Existence of Non-Measurable Subset of Real Numbers

Theorem
There exists a subset of the real numbers which is not measurable.

Proof
We construct such a set.

For $x, y \in \hointr 0 1$, define the sum modulo 1:


 * $x +_1 y = \begin {cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end {cases}$

Let $E \subset \hointr 0 1$ be a measurable set.

Let $E_1 = E \cap \hointr 0 {1 - x}$ and $E_2 = E \cap \hointr {1 - x} 1$.

By Measure of Interval is Length, these disjoint intervals are measurable.

By Measurable Sets form Algebra of Sets, so are these intersections $E_1$ and $E_2$.

So:
 * $\map m {E_1} + \map m {E_2} = \map m E$

We have:
 * $E_1 +_1 x = E_1 + x$

By Lebesgue Measure is Translation-Invariant:
 * $\map m {E_1 +_1 x} = \map m {E_1}$

Also:
 * $E_2 +_1 x = E_2 + x - 1$

and so:
 * $\map m {E_2 +_1 x} = \map m {E_2}$

Then we have:
 * $\map m {E +_1 x} = \map m {E_1 +_1 x} + \map m {E_2 +_1 x} = \map m {E_1} + \map m {E_2} = \map m E$

So, for each $x \in \hointr 0 1$, the set $E +_1 x$ is measurable and:
 * $\map m {E + x} = \map m E$

Taking, as before, $x, y \in \hointr 0 1$, define the relation:
 * $x \sim y \iff x - y \in \Q$

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

By Difference is Rational is Equivalence Relation, $\sim$ is an equivalence relation.

As this is an equivalence relation we can invoke the fundamental theorem on equivalence relations.

Hence $\sim$ partitions $\hointr 0 1$ into equivalence classes.

By the axiom of choice, there is a set $P$ which contains exactly one element from each equivalence class.

$P$ is measurable.

Let $\set {r_i}_{i \mathop = 0}^\infty$ be an enumeration of the rational numbers in $\hointr 0 1$ with $r_0 = 0$.

Let $P_i := P +_1 r_i$.

Then $P_0 = P$.

Let $x \in P_i \cap P_j$.

Then:
 * $x = p_i + r_i = p_j + r_j$

where $p_i, p_j$ are elements of $P$.

But then $p_i - p_j$ is a rational number.

Since $P$ has only one element from each equivalence class:
 * $i = j$

The $P_i$ are pairwise disjoint.

Each real number $x \in \hointr 0 1$ is in some equivalence class and hence is equivalent to an element of $P$.

But if $x$ differs from an element in $P$ by the rational number $r_i$, then $x \in P_i$ and so:
 * $\ds \bigcup P_i = \hointr 0 1$

Since each $P_i$ is a translation modulo $1$ of $P$, each $P_i$ will be measurable if $P$ is, with measure $\map m {P_i} = \map m P$.

But if this were the case, then:
 * $\ds m \hointr 0 1 = \sum_{i \mathop = 1}^\infty m \paren {P_i} = \sum_{i \mathop = 1}^\infty \map m P$

Therefore:
 * $\map m P = 0$ implies $m \hointr 0 1 = 0$

and:
 * $\map m P \ne 0$ implies $m \hointr 0 1 = \infty$

This contradicts Measure of Interval is Length.

So the set $P$ is not measurable.