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 $x, y \in \left[{0. . 1}\right)$, 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 $E \subset \left[{0. . 1}\right)$ be a measurable set.

Let $E_1 = E \cap \left[{0. . 1 - x}\right)$ and $E_2 = E \cap \left[{1 - x . . 1}\right)$.

These disjoint intervals are necessarily measurable and hence so are these intersections, so $m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$.

We have $E_1 +_1 x = E_1 + x$, and so by the translation invariance of Lebesgue measure:
 * $m \left({E_1 +_1 x}\right) = m \left({E_1}\right)$

Also:
 * $E_2 +_1 x = E_2 + x - 1$, and so $m \left({E_2 +_1 x}\right) = m \left({E_2}\right)$

Then we have:
 * $m \left({E +_1 x}\right) = m \left({E_1 +_1 x}\right) + m \left({E_2 +_1 x}\right) = m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$

So, for each $x \in \left[{0. . 1}\right)$, the set $E +_1 x$ is measurable and:
 * $m \left({E + x}\right) = m \left({E}\right)$

Taking, as before, $x, y \in \left[{0. . 1}\right)$, define an equivalency $x \sim y$ iff $x - y \in \Q$, the set of rationals.

This is an equivalence relation and hence partitions $\left[{0. . 1}\right)$ into equivalence classes.

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

Let $\left\{{r_i}\right\}_{i=0}^\infty$ be an enumeration of the rational numbers in $\left[{0. . 1}\right)$ with $r_0 = 0$ and define $P_i = P +_1 r_i$. 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, and since $P$ has only one element from each equivalence class, $i = j$.

The $P_i$ are pairwise disjoint.

Each real number $x \in \left[{0. . 1}\right)$ 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 $\displaystyle \bigcup P_i = \left[{0. . 1}\right)$.

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

But if this were the case, then:
 * $\displaystyle m \left[{0 . . 1}\right) = \sum_{i=1}^\infty m \left({P_i}\right) = \sum_{i=1}^\infty m \left({P}\right)$

Therefore:
 * $m \left({P}\right) = 0$ implies $m \left[{0 . . 1}\right) = 0$

and:
 * $m \left({P}\right) \ne 0$ implies $m \left[{0 . . 1}\right) = \infty$

This contradicts basic results regarding Lebesgue measure, and so the set $P$ is not measurable.