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Measurability in Product Space

  1. Sep 2, 2011 #1
    Hello all,

    I have some difficulty in determining the measurability in product space. Suppose the product space is [itex]T \times \Omega [/itex] equipped with [itex] \mathcal{T} \otimes \mathcal{F}[/itex] where [itex] ( T , \mathcal{T} , \mu ), ( \Omega , \mathcal{F} , P)[/itex] are themselves measurable spaces.

    Now, if there exists a set [itex] T_0 [/itex] in [itex] T [/itex] with [itex] \mu(T_{0}^{c}) =0[/itex] and, for each fixed [itex] t \in T_0 [/itex], a property holds almost everywhere in [itex] \Omega [/itex], so this means there exists a [itex] \Omega_{t} [/itex] such that [itex] P(\Omega_{t}^{c}) = 0 [/itex] and that property holds on this set.

    How can we conclude that the property will holds almost everywhere in the product space [itex]T \times \Omega [/itex]? Are they saying the set [itex]T_0 \times\Omega_{t} [/itex] is measurable?

    Or in other words, when does the measurability hold if the second set [itex]\Omega_{t} [/itex] is a function of the first set [itex]T_0 [/itex]?

    Thanks very much.

  2. jcsd
  3. Sep 2, 2011 #2
    The set [itex]T_0\times \Omega_t[/itex] will always be measurable. By definition, since the product of measurable sets is always measurable.

    But this is not what you're asking. You need to show that there is a set A of [itex]T\times \Omega[/itex] such that [itex](\mu\times P)(A^c)=0[/itex] and such that "the property" holds on A.

    But what property are we talking about?? Surely this isn't true for every property...
  4. Sep 2, 2011 #3
    Thanks for the reply.

    First of all, I want to know if the second component [itex] \Omega_t [/itex] depends on the first component [itex] T_0 [/itex], how can we show/prove its measurability?

    Secondly, indeed I read this from a book, and the aurthor simply stated that "as there is [itex] T_0 [/itex] such that for each fixed [itex] t\in T_0 [/itex], the property of [itex] a(t,\omega) = b(t,\omega)"[/itex] holds almost surely, then this also holds almost everywhere on [itex] T \times \Omega [/itex], how can he jump to this conclusion?

    Thanks very much.

  5. Sep 2, 2011 #4
    What book are you reading?

    Well, for fixed t, the set [itex]T_0\times \Omega_t[/itex] is measurable by the definition of the product sigma algebra. There's nothing much else to prove.

    Recall that a=b if for each measurable set A holds that

    [tex]\iint_A{(a-b)dP d\mu}=0[/tex]

    Now, by Fubini's theorem, this amounts to

    [tex]\int_T \int_\Omega I_A (a-b)dP d\mu = \int_{T_0} \int_\Omega (a(t,\omega)-b(t,\omega))dP d\mu[/tex]

    Since for each fixed t, we have that [itex]a(t,\omega)=b(t,\omega)[/itex]. It follows that the integral is 0. Thus a=b almost everywhere.
  6. Sep 2, 2011 #5
    Thanks so much for the explanation!

    It is the book by Doob on stochastic process, it tried to regard a stochastic process as a function of two variables.

    So it is true that the dependence of second component on the first one does not affect the condition of measurability in a product space as long as for each fixed point, the second component is measurable? or in other words, the collection of [itex] ( t , f(t,\omega) ) [/itex] is a measurable set iff, for each fixed [itex] t [/itex], [itex] f(t,\omega)[/itex] is measurable in [itex] \Omega [/itex]?


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