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Random Variable

  1. Jun 20, 2010 #1
    Hello all,

    I have the following question:

    Assume [tex](\Omega, \mathcal{F},P) = ([0,1],\mathcal{B}([0,1]),\lambda)[/tex], where [tex]\lambda[/tex] is Lebesgue mesure, so is [tex]X(\omega) = \frac{1}{\omega}[/tex] a random variable defined on this probability space?

    If yes, then can I say that [tex]X[/tex] is bounded a.s. because the set for unboundedness is [tex]{0}[/tex] which is of measure 0?


  2. jcsd
  3. Jun 20, 2010 #2

    Because there is no set of measure 0 on the compliment of which the function is bounded.
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