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Proof of existence of nonmeasurable sets

  1. Dec 22, 2012 #1
    Hi,

    I'm reading through a proof of the existence of a nonmeasurable set. I've copied down the proof below more or less verbatim:



    In particular, I am trying to understand the significance of why ##\alpha## has to be an irrational number. Would the proof not hold if we used any other number ##\alpha \in \mathbb{R}##? For your reference, this proof was given in Kolmogorov's "Elements of the Theory of Functions and Functional Analysis". I'd really appreciate if someone could explain this for me. Thanks!
     
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  3. Dec 22, 2012 #2

    HallsofIvy

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    If [itex]\alpha[/itex] were a rational number, say [itex]\alpha= p/q[/itex] for integers p and q, then [itex]n\apha[/itex] would be an integer for some n. The sets would not be "pairwise disjoint".
     
  4. Dec 22, 2012 #3
    Ah okay I see. In the case that ##\alpha \in \mathbb{Q}##, each ##\Phi _n## is either pairwise disjoint or exactly equal to ##\Phi _m## (for example ##\Phi _0 = \Phi _{360}## if ##\alpha = 1##, assuming we are using degree angles), so that consequently we wouldn't have the the infinite union ##C = \bigcup _{n=-\infty}^\infty \Phi _n## but rather a finite union ##C = \bigcup _{n=1}^k \Phi _n ## after removing redundant sets, where the contradiction would no longer hold. Is this more or less what you are hinting at? Thanks!
     
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