# proof of Vitali’s Theorem

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

 $x\sim y\quad\Leftrightarrow\quad x-y\in\mathbb{Q}$

and let $\mathcal{F}$ be the family of all equivalence classes of $\sim$. Let $V$ be a of $\mathcal{F}$ 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 $\mu([0,1))=+\infty$.

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

Title proof of Vitali’s Theorem ProofOfVitalisTheorem 2013-03-22 13:45:50 2013-03-22 13:45:50 paolini (1187) paolini (1187) 7 paolini (1187) Proof msc 28A05 ProofOfPsuedoparadoxInMeasureTheory