# Levy's construction of Brownian Motion

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.

• The $B_i$ are defined as the uniform limit of piecewise linear functions. See e.g. "Fig. 1.2. The first three steps in the construction of Brownian motion" in the text. Commented Apr 8, 2018 at 8:52

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.
• Thanks! Now I have another question. Exercise 1.2 of the book asks us to show that $B(t)$ is jointly measurable w.r.t. $t$ and $\omega$. This is easy if we just look at the original definition of $B:[0,1]\to \mathbb R$. But how could we show joint measurability for $B:[0,\infty)\to \mathbb R$? Commented Apr 8, 2018 at 19:55
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.