# Vitali set

In

mathematics , a**Vitali set**is an elementary example of a set ofreal number s that is not Lebesgue measurable. The**Vitali theorem**is theexistence theorem that there are such sets. It is a non-constructive result. The naming is forGiuseppe Vitali .Despite the terminology, there are many Vitali sets. Their existence is proved using the

axiom of choice , and for reasons too complex to discuss here, Vitali sets are impossible to describe explicitly.**The importance of non-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, they likewise have well-defined masses. If the [0, 1] rod weighs 1 kilogram, then the [3, 9] rod weighs 6 kilograms. 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'd 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 'length'? As an example, we might ask what is the mass of the set of

rational number s. They are very finely spread over all of thereal line , so any answer may appear reasonable at first pass.As it turns out, the physically relevant solution is to use measure theory. In this setting, the

Lebesgue measure , which assigns weight "b" − "a" to the interval ["a", "b"] , will assign weight 0 to the set of rational numbers. Any set which has a well-defined weight is said to be "measurable." The construction of the Lebesgue measure (for instance, using theouter measure ) does not make obvious whether there are non-measurable sets.**Construction and proof**If "x" and "y" are

real number s and "x" − "y" is arational number , then we write "x" ~ "y" and we say that "x" and "y" are "equivalent"; ~ is anequivalence relation . For each "x", there is a subset ["x"] = {"y" in**R**: "x" ~ "y"} of**R**called the "equivalence class" of "x". The set of these equivalence classes partitions**R**. By theaxiom of choice , we are able to choose a set $V\; subset\; [0,\; 1]$ containing exactly one representative out of each equivalence class (for any equivalence class ["x"] , the set "V" ∩ ["x"] is a singleton). We say that "V" is a Vitali set.A Vitali set is non-measurable. To show this, we assume that "V" is measurable. From this assumption we carefully work and prove something absurd: namely that "a" + "a" + "a" + ... (an infinite sum of identical numbers) is between 1 and 3. Since an absurd conclusion is reached, it must be that the only unproved hypothesis ("V" is measurable) is at fault.

First we let "q"

_{1}, "q"_{2}, ... be an enumeration of the rational numbers in [−1, 1] (recall that the rational numbers arecountable ). From the construction of "V", note that the sets $V\_k=\{v+q\_k\; :\; v\; in\; V\}$, "k" = 1, 2, ... are pairwise disjoint, and further note that $[0,1]\; subseteqigcup\_k\; V\_ksubseteq\; [-1,2]$. (To see the first inclusion, consider any real number "x" in [0,1] and let "v" be the representative in "V" for the equivalence class ["x"] ; then "x" −"v" = "q" for some rational number in [-1,1] (say "q" = "q"_{l}) and so "x" is in "V"_{l}.)From the definition of Lebesgue measurable sets, it can be shown that all such sets have the following two properties:

1. The measure is

countably additive , that is if $A\_i$ is a set of at most a countable number of pairwise-disjoint sets, then $mu\; left(igcup\_\{i=1\}^\{infty\}A\_i\; ight)=sum\_\{i=1\}^\{infty\}mu(A\_i)$.2. The measure is

translation invariant , that is, for any real number x, $mu(A)=mu(A+x)$.Consider now the measure μ of the union given above. Because μ is countably additive, it must also have the property of being "monotone"; that is, if "A"⊂"B", then μ("A")≤μ("B"). Hence, we know that

:$1\; leq\; muleft(igcup\_k\; V\_k\; ight)\; leq\; 3.$

By countable additivity, one has

:$muleft(igcup\_k\; V\_k\; ight)\; =\; sum\_\{k=1\}^infty\; mu(V\_k)$

with equality following because the "V"

_{"k"}are disjoint. Because of translation invariance, we see that for each "k" = 1, 2, ..., μ("V"_{"k"}) = μ("V"). Combining this with the above, one obtains:$1\; leq\; sum\_\{k=1\}^infty\; mu(V)\; leq\; 3.$

The sum is an infinite sum of a single real-valued constant, non-negative term. If the term is zero, the sum is likewise zero, and hence it is certainly not greater than or equal to one. If the term is nonzero then the sum is infinite, and in particular it isn't smaller than or equal to 3.

This conclusion is absurd, and since all we've used is translation invariance and countable additivity, it must be true that "V" is non-measurable.

**ee also***

Non-measurable set

*Banach–Tarski paradox **References*** Herrlich, Horst: "Axiom of Choice", page 120. Springer, 2006.

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