brownian motion continuous property

Question

In Stein and Shakarchi's functional analysis,

Brownian motion $B_t$ is defined in terms of a probability space $(\Omega, P)$ with $P$ its probability measure and $\omega$ denoting a typical point in $\Omega$. We suppose that for each $t$, $0 \le t < \infty$, the function $B_t$ is defined on $\Omega$ and takes the values in $\mathbb{R}^d$. The Brownian motion process $B_t = B_t(\omega)$ satisfy $B_0(\omega) = 0$ almost everywhere and

1. The increments are independent.
2. The increment $B_{t+h} - B_t$ are Gaussian with covariance $hI$ and mean zero for each $0 \le t < \infty$.
3. For almost every $\omega \in \Omega$, the path $t \mapsto B_t(\omega)$ is continuous for $0 \le t < \infty$

Now it will turn out that this process can be realized in a canoncial way in terms of a natural choice of the probability space $\Omega$. This probability space, denoted by $\mathcal{P}$, is the space of continuous paths in $\mathcal{R}^d$ starting at the origin: it consists of the continuous functions $t \mapsto p(t)$ from $[0, \infty)$ to $\mathcal{R}^d$ with $p(0) = 0$.

Since, by assumption 3, for almost every $\omega \in \Omega$ the function $t \mapsto B_t(\omega)$ is such a continuous path, we get an inclusion $i : \Omega \to \mathcal{P}$ and then the probability measure $P$ gives us, as we will see, a corresponding measure $W$ (the "Wiener measure") on $\mathcal{P}$.

The first two are straightforward. The third property however confuses me slightly. It seems like we are $t \mapsto B_t(\omega)$ is continuous for a fixed $\omega \in \Omega$. Does that mean if we have $\omega_1, \omega_2 \in \Omega$, $\begin{cases} B_t(\omega_1) \quad \text{ if } 0 \le t \le N \\ B_t(\omega_2) \quad \text{ if } t > N \end{cases}$ is not necessarily continuous?

EDIT: So if I understood this correctly, $(\Omega, P)$ is not a product space but just an ordinary sample space with measure $P$. Now, we can define $\mathcal{P} := \{ p \in \mathbf{C}([0, \infty)): p(0) = 0\}$. Since by assumption 3, for any fixed $\omega \in \Omega$, $t\mapsto B_t(\omega)$ is continuous, we can get a new mapping $i : \Omega \to \mathcal{P}$, i.e, for any fixed $\omega$, there is a corresponding continuous function $p$. However, what confuses me is, again, the fixed $\omega$. If $\omega$ is fixed, wouldn't $B_t(\omega)$ be a constant function for all $t$?

Answer 1

I think your confusion relates to the notion of fixing $\omega \in \Omega.$ It should be understand that $\Omega$ is really $\Omega \times [0,\infty)$ in this case. Fixing $\omega \in \Omega$ means to fix a sequence of $\omega$ in $\Omega \times [0,\infty)$, which is what is meant by fixing a path.

Answer 2

After reading all the insightful comments, I think I finally understand what the book is trying to say. I think these are the subtle points that I often miss when going through math textbooks. I initially assumed that $\Omega$ is the space of continuous functions. While this may be the most intuitive and canonical way of constructing a Brownian motion, it doesn't necessarily have to be this way.

In fact, later in the book, it says that

In general it is useful to think of this process as either in terms of an abstract realization $B_t$ on $(\Omega, P)$ satisfying conditions B-1, B-2, and B-3, or its concrete realization on $(\mathcal{P}, W)$ with $W$ the Wiener measure, given in terms of $B_t(\omega) = p(t)$, where $\omega$ is identified with $p$.

So the continuity property of a Brownian motion is with regard to a fixed $\omega \in \Omega$. If we set $\Omega = C([0, \infty))$ and $B_t = \omega(t)$ (the book uses $p$ instead of $\omega$ and $\mathcal{P}$ instead of $\Omega$), we get a more intuitive construction process.