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 $$ \bigcup P_i = \left[{0 \,. \, . \, 1}\right) \ $$.

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:
 * $$m \left[{0 \, . \, . \, 1}\right) = \sum_{i=1}^\infty m \left({P_i}\right) = \sum_{i=1}^\infty m \left({P}\right)$$

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.