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Concrete non-measurable sets

  1. Aug 18, 2011 #1
    "Concrete" non-measurable sets

    I've had Vitali's proof of the existence of non-(Lebesgue) measurable sets branded into the side of my brain over the years. However, the proof always critically relies on evoking the axiom of choice. Has anybody every demonstrated a non-AoC construction of a non-measurable set? Or do the intuitionist logicians just avoid measure theory all together?
     
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  3. Aug 18, 2011 #2

    micromass

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    Re: "Concrete" non-measurable sets

    Hi Kreizhn! :smile:

    The axiom of choice is necessary for the existence of non-measurable sets. If you don't adopt the axiom of choice (or a similar principle), then it can happen that all sets are measurable! There is a current investigation of some theories in which all sets are measurable, so it certainly can happen!
     
  4. Aug 18, 2011 #3
    Re: "Concrete" non-measurable sets

    Good to know. Then does using AoC necessarily preclude the ability to construct non-measurable sets that are anything but "existential?"

    For example, (unless it's just been so long since I've done set theory) I imagine there are times when we must use the AoC in general, but for special cases we could explicitly construct a choice function. Are there special cases wherein this can be done for non-measurable sets? Maybe that's just a silly question.
     
  5. Aug 18, 2011 #4

    micromass

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    Re: "Concrete" non-measurable sets

    I don't quite understand what you're asking. Do you want to construct a choice function for a non-measurable set?? This is not possible, everything you do with non-measurable sets involves the axiom of choice. It is not possible to construct a choice function for them.
     
  6. Aug 18, 2011 #5
    Re: "Concrete" non-measurable sets

    I guess what I'm thinking is that AoC always guarantees the existence of a choice function, without explicitly defining it. However, there are instances in which AoC is not required to give the choice function.

    But then I guess the answer is that if we must always use AoC, then it is impossible to explicitly construct a choice function, since then AoC would not be necessary to construct nonmeasurable sets, and you have just told me that it is.
     
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