Inmathematics, aVitali set is an elementary example of a set ofreal numbers that is notLebesgue measurable, found byGiuseppe Vitali in 1905.^[1] TheVitali theorem is theexistence theorem that there are such sets. Each Vitali set isuncountable, and there are uncountably many Vitali sets. The proof of their existence depends on theaxiom of choice.

Certain sets have a definite 'length' or 'mass'. For instance, theinterval${\displaystyle [0,1]}$ is deemed to have length${\displaystyle 1}$; more generally, an interval${\displaystyle [a,b],a\le b}$, is deemed to have length${\displaystyle b-a}$. If we think of such intervals as metal rods with uniform density, they likewise have well-defined masses. The set${\displaystyle [0,1]\cup [2,3]}$ is composed of two intervals of length one, so we take its total length to be${\displaystyle 2}$. In terms of mass, we have two rods of mass${\displaystyle 1}$, so the total mass is${\displaystyle 2}$.

There is a natural question here: if${\displaystyle E}$ 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 ofrational numbers between${\displaystyle 0}$ and${\displaystyle 1}$, given that the mass of the interval${\displaystyle [0,1]}$ is${\displaystyle 1}$. The rationals aredense in the reals, so any value between and including${\displaystyle 0}$ and${\displaystyle 1}$ may appear reasonable.

However the closest generalization to mass issigma additivity, which gives rise to theLebesgue measure. It assigns a measure of${\displaystyle b-a}$ to the interval${\displaystyle [a,b]}$, but will assign a measure of${\displaystyle 0}$ to the set of rational numbers because it iscountable. Any set which has a well-defined Lebesgue measure is said to be "measurable", but the construction of the Lebesgue measure (for instance usingCarathéodory's extension theorem) does not make it obvious whether non-measurable sets exist. The answer to that question involves theaxiom of choice.

A Vitali set is a subset${\displaystyle V}$ of theinterval${\displaystyle [0,1]}$ ofreal numbers such that, for each real number${\displaystyle r}$, there is exactly one number${\displaystyle v\in V}$ such that${\displaystyle v-r}$ is arational number. Vitali sets exist because the rational numbers${\displaystyle \mathbb{Q}}$ form anormal subgroup of the real numbers${\displaystyle \mathbb{R}}$ underaddition, and this allows the construction of the additivequotient group${\displaystyle \mathbb{R}/\mathbb{Q}}$ of these two groups which is the group formed by thecosets${\displaystyle r+\mathbb{Q}}$ of the rational numbers as a subgroup of the real numbers under addition. This group${\displaystyle \mathbb{R}/\mathbb{Q}}$ consists ofdisjoint "shifted copies" of${\displaystyle \mathbb{Q}}$ in the sense that each element of this quotient group is a set of the form${\displaystyle r+\mathbb{Q}}$ for some${\displaystyle r}$ in${\displaystyle \mathbb{R}}$. Theuncountably many elements of${\displaystyle \mathbb{R}/\mathbb{Q}}$partition${\displaystyle \mathbb{R}}$ into disjoint sets, and each element isdense in${\displaystyle \mathbb{R}}$. Each element of${\displaystyle \mathbb{R}/\mathbb{Q}}$ intersects${\displaystyle [0,1]}$, and theaxiom of choice guarantees the existence of a subset of${\displaystyle [0,1]}$ containing exactly onerepresentative out of each element of${\displaystyle \mathbb{R}/\mathbb{Q}}$. A set formed this way is called a Vitali set.

Every Vitali set${\displaystyle V}$ is uncountable, and${\displaystyle v-u}$ is irrational for any${\displaystyle u,v\in V,u\ne v}$.

A Vitali set is non-measurable. To show this, we assume that${\displaystyle V}$ is measurable and we derive a contradiction. Let${\displaystyle q_1,q_2,\dots }$ be an enumeration of the rational numbers in${\displaystyle [-1,1]}$ (recall that the rational numbers arecountable). From the construction of${\displaystyle V}$, we can show that the translated sets${\displaystyle V_k=V+q_k=\{v+q_k:v\in V\}}$,${\displaystyle k=1,2,\dots }$ are pairwise disjoint. (If not, then there exists distinct${\displaystyle v,u\in V}$ and${\displaystyle k,\ell \in \mathbb{N}}$ such that${\displaystyle v+q_k=u+q_{\ell }⟹v-u=q_{\ell }-q_k\in \mathbb{Q}}$, a contradiction.)

Next, note that

${\displaystyle [0,1]\subseteq \bigcup _k{V}_k\subseteq [-1,2].}$

To see the first inclusion, consider any real number${\displaystyle r}$ in${\displaystyle [0,1]}$ and let${\displaystyle v}$ be the representative in${\displaystyle V}$ for the equivalence class${\displaystyle [r]}$; then${\displaystyle r-v=q_i}$ for some rational number${\displaystyle q_i}$ in${\displaystyle [-1,1]}$ which implies that${\displaystyle r}$ is in${\displaystyle V_i}$.

Apply the Lebesgue measure to these inclusions usingsigma additivity:

${\displaystyle 1\le \sum _{k=1}^{\mathrm{\infty }}\lambda (V_k)\le 3.}$

Because the Lebesgue measure is translation invariant,${\displaystyle \lambda (V_k)=\lambda (V)}$ and therefore

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

But this is impossible. Summing infinitely many copies of the constant${\displaystyle \lambda (V)}$ yields either zero or infinity, according to whether the constant is zero or positive. In neither case is the sum in${\displaystyle [1,3]}$. So${\displaystyle V}$ cannot have been measurable after all, i.e., the Lebesgue measure${\displaystyle \lambda }$ must not define any value for${\displaystyle \lambda (V)}$.

No Vitali set has theproperty of Baire.^[2]

By modifying the above proof, one shows that each Vitali set hasBanach measure${\displaystyle 0}$. This does not create any contradictions since Banach measures are not countably additive, but only finitely additive.

The construction of Vitali sets given above uses theaxiom of choice. The question arises: is the axiom of choice needed to prove the existence of sets that are not Lebesgue measurable? The answer is yes, provided thatinaccessible cardinals are consistent with the most common axiomatization of set theory, so-calledZFC.

In 1964,Robert Solovay constructed a model of Zermelo–Fraenkel set theory without the axiom of choice where all sets of real numbers are Lebesgue measurable. This is known as theSolovay model.^[3] In his proof, Solovay assumed that the existence of inaccessible cardinals isconsistent with the other axioms of Zermelo-Fraenkel set theory, i.e. that it creates no contradictions. This assumption is widely believed to be true by set theorists, but it cannot be proven in ZFC alone.^[4]

In 1980,Saharon Shelah proved that it is not possible to establish Solovay's result without his assumption on inaccessible cardinals.^[4]

• Banach–Tarski paradox – Geometric theorem
• Carathéodory's criterion – Part of measure theory mathematics
• Non-measurable set – Set which cannot be assigned a meaningful "volume"
• Outer measure – Mathematical function
• Infinite parity function

• Herrlich, Horst (2006).Axiom of Choice. Springer. p. 120.ISBN 9783540309895.
• Vitali, Giuseppe (1905). "Sul problema della misura dei gruppi di punti di una retta".Bologna, Tip. Gamberini e Parmeggiani.

• Sets of real numbers
• Measure theory
• Axiom of choice

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