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Are F-measurable functions 1-to1?

  1. Jul 24, 2012 #1
    Hi everybody. Can anyone help me to clarify these things? The definition of F-measurable function is as this:

    f:Ω→ℝ defined on (Ω,F,P) probability space is F-measurable if f-1(B)={ω∈Ω: f(ω)∈B} ∈ F for all B∈B(ℝ)

    where B(ℝ) is Borel field over ℝ and B is any Borel subset of the Borel field.

    My confusions are:

    1-Is the function f:Ω→ℝ 1-to-1?
    2-Is f-1(B):B(ℝ)→F mapping to mutually exclusive and collectively exhaustive subsets of F?

    Thank you for any contributions.
     
  2. jcsd
  3. Jul 24, 2012 #2

    micromass

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    No, not necessarily. The [itex]f^{-1}(B)[/itex] is just an (unfortunate) notation for the preimage and has nothing to do here with the inverse of f (which does not exist necessarily).

    But [itex]f^{-1}(B)[/itex] is a set, not a map.
     
  4. Jul 24, 2012 #3
    The word mapping could be wrong maybe. I mean any Borel set in real line should go to the an element of the field F (an element is any subset of Ω). What I wonder is if that elements of the field F, that are assigned by Borel sets from real line, should be mutually exclusive and collectively exhaustive?
     
  5. Jul 24, 2012 #4

    micromass

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    Not necessarily. It's not necessarily true that [itex]f^{-1}(B)\cap f^{-1}(B^\prime)=\emptyset[/itex]. It is true if [itex]B\cap B^\prime=\emptyset[/itex] though.

    Likewise, if [itex]\bigcup_{i\in I}B_i=\mathbb{R}[/itex], then [itex]\bigcup_{i\in I} f^{-1}(B_i)=\Omega[/itex].
     
  6. Jul 25, 2012 #5
    I see now, thank you for help. From this point, should I also assume that the field F to constitute a probability measure P:F→[0,1] have elements (subsets of Ω) which are not necessarily mutually disjoint?
     
  7. Jul 25, 2012 #6

    micromass

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    Right. In general, elements of [itex]\mathcal{F}[/itex] are not necessarily disjoint.
     
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