The Vitali Set | Real Analysis

Using the axiom of choice to pick one representative from each class of

\mathbb{R}/\mathbb{Q}

births a non-measurable set. Countably many disjoint translates cover

[0,1]

inside

[-1,2]

— yet measure 0 gives total 0 and positive measure gives infinity. Both impossible.

Examples & Exercises

Put the equivalence relation $x \sim y \iff x - y \in \mathbb{Q}$ on $[0, 1]$ and, by the axiom of choice, build $V \subset [0, 1]$ by picking one representative from each class. Prove completely that the translates $V + q_1$ and $V + q_2$ are disjoint for distinct rationals $q_1 \ne q_2$ .

x \in (V + q_1) \cap (V + q_2) \;\Longrightarrow\; q_1 = q_2

d = 3

9

rationals with denominator at most

3

— each row a translate of the mock representative set. The real proof covers all countably many rationals.

Show proof

[Use of AC — the only one] $x \sim y \iff x - y \in \mathbb{Q}$ is an equivalence relation on $[0, 1]$ (reflexive, symmetric, transitive because $\mathbb{Q}$ is an additive group). Invoke the axiom of choice exactly once to pick one representative from each class meeting $[0, 1]$ . The collection of picks is $V$ — so $V \subset [0, 1]$ and $V$ holds exactly one element from each class.
Let $x \in (V + q_1) \cap (V + q_2)$ ; then $x = v_1 + q_1 = v_2 + q_2$ for some $v_1, v_2 \in V$ . Rearranging, $v_1 - v_2 = q_2 - q_1 \in \mathbb{Q}$ — i.e. $v_1 \sim v_2$ .
Since $V$ holds only one element per class, $v_1 \sim v_2$ forces $v_1 = v_2$ . Then $q_1 = x - v_1 = x - v_2 = q_2$ — contradiction. Hence translates by distinct rationals are disjoint.

For the $V$ of problem 1, prove completely the sandwich $[0, 1] \subset \bigcup_q (V + q) \subset [-1, 2]$ , where $q$ ranges over the rationals in $\mathbb{Q} \cap [-1, 1]$ .

[0, 1] \subset \bigcup_{q \in \mathbb{Q} \cap [-1,1]} (V + q) \subset [-1, 2]

d = 6

25

rationals — the mock translates pack the

[-1, 2]

box and cover the

[0, 1]

band.

Show proof

[The axiom of choice — used as in problem 1] $V$ is the set of representatives chosen via the axiom of choice, one per $\sim$ -class, with $V \subset [0, 1]$ . Both inclusions below are direct consequences of this construction.
Left inclusion: let $x \in [0, 1]$ . Its class has a representative $v \in V$ . Then $q = x - v \in \mathbb{Q}$ , and since $x, v \in [0, 1]$ , $q \in [-1, 1]$ . So $x = v + q \in V + q$ — i.e. $[0, 1] \subset \bigcup_q (V + q)$ .
Right inclusion: $V \subset [0, 1]$ and $q \in [-1, 1]$ , so $V + q \subset [0, 1] + [-1, 1] = [-1, 2]$ . The same holds for the union over all $q$ . The scene's $[-1, 2]$ box and $[0, 1]$ band are exactly this sandwich.

Assume $V$ is Lebesgue measurable and derive a contradiction: from countable additivity and translation invariance, prove completely that both $m(V) > 0$ and $m(V) = 0$ are impossible.

m\left(\bigcup_q (V + q)\right) = \sum_q m(V + q) = \sum_q m(V) \quad\text{yet}\quad 1 \le m\left(\bigcup_q\right) \le 3

d = 2

: even with just

5

rationals shown — with

m = 1

the measure total

5 > 3

already overflows the box (the finite shadow of the contradiction).

Show proof

[The axiom of choice — used as in problem 1] $V$ is the representative set built with the axiom of choice. Suppose $V$ is measurable and put $m = m(V)$ . By translation invariance, $m(V + q) = m$ for every rational $q$ .
By problem 1 the translates are disjoint, so countable additivity gives $m\left(\bigcup_q (V + q)\right) = \sum_q m(V + q) = \sum_q m$ — countably many copies of $m$ summed. Monotonicity on problem 2's sandwich gives $1 = m([0, 1]) \le m\left(\bigcup_q\right) \le m([-1, 2]) = 3$ .
If $m > 0$ then $\sum_q m = \infty$ , contradicting $\le 3$ . If $m = 0$ then $\sum_q m = 0$ , contradicting $\ge 1$ . No $m$ works, so the assumption (measurability) is false — $V$ is not Lebesgue measurable. AC was used exactly once, to choose the representatives, yet out comes a non-measurable set.