proof of Vitali’s Theorem

Consider the equivalence relationMathworldPlanetmath in [0,1) given by


and let be the family of all equivalence classesMathworldPlanetmath of . Let V be a of i.e. put in V an element for each equivalence class of (notice that we are using the axiom of choiceMathworldPlanetmath).

Given q[0,1) define


that is Vq 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[0,1) there exists yV such that xy (because V is a section of ) and hence there exists q[0,1) such that xVq. So


Moreover all the Vq are disjoint. In fact if xVqVp 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 Vq are measurable and μ(Vq)=μ(V). Moreover by the countable additivityMathworldPlanetmath of μ we have


So if μ(V)=0 we had μ([0,1))=0 and if μ(V)>0 we had μ([0,1))=+.

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