Existence of Non-Measurable Sets

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.

$\blacksquare$

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.