Proposition 1.2.18 (Non-measurable sets)

We follow the steps of the Vitali set construction.

(i)

$\forall <!--\; FUNCTION\; APPLICATION\; -->C\in ℝ∕\mathbb{Q}\exists <!--\; FUNCTION\; APPLICATION\; -->x_c\in C:x_C\in [0,1]$
Pick an arbitrary coset $C\in ℝ∕\mathbb{Q}$, i.e., an arbitrary $x\in ℝ$ with $C=x+\mathbb{Q}$. Then, we by the denseness of rationals in reals can find a rational number $q\in [-x,-x+1]$, and thus $x+q\in C$ and $x+q\in [0,1]$ simultaneously.

(ii)

For each $C\in ℝ∕\mathbb{Q}$ pick an element $x_C\in C\cap [0,1]$ and put it in a set $E$.
This requires us to use the uncountable axiom of choice. Furthermore $E\subseteq [0,1]$ by construction.

(iii)

We now look at the the measure of the following set $$X:=\bigcup _{q\in \mathbb{Q}\cap [0,1]}E+q$$

Assume for the sake of contradiction that $X$ were Lebesgue measurable. We then would have by the additivity

$$m(X)=\sum _{q\in \mathbb{Q}\cap [0,1]}m(E+q)=\sum _{q\in \mathbb{Q}\cap [0,1]}m(E)=\{\begin{array}{cc}0\hfill & \text{if }m(E)=0\hfill \\ \infty \hfill & \text{if }m(E)>0\hfill \end{array}$$

(iv)

We now demonstrate that both cases lead to a contradiction. For this we establish an upper and a lower bound for $X$.

• Lower bound:

  $$[0,1]\subseteq X$$

  Pick an arbitrary $y\in [0,1]$, and choose some coset in which $y$ is guaranteed contained in (for instance $y\in y+\mathbb{Q}$). From this coset we have already taken associated $x_C$. Notice that the quantity $x_C-y\in \mathbb{Q}$ must be rational and is contained in $[-1,1]$ Therefore $y=x_C+q\in E+q$ for some $q\in [-1,1]$ and $y\in X$ follows.

• Upper bound:

  $$X\subseteq [-1,2]$$

  Pick an arbitrary $y\in X$. Then $y=x+q$ for some $x\in E\subseteq [0,1]$ and $q\in [-1,1]$. Thus, $y\in [-1,2]$.

Thus, by monotonicity $1\le m(X)\le 3$ and the two facts contradict each other.