Vitali Set Existence Theorem

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

Lemma
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, $x \sim y$ their difference is rational.

By Vitali Theorem: Lemma, $\sim$ is an equivalence relation.

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

By the Axiom of Choice, it is possible to choose $1$ element from each equivalence class.

Thus is created a set $M \subset \mathbb I$ such that:
 * 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:
 * $(1): \quad \ds \R = \bigcup \set {M_r: r \in \Q}$

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:

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

Hence $M$ is not Lebesgue measurable.

Also see
The set $M$ described here is an example of a Vitali set: a subset of the real numbers which has no Lebesgue measure.