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Measure theory

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Let D be the collection of all finite subsets ( including the empty set) of [0,1].
    Prove that D is a semi-ring. What is [tex]\sigma(D)[/tex] ? Define on D: [tex]\mu (A)[/tex]=#A . Prove that [tex]\mu[/tex] is a premeasure and identify [tex]\mu_{e}[/tex] and
    [tex]\Sigma_{mu_{e}}[/tex] . Is ([0,1],[tex]\sigma (D)[/tex], [tex]\mu_{e}[/tex]) complete?
    Prove that ([0,1],[tex]\sigma (D)[/tex], [tex]\mu_{e}[/tex]) [tex]\neq[/tex]
    ([0,1],[tex]\Sigma_{mu_{e}}[/tex],[tex]\mu_{e}[/tex]).
    2. Relevant equations
    [tex]\mu_{e}[/tex] is the outer measure,
    [tex]\Sigma_{mu_{e}}[/tex] is the collection of all [tex]\mu_{e}[/tex] measurable sets.

    [tex]\sigma (D)[/tex] is the sigma algebra generated by D
    3. The attempt at a solution
    showing that D is a semi ring is clear.
    but [tex]\sigma (D)[/tex] is a little unclear, since it must be closed under complementation, so if A [tex]\in[/tex] D, then A is a finite set, but A[tex]^{c}[/tex]
    may not be a finite set.
    showing that [tex]\mu[/tex] is a pre-measure is clear.
    any comments for the rest is appreciated.
     
    Last edited: Jan 21, 2008
  2. jcsd
  3. Jan 21, 2008 #2
    Um, the complement of a finite set in [0,1] will definitely not be finite. But why is that a problem? Doesn't it just make the sigma algebra generated by D the collection of all sets with finite complement in [0,1] and their complements?

    Edit: Oh, sorry, it has to be closed under countable unions, so I guess it's not that simple.
     
    Last edited: Jan 21, 2008
  4. Jan 21, 2008 #3

    morphism

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    Why stick to finiteness? sigma algebras work well with countability. The sigma algebra generated by D certainly contains all countable sets and sets whose complement is countable (i.e. cocountable sets); can it contain anything else?
     
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