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Complete measure space

  1. Jan 22, 2008 #1
    [SOLVED] complete measure space

    1. The problem statement, all variables and given/known data

    Assume that ([tex]\Omega[/tex],[tex]\Sigma[/tex],[tex]\mu[/tex]) is a complete
    measure space, let [tex]\mu_{e}[/tex] be the outer measure defined by [tex]\mu[/tex]
    . Prove that if [tex]\mu_{e}[/tex](S)=0 [tex]\Rightarrow[/tex] S[tex]\in[/tex][tex]\Sigma[/tex] .

    I know that [tex]\mu_{e}[/tex] = [tex]\mu[/tex] when restricted on [tex]\Sigma[/tex]

    and that if [tex]\mu[/tex](A)=0, then every subset of A has measure zero from completeness.

    I intuitively can see this to be true, since if S where not in [tex]\Sigma[/tex] and
    [tex]\mu_{e}[/tex] = 0 , then this would contradict the fact that ([tex]\Omega[/tex],[tex]\Sigma[/tex],[tex]\mu[/tex]) is complete, since the fact that
    [tex]\mu_{e}(A_{1})[/tex]=[tex]\mu_{e}(A_{2})[/tex]...=[tex]\mu_{e}(A_{n})[/tex]=0, for [tex]A_{1}\subseteq A_{2}\subseteq.......\subseteq A_{n}\subseteq S[/tex]
    would imply that [tex]\Sigma[/tex] is incomplete, but not sure how to rigorously prove
    Last edited: Jan 22, 2008
  2. jcsd
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