# 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 $\mathcal{F}$ 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 $\mu ([0,1))=+\mathrm{\infty}$.

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 |