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Math Help - Measurability question

  1. #1
    Junior Member TheProphet's Avatar
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    Measurability question

    Let  X \, : \, (\Omega, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})) be a random variable. Let
     \mathcal{F} = \{A \, : \, A = X^{-1}(B), \mbox{ some } B \in \mathcal{B}(\mathbb{R})\} .

    Show that  X is measurable as a function from  (\Omega, \mathcal{F}) \mbox{ to } (\mathbb{R},\mathcal{B}(\mathbb{R})) .

    Here  \mathcal{B}(\mathbb{R}) is the Borel  \sigma-algebra on  \mathbb{R} .

    Is it sufficient just to say that for every  B \in \mathcal{B}(\mathbb{R}), \, A = X^{-1}(B) \in \mathcal{F} , so X is measurable from  (\Omega, \mathcal{F}) \mbox{ to } (\mathbb{R},\mathcal{B}(\mathbb{R})) by definition?
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  2. #2
    Super Member girdav's Avatar
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    Re: Measurability question

    It's sufficient. Maybe the most interesting part of this exercise is to show that \mathcal F is a \sigma-algebra.
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Measurability question

    Thank you.

    I guess  \mathcal{F} is a \sigma follows from that X^{-1} commutes with complements, countable intersections and unions.
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  4. #4
    Super Member girdav's Avatar
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    Re: Measurability question

    Quote Originally Posted by TheProphet View Post

    I guess  \mathcal{F} is a \sigma follows from that X^{-1} commutes with complements, countable intersections and unions.
    Yes.
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