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 $$x,y \in [0,1) \ $$, define the sum modulo 1

$$x+_1y = \begin{cases} x+y, & \mbox{if } x+y<1 \\ x+y-1,  & \mbox{if } x+y \geq 1 \end{cases}$$

Let $$E \subset [0,1) \ $$ be a measurable set.

Let $$E_1 = E \cap [0,1-x) \ $$ and $$E_2 = E \cap [1-x,1) \ $$.

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

We have $$E_1+_1x =E_1 + x \ $$, and so by the translation invariance of Lebesgue measure, $$m(E_1 +_1 x)=m(E_1) \ $$.

Also, $$E_2 +_1 x = E_2 + x -1 \ $$, and so $$m(E_2 +_1 x) = m(E_2) \ $$.

Then we have $$m(E+_1x)=m(E_1+_1x)+m(E_2+_1x)=m(E_1)+m(E_2)=m(E) \ $$

So, for each $$x \in [0,1) \ $$, the set $$E +_1 x \ $$ is measurable and $$m(E+x)=m(E) \ $$

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

This is an equivalence relation and hence partitions $$[0,1) \ $$ 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 $$[0,1) \ $$ 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 $$x \in [0,1) \ $$ 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 $$ \bigcup P_i = [0,1) \ $$.

Since each $$P_i \ $$ is a translation modulo 1 of $$P \ $$, each $$P_i \ $$ will be measurable if $$P \ $$ is, with measure $$m(P_i) = m(P) \ $$. But if this were the case,

$$m[0,1) = \sum_{i=1}^\infty m(P_i) = \sum_{i=1}^\infty m(P)$$

Therefore, $$m(P)=0 \ $$ implies $$m[0,1)=0 \ $$ and $$m(P) \neq 0 \ $$ implies $$m[0,1) = \infty$$.

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