Vitali Set Existence Theorem

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

Proof
Let $\mu \left({X}\right)$ denote the Lebesgue measure of a set $X$ of real numbers.

We have that:


 * $(1): \quad$ $\mu \left({X}\right)$ is a countably additive function
 * $(2): \quad$ $\mu \left({X}\right)$ is translation-invariant
 * $(3): \quad$ From Measure of Interval is Length, $\mu \left[{a \,.\,.\, b}\right] = b - a$ for every closed interval $\left[{a \,.\,.\, b}\right]$.

For all real numbers in the closed unit interval $\mathbb I = \left[{0 \,.\,.\, 1}\right]$, 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 Non-Measurable Set of Real Numbers: Lemma 1, $\sim$ is an equivalence relation.

For each $x \in \mathbb I$, let $\left[\!\left[{x}\right]\!\right]$ 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:
 * $M_r = \left\{{y + r: y \in M}\right\}$

for each $r \in \Q$.

Thus $\R$ is partitioned into countably many disjoint sets:
 * $\displaystyle (1): \quad \R = \bigcup \left\{{M_r: r \in \Q}\right\}$

Suppose $M$ were Lebesgue measurable.

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

Suppose $\mu \left({M}\right) > 0$.

Then:

So $\mu \left({M}\right) > 0$ is also impossible.

Hence $M$ is not Lebesgue measurable.

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