Vitali Set

Contents (hide)

  1. 1. Introduction
  2. 2. Construction of the set
  3. 3. Proof that the Vitali set is not measurable

1.  Introduction

The Vitali set is an example of a set that is not measurable. As we expect, the structure of the Vitali set is very complicated for it to be constructed explicitely, but we can show its existence and the fact that it cannot be measurable.

2.  Construction of the set

Consider the interval \fs4 \left[0,1\right], and assume the simple measure of length (the Lesbegue measure)

This maps each subinterval to its length, for example \fs4 L([1/2, 3/4]) = 1/4 or \fs4 L([1/3,2/3] \cup [8/9,1]) = 4/9, etc. In terms of probabilities, this would be equivalent to the uniform probability measure, where the probability that we pick a number in \fs4 \left[a,b\right] is equal to the length \fs4 |b-a|.

The question is: can we say that any arbitrary subset of the unit interval has a length? Or equivalently, does any subset of the unit interval has a probability (which might be zero of course)? Intuition would tell us that the answer should be yes, but the Vitali set shows otherwise.

We will say that two numbers \fs4 x,y \in \mathbb{R} are equivalent, if their difference is a rational number. Then we can split all real numbers in classes where they are all equivalent. Then, from each class we select one representative element that lies in the unit interval (we can do that due to the axiom of choice).

For example

  • The numbers \fs4 \sqrt{1/2}, \fs4 \sqrt{1/2} + 3, \fs4 \sqrt{1/2},... are all in the same class. We can choose \fs4 \sqrt{1/2} as the representative
  • The numbers \fs4 \pi/2, \fs4 \pi/2 - 1, \fs4 \pi/2+6,... are all in the same class. We can choose \fs4 \pi/2 - 1 as the representative
  • The numbers \fs4 \sqrt{3}, \fs4 \sqrt{3} - 1, \fs4 \sqrt{3} - 6/5,... are all in the same class. We can choose \fs4 \sqrt{3}-6/5 as the representative (although we could have chosen \fs4 \sqrt{3} - 1 if we wanted to, but we only select one!)
  • All rational numbers are in the same class (since their difference will be rational!) and we can select \fs4 0 as their representative

The collection of all these representatives will be the Vitali set, denoted with \fs4 V. Now what is the length of this set? Or, what is the probability that a random number in \fs4 \left[0,1\right] will belong to this set?

3.  Proof that the Vitali set is not measurable

Suppose that the Vitali set is measurable, and say that its length is \fs4 P=L(V). \fs4 P will also be the probability that we randomly select a member of the Vitali set.

The measure is an operation that is not affected by translation: that is to say if I add a constant to each element of a set, its length will be the same. For any number \fs4 x the set

\fs4 V(x) = \{v+x : v \in V\}

will have the same length as \fs4 V: \fs4 L(V(x)) = L(V) = P.

Now assume an ordering of the rational numbers in the unit interval (which we know exists), say

\fs4 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, \ldots

and constuct the sets \fs4 V_i = V(q_i) for the \fs4 i-th rational according to this ordering. It is easy to show that for different rationals \fs4 q_i, q_j \in \mathbb{Q} their corresponding sets are disjoint, \fs4 V_i \cap V_j = \emptyset [Proof: If they had a common element, say \fs4 y then \fs4 y = x_i + q_i = x_j + q_j, with \fs4 x_i, x_j \in V. That implies \fs4 x_i - x_j = q_i - q_j, which is a rational. Therefore \fs4 x_i and \fs4 x_j are equivalent, and they belong to the same class. Therefore they cannot belong to \fs4 V, since \fs4 V has only one representative of each class. QED]

The measure of a countable union of disjoint sets is just the sum of their measures (a measure property), and therefore

\fs4 L(V_1 \cup V_2 \cup \cdots) = L(V_1) + L(V_2) + \cdots = P + P + \cdots

This infinite sum can be either zero (if the measure of the Vitali set \fs4 P is zero) of infinity (if the measure of the Vitali set is non zero).

But we also have that

\fs4 [0, 1] \subseteq V_1 \cup V_2 \cup \cdots \subseteq [0, 2]

which indicates that

\fs4 1 = L([0,1]) \leq L(V_1 \cup V_2 \cup \cdots) \leq L([0,2]) = 2

Thus the infinite sum cannot be zero nor infinity and the measure \fs4 L(V) simply does not exist!

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