# Measurability in Product Space

1. Sep 2, 2011

### 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

2. Sep 2, 2011

### micromass

Staff Emeritus
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...

3. Sep 2, 2011

### wayneckm

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

4. Sep 2, 2011

### micromass

Staff Emeritus

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.

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.

5. Sep 2, 2011

### 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