When Does Measurability Hold in Product Space?

[wayneckm]

Hello all,


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

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

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

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

Thanks very much.


Wayne

 

[micromass]

The set T_0\times \Omega_t 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 T\times \Omega such that (\mu\times P)(A^c)=0 and such that "the property" holds on A.

But what property are we talking about?? Surely this isn't true for every property...

 

[wayneckm]

Thanks for the reply.

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

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

Thanks very much.Wayne

 

[micromass]

What book are you reading?

Thanks for the reply.
First of all, I want to know if the second component \Omega_t depends on the first component T_0, how can we show/prove its measurability?

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

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

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

\iint_A{(a-b)dP d\mu}=0

Now, by Fubini's theorem, this amounts to

\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

Since for each fixed t, we have that a(t,\omega)=b(t,\omega). It follows that the integral is 0. Thus a=b almost everywhere.

 

[wayneckm]

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 ( t , f(t,\omega) ) is a measurable set iff, for each fixed t, f(t,\omega) is measurable in \Omega?

Thanks!Wayne