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单词 ProofOfVitalisTheorem
释义

proof of Vitali’s Theorem


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

xyx-y

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

Given q[0,1) define

Vq=((V+q)[0,1))((V+q-1)[0,1))

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

q[0,1)Vq=[0,1).

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

μ([0,1))=q[0,1)μ(Vq)=qμ(V).

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.

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更新时间:2025/5/4 18:44:06