Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]?


Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Measurability Product Space Date
Get equation that describes set of measured values Dec 19, 2016
I Lebesgue measure and Fourier theory May 6, 2016
Stationary point for convex difference measure Nov 14, 2015
Product measure Dec 5, 2008
Product of non measurable sets Mar 28, 2007