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Measure theory Question - Help

  1. May 17, 2012 #1
    Hi all,
    I am reading Probability and Measure by Patrick Billingsley, and I am stuck at one example, please help me understanding it.
    http://desmond.imageshack.us/Himg201/scaled.php?server=201&filename=30935274.jpg&res=landing [Broken]
    Ω=(0,1]
    My question is that how come the A^c = (0,a_1]U(a'_1, a_2]U.....U(a'_m-1, a_m]U(a'_m, 1] ?????? because lets say that A= {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]} then
    A^c = Ω - A
    A^c = (0, 1] - {(0,0.1], (0.2, 0.3], (0.4, 0.5], (0.6, 0.7], (0.8, 1]}
    A^c = ∅ .........an empty set??????????

    You can see this example at http://books.google.co.uk/books?id=...q=probability and measure billingsley&f=false
    Example no 2.2 (section: Probability Measure), page 21.

    Thanks in advance.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 17, 2012 #2

    micromass

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    What does this even mean??? A is a set with intervals as elements?? That's not what Billingsley means.
     
  4. May 17, 2012 #3

    micromass

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    Please ignore this post, as it is wrong

    If you mean

    [tex]A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1][/tex]

    then yes, Ac is the empty set, because A=(0,1].
     
    Last edited: May 24, 2012
  5. May 17, 2012 #4
    Sorry for the typing mistake, yes I mean union. but then why text says A^c = (0,a_1]U(a'_1, a_2]U.....U(a'_m-1, a_m]U(a'_m, 1] , why does the complement have 0 & 1 in text ????

    [STRIKE]Perhaps I am also wrong of being making the A^c = empty set because Ω contains only 0 & 1 so shouldn't it be A^c = {(0.1, 0.2], (0.3, 0.4], (0.5, 0.6], (0.7, 0.8]} ????[/STRIKE]
     
    Last edited: May 17, 2012
  6. May 17, 2012 #5
    If [itex]A=(0,0.1]\cup (0.2, 0.3]\cup (0.4, 0.5]\cup (0.6, 0.7]\cup (0.8, 1][/itex], then the formula given states that [itex]A^{c} = (0, 0] \cup (.1, .2] \cup (.3, .4] \cup (.5, .6] \cup (.7, .8] \cup (1, 1][/itex], which is correct.
     
  7. May 18, 2012 #6
    OK
    So,
    If a1=0, then
    A^c=(0,0]∪(a1′,a2]∪...∪(am′,1]=(a1′,a2]∪...∪(am′,1]
    since (0,0]=∅. Likewise, if am′=1 we have
    A^c=(0,a1]∪(a1′,a2]∪...∪(1,1]=(0,a1]∪...∪(am−1′,am]
    since (1,1]=∅
     
  8. May 24, 2012 #7
    How comes? Shouldn't
    Ac = {0} U (0.1, 0.2] U (0.3, 0.4] U (0.5, 0.6] U (0.7, 0.8] ?
     
  9. May 24, 2012 #8
    I think (0,0]= {0}.
     
  10. May 24, 2012 #9

    D H

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    No. (0,0] is the empty set. Other than that, you are correct, as is Citan Uzuki in post #5.
     
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