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Properties of probability measures

  1. Sep 3, 2004 #1
    I hate it when a fact is so obvious that it isn't obvious how to prove it. Like showing that a subset of a finite set is finite. So ... here goes:

    A probability measure [itex]P[/itex] on a [itex]\sigma[/itex]-field [itex]\mathcal{F}[/itex] of subsets of a set [itex]\Omega[/itex] is a function from [itex]\mathcal{F}[/itex] to the unit interval [itex][0,1][/itex] such that [itex]P(\Omega)=1[/itex] and

    [tex]
    P\left(\bigcup_{m=1}^{\infty}A_m\right)=\sum_{m=1}^{\infty}P\left(A_m\right)
    [/tex]

    for each pairwise disjoint sequence [itex](A_m:m=1,2,3,\ldots)[/itex] of members of [itex]\mathcal{F}[/itex]. Because [itex]P[/itex] satisfies this summation condition it is said to be countably additive.

    The problem is to show that [itex]P[/itex] is finitely additive, that is:

    [tex]
    P\left(\bigcup_{m=1}^{n}A_m\right)=\sum_{m=1}^{n}P\left(A_m\right)
    [/tex]

    for each pairwise disjoint finite sequence [itex](A_1,\ldots,A_n)[/itex] of members of [itex]\mathcal{F}[/itex].

    Anyone have any hints to toss my way? Thanks.

    Doug
     
  2. jcsd
  3. Sep 3, 2004 #2

    Hurkyl

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    Well, you might look at the LHS, and try to consider sets that act as an identity under union, so you can set an infinite union equal to a finite union.

    Or, you might look at the RHS and consider numbers that act as an identity under addition, so that you can set an infinite sum equal to a finite sum.
     
  4. Sep 3, 2004 #3
    The proof would be trivial if [itex]P(\emptyset)=0[/itex] but that too is a fact that must be proved.

    Doug
     
  5. Sep 3, 2004 #4

    Hurkyl

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    Hrm, do you know any disjoint sequences of sets whose union is the empty set?
     
  6. Sep 3, 2004 #5
    I would say there's only one: [itex](\emptyset,\emptyset,\ldots)[/itex].

    Doug
     
  7. Sep 3, 2004 #6

    Hurkyl

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    So what happens if [itex]P(\emptyset) \neq 0[/itex]?
     
  8. Sep 3, 2004 #7
    Aha. A contradiction. :smile:
     
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