Measurability conditions in the definition of Brownian motion

Question

I am trying to study stochastic integration from Kuo, Introduction to stochastic integration.

I have two small questions about the definition of Brownian motion.

First question:

Kuo presents it as a measurable function defined on the product space $[0,\infty)\times \Omega=T\times \Omega$, where $\Omega$ is a measuable space $(\Omega, \mathbf{F},P)$, with several characteristics. I guess $T$ comes endowed with the Borel sigma algebra and the Lesbegue measure (I call it here $\mathbf{B}$ the Borel sigma algebra).

Kuo observes (even if not in the definition) that:

1. for each $\omega \in \Omega, X(\cdot,\omega)$ is a measurable function (called a sample path)

I was wondering if this is an assumption or not. E.g. to have $X(\cdot,\omega)$ $\mathbf{B}$-measurable, we should have, taken a borel set $B$, that the set $\{t|X(t,\omega)\in B \}\subset T$ should be $\mathbf{B}$-measurable for a fixed $\omega$. But how we can deduce if from the fact that $\{(t,\omega)|X(t,\omega)\in B \}\subset T \times \Omega$ is $\mathbf{B} \times \mathbf{F}$-measurable ? (this we know because $X$ is defined to be measurable in the product space)

Maybe I am missing some basic property of product sigma algebras ?

Second question:

One of the conditions is that:

$P(\omega | X(\cdot,t) \text{is continuous})=1$

I guess here the probability measure is $P$. But how can we be sure that the set $\{ \omega | X(\cdot,t) \text{is continuous} \}$ is $\mathbf{F}$-measurable, to be able to declare its probability? Is it an other assumption or follows from some argument?

I hope the questions make sense, otherwise feel free to point out inconsistencies.

Answer 1

First answer:

We need to show that for any $\omega \in \Omega$, function $Y_\omega(t)= X(t,\omega)$ is $\mathcal B([0,\infty))$ measurable. Note that $Y_\omega = X \circ h_\omega, $ where $h_\omega(t) = (t,\omega)$. If we show that $h$ is measurable, then $Y_\omega$ as a composition of measurable maps will be measurable. Since any set $E \in \mathcal B([0,\infty)) \otimes \mathcal F$ is generated by sets of form $A \times B$, where $A \in \mathcal B([0,\infty))$ and $B \in \mathcal F$, we need to show that $h_\omega^{-1}[A \times B]$ is $\mathcal B([0,\infty))$ measurable. But $$h_\omega^{-1}[A \times B] = \begin{cases} A & \omega \in B \\ \emptyset & \omega \not \in B\end{cases}, $$ and both $A,\emptyset$ are $\mathcal B([0,\infty))$ measurable. It follows that $Y_\omega$ is $\mathcal B([0,\infty))$ measurable for any $\omega \in \Omega$.

Second answer:

It's subtle. In general, $\mathcal F$ need not contain the set $A=\{\omega \in \Omega : t \mapsto X(\omega,t)$ is continuous $\}$. There are two ways of dealing with it.

1. Either instead of saying $\mathbb P(A)=1$, we say that there is an event $B \in \mathcal F$ such that $A \supset B$ and $\mathbb P(B) = 1$ (i.e, the set $A$ (which need not be an event) is larger than some event $B$ which has measure $1$)

2. We just require that $\sigma-$field $\mathcal F$ is complete, i.e if $E \in \mathcal F$ and $\mathbb P(E)=0$, then for any subset $C \subset E$ we have $\mathbb P(C) = 0$, too (i.e, with all event of measure $0$, $\mathcal F$ also contains all of its subsets). It can be shown that all $\sigma-$fields can be completed in such a way (extending a measure, to actually measure those sets).

Answer 2

Kuo's definition of "stochastic process" is unconventional; most sources would use the term "measurable stochastic process". The measurability of the random variables $\omega\mapsto X(t,\omega)$ ($t$ fixed) and sample paths $t\mapsto X(t,\omega)$ ($\omega$ fixed) follows from the assumed joint measurability and Fubini's theorem.

The assertion about "continuous realization" (second paragraph on page 8 of Kuo's text) is wrong. It's possible to have a (measurable) process satisfying the first three properties of a Brownian motion (initial value $0$, stationary independent Gaussian increments) such that no sample path is continuous.

Example. Let $B=(B_t)$ be a standard Brownian motion, with all sample paths continuous, defined on a probability space on which is also defined a random variable $U$ that is uniformly distributed on $(0,1)$ and independent of $B$. This can be arranged by a product measure construction ensuring that $(t,\omega)\mapsto (B_t(\omega),U(\omega))$ is measurable. For convenience, I write $X_t(\omega)$ for $X(t,\omega)$. Define $$ X_t(\omega):=\cases{B_t(\omega),&$t\not=U(\omega)$,\cr B_t(\omega)+1,&$t=U(\omega)$.\cr} $$ Notice that $P[X_t\not=B_t]=P[U=t]=0$ for all $t\ge0$. This process $X$ is jointly measurable in $(t,\omega)$, so it is a "stochastic process" in Kuo's sense, and has the distributional properties of Brownian motion, but the sample path $t\mapsto X_t(\omega)$ is discontinuous at $t=U(\omega)$, for each $\omega$. [Kolmogorov's theorem as stated by Kuo in Theorem 3.3.8, page 31, requires a separable process, which the process $X$ of the example is not.]