representative of the equivalence class , then there exists a rational number from , so that for a is, therefore is .) ${\displaystyle x}$ ${\displaystyle [0,1]}$ ${\displaystyle v\in V}$ ${\displaystyle [x]}$ ${\displaystyle q_{k}}$ ${\displaystyle [-1,1]}$ ${\displaystyle x-v=q_{k}}$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle x\in V_{k}}$

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

1. The Lebesgue measure is σ-additive , that means it holds for a countable number of pairwise disjoints ${\displaystyle A_{i}}$

${\displaystyle \lambda \left(\bigcup _{i=1}^{\infty }A_{i}\right)=\sum _{i=1}^{\infty }\lambda (A_{i}).}$

2. The Lebesgue measure is translation-invariant, that is, it holds for real numbers ${\displaystyle x}$

${\displaystyle \lambda (A)\,=\,\lambda (A+x)}$.

Now consider the measure of the union given above. Since σ is additive, it is also monotonic, that is for . It follows: ${\displaystyle \lambda }$ ${\displaystyle \lambda }$ ${\displaystyle \lambda (A)\leq \lambda (B)}$ ${\displaystyle A\subseteq B}$

${\displaystyle 1\leq \lambda \left(\bigcup _{k=1}^{\infty }V_{k}\right)\leq 3.}$

Because of σ-additivity it follows, since they are disjoint: ${\displaystyle V_{k}}$

${\displaystyle \lambda \left(\bigcup _{k=1}^{\infty }V_{k}\right)=\sum _{k=1}^{\infty }\lambda (V_{k})}$

Because of translation invariance holds for each . Together with the above you get: ${\displaystyle k\in \mathbb {N} :\lambda (V_{k})=\lambda (V)}$

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

But this is a contradiction because it is not possible and implies that it is true. ${\displaystyle \lambda (V)=0}$ ${\displaystyle \lambda (V)>0}$ ${\displaystyle \sum _{k=1}^{\infty }\lambda (V)=\infty }$

Therefore it is not measurable. ${\displaystyle V}$

See also

• Banach-Tarski paradox
• Dimensional problem

literature

• Horst Herrlich: Axiom of Choice (= Lecture Notes in Mathematics. Vol. 1876). Springer, Berlin et al. 2006, ISBN 3-540-30989-6 , p. 120.