Levy's construction of Brownian Motion

Question

In M\"orters and Peres' book Brownian Motion, in constructing Brownian motions, they wrote:

We have thus constructed a continuous process $B: [0, 1] → \mathbb R$ with the same finite dimensional distributions as Brownian motion. Take a sequence $B_0,B_1, . . .$ of independent $C[0, 1]$-valued random variables with the distribution of this process, and define $\{B(t): t \geq 0\}$ by gluing together the parts.

My question is: Why does the sequence $B_0,B_1,\dots$ exist? I only know that we can construct an i.i.d sequence of real valued random variables, but I don't know that can be done with $C[0,1]$-valued random variables.

Answer 1

If $\mu$ is the Borel measure on $C[0,1]$ induced by $B:[0,1] \to \mathbb R$ then the coordinate maps on the infinite product $(C[0,1],\mu) \times (C[0,1],\mu) \cdots $ will serve as the sequence $\{B_n\}$. There is no need to define $B_n$'s on the same space as the one on which $B$ is defined.

Answer 2

Actually, there is another, arguably more concrete way to think about it. It shows that there is Brownian motion on $[0,1]$ with Lebesgue measure.

The Lévy construction can be carried out on $[0,1]$ with Lebesgue measure. By the Lévy construction, you obtain a Brownian motion with time index set $[0,1]$. In other words, there is a measurable map $f: [0,1] \rightarrow \mathbb R^{[0,1]}$ such that $f$, seen as random variable, is a Brownian motion. But then, for any uniformly distributed random variable $U$ on $[0,1]$, $f(U)$ is a Brownian motion with time index set $[0,1]$.

Now, it is a well-known fact that on a given prob. space there is a uniform iff there is an i.i.d. sequence of Bernoullis, and each infinite countable set admits a countable partition of countable subsets. Hence, on $[0,1]$ with Lebesgue measure there is an infinitely countable i.i.d. sequence of uniformly distributed random variables, say $(U_n)_{n\in\mathbb N}$. Hence, $B_n := f(U_n)$, $n\in\mathbb N$, defines the required sequence of i.i.d. Brownian motions on a compact time interval.

Finally, for $n\in\mathbb N$, and $t\in [n,n+1)$, set $$B(t):= \sum_{k=0}^n B_k(\min(t-k,1)) = B_0(1) + \dots + B_{n-1}(1) + B_n(t-n).$$ $B$ is a Brownian motion with time index set $\mathbb R_+$ on $[0,1]$ with Lebesgue measure.