Vitali set

The term Vitali set describes any set obtained by a particular mathematical construction. The construction uses the axiom of choice and its result given by an existence theorem is not uniquely determined. Vitali sets have many important applications in mathematics, most notable being a proof of existence of Lebesgue non-measurable sets in the measure theory. The name was given after the Italian mathematician Giuseppe Vitali.

Formal construction

We begin by defining the following relation on the real line. Two real numbers x and y are said to be equivalent if and only if the difference x-y is rational. In symbols,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \sim y \iff x-y \in \mathbb{Q}.}

It is easy to verify that it is in fact an equivalence relation. Thus, it yields the partition the set of reals into its equivalence classes. By the axiom of choice we can select a representative of each single class. The Vitali set V is defined to be the union all selected representatives.

Since for any real x the elements of the set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ x+i:\;\;i\in \mathbb{Z} \}} are in the same equivalence class, me may (and do) additionaly require that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\subset [0,1]} (indeed, x+i belong to [0,1] for some i).

Application to measure theory

A Vitali set can not be included in the family of measurable sets for any translation invariant measure. In particular it is not Lebesgue measurable.

More precisely, suppose that a measure μ defined over a σ-algebra Σ of subsets of the real line satisfies

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(A) = \mu(A+x),\quad A\in\Sigma,\, x\in \mathbb{R}.}

In particular, this presumes that any translation of any measurable set is measurable. We will show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\notin \Sigma.}

Observe that for any rational Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q\not=0} the sets V+q and V are disjoint. Indeed, if there is any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in (V+q)\cap V} then at the same time Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x-q\in V} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in V} . In other words V contains two distinct representatives of the same equivalence class, which contradicts the definition of V.

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_1,q_2,...} be an enumeration of the rationals from [-1,1] and define

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W=\bigcup_{i=n}^\infty (V+q_n).}

Then, clearly

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]\subset W\subset [-1,2].}

Indeed, the second inclusion follows directly from the fact that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\subset[0,1]} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_n\in[-1,1].} For the first one, observe that for any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in [0,1]} , the class of equivalence of x has its uniqe representative y in V. Therefore x-y is rational and, by the fact that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\subset[0,1]} we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x-y\in [-1,1]} . In other words, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x-y=q_n} for some n and so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in V+q_n\subset W} .

Now suppose that the set V is measurable (and so are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V+q_n} ). Recall that by definition any measure is supposed to be countably additive. It follows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(W) = \mu (\bigcup_{n=1}^\infty (V+q_n)) = \sum_{n=1}^\infty \mu(V+q_n).}

since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (V+q_n)_{n=1..\infty}} is a family of pairwise disjoint sets. Further, by tranlation invariance of μ this is equal to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \mu(V).} Since W contains the interval [0,1] we clearly have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(W) \ge \mu([0,1])=1,} which implies that $\mu (V)$ is strictly positive. At the same time, the infinite sum is bounded by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu([-1,2]) = 3} , which is a contradiction. Consequently, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V\notin \Sigma.}