Vitali Set Existence Theorem/Proof 1

Theorem

There exists a set of real numbers which is not Lebesgue measurable.

Proof

Lemma

For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that:

$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$

where $\Q$ is the set of rational numbers.

That is, $x \sim y$ if and only if their difference is rational.

Then $\sim$ is an equivalence relation.

$\Box$

Let $\map \mu X$ denote the Lebesgue measure of a set $X$ of real numbers.

We have that:

$(1): \quad$ $\map \mu X$ is a countably additive function

$(2): \quad$ $\map \mu X$ is translation invariant

$(3): \quad$ From Measure of Interval is Length, $\map \mu {\closedint a b} = b - a$ for every closed interval $\closedint a b$.

For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that:

$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$

where $\Q$ is the set of rational numbers.

That is, $x \sim y$ if and only if their difference is rational.

By the Lemma, $\sim$ is an equivalence relation.

For each $x \in \mathbb I$, let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$.

Let us use the Axiom of Choice to create a set $M \subset \mathbb I$ containing exactly $1$ element from each equivalence class.

Hence:

for each $x \in \R$ there exists a unique $y \in M$ and $r \in \Q$ such that $x = y + r$.

Let:

$M_r = \set {y + r: y \in M}$

for each $r \in \Q$.

Thus $\R$ is partitioned into countably many disjoint sets:

$(1): \quad \ds \R = \bigcup \set {M_r: r \in \Q}$

Aiming for a contradiction, suppose $M$ were Lebesgue measurable.

First, $\map \mu M = 0$ is impossible, because from $(1)$ this would mean $\map \mu \R = 0$.

Suppose $\map \mu M > 0$.

Then:

$\ds \closedint 0 2$

$\supseteq$

$\ds \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}$

$\ds \leadsto \ \ $

$\ds \map \mu {\closedint 0 2}$

$\ge$

$\ds \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$

Measure is Monotone

$\ds $

$=$

$\ds \sum_{\substack {r \mathop \in \Q \\ 0 \mathop \le r \mathop \le 1} } \map \mu {M_r}$

Axiom $(2)$ of Measure

$\ds $

$=$

$\ds \infty$

as each $M_r$ would have to have the same measure as $M$

So $\map \mu M > 0$ is also impossible.

Hence $M$ is not Lebesgue measurable.

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Source of Name

This entry was named for Giuseppe Vitali.