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Show that the complement of a non-measurable set is also n.m

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the complement of a non-measurable set is also non-measurable.
    3. The attempt at a solution
    Let A be a set that is non measurable. Let B be the complement of A.
    Lets assume for contradiction that B has a measure. Now from the axioms of measure theory they have countable additive of measure. case 1: A+B has a measure.
    If B has a measure and A+B has a measure then A+B-B should have a measure. But A has no measure so this is a contradiction therefore B has no measure.
    Case 2: A+B has no measure.
    Lets assume B has a measure, If B has a measure then A+B has a measure
    But this is a contradiction, therefore B has no measure .
    Not sure If on case 2 I can talk about A+B having a measure
     
  2. jcsd
  3. Sep 25, 2016 #2

    andrewkirk

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    It depends on the definition of measurable used. The usual definition is as shown here. Under that definition, all sets in the sigma-algebra ##\Sigma## are measurable and, since a set is in ##\Sigma## iff its complement is, the result follows trivially.

    I suspect you are using a different definition of measurable - perhaps a specific measure like Lebesgue measure. There is a list of possible measures here. Do you mean one of them? If so, which one?
     
  4. Sep 26, 2016 #3
    When my teacher told us about the complement of a non-measurable set is non-measurable. I wanted to prove this for myself, he talked about it before Lebesgue measure. And before sigma-algebra. We just had the basic axioms of measure theory like countably additive, the measure of a point is zero. the measure of an interval is the difference between the endpoints. Like for example the measure of the irrationals is infinite.
     
  5. Sep 26, 2016 #4

    micromass

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    A measure is a function. What is the domain of the measure defined like?
     
  6. Sep 26, 2016 #5
    It can be no more than a countable union of disjoint sets. If the measure exists it will be assigned a non-negative real number. And it is closed under compliment.
    If its closed under compliment then the set and its compliment will have a measure. Because the set and its compliment is just one partition of the set. thanks for the responses
     
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