proof of Vitali’s Theorem

Consider the equivalence relation in $[0,1)$ given by

$$x\sim y\mathit{\hspace{1em}}\iff \mathit{\hspace{1em}}x-y\in \mathbb{Q}$$

and let $ℱ$ be the family of all equivalence classes of $\sim$. Let $V$ be a of $ℱ$ i.e. put in $V$ an element for each equivalence class of $\sim$ (notice that we are using the axiom of choice).

Given $q\in \mathbb{Q}\cap [0,1)$ define

$$V_q=((V+q)\cap [0,1))\cup ((V+q-1)\cap [0,1))$$

that is $V_q$ is obtained translating $V$ by a quantity $q$ to the right and then cutting the piece which goes beyond the point $1$ and putting it on the left, starting from $0$.

Now notice that given $x\in [0,1)$ there exists $y\in V$ such that $x\sim y$ (because $V$ is a section of $\sim$) and hence there exists $q\in \mathbb{Q}\cap [0,1)$ such that $x\in V_q$. So

$$\bigcup _{q\in \mathbb{Q}\cap [0,1)}{V}_q=[0,1).$$

Moreover all the $V_q$ are disjoint. In fact if $x\in V_q\cap V_p$ then $x-q$ (modulus $[0,1)$) and $x-p$ are both in $V$ which is not possible since they differ by a rational quantity $q-p$ (or $q-p+1$).

Now if $V$ is Lebesgue measurable, clearly also $V_q$ are measurable and $\mu (V_q)=\mu (V)$. Moreover by the countable additivity of $\mu$ we have

$$\mu ([0,1))=\sum _{q\in \mathbb{Q}\cap [0,1)}\mu (V_q)=\sum _q\mu (V).$$

So if $\mu (V)=0$ we had $\mu ([0,1))=0$ and if $\mu (V)>0$ we had $\mu ([0,1))=+\mathrm{\infty }$.

So the only possibility is that $V$ is not Lebesgue measurable.

Title	proof of Vitali’s Theorem
Canonical name	ProofOfVitalisTheorem
Date of creation	2013-03-22 13:45:50
Last modified on	2013-03-22 13:45:50
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	7
Author	paolini (1187)
Entry type	Proof
Classification	msc 28A05
Related topic	ProofOfPsuedoparadoxInMeasureTheory