I'm reading about the construction of the classical Wiener space in Schilling-Partzsch, Chapter 4.

The gist of the construction is as follows. Consider the space

$$ C_0 = \{ f: [0, \infty) \to \mathbb{R}^d \mid f \text{ is continuous, } f(0) = 0 \} $$

equipped with the Borel $\sigma$-algebra $\mathcal{B}(C_0)$, where the topology on $C_0$ is given by the metric of locally uniform convergence.

We assume that we have constructed a Brownian motion $(B_t)_{t \geq 0}$ on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$. We define the mapping $\psi: \Omega \to C_0$ via $\psi(\omega) = B(-, \omega)$, i.e. mapping an $\omega \in \Omega$ to the corresponding sample path of the Brownian motion. One can show that the mapping $\psi$ is $\mathcal{F} / \mathcal{B}(C_0)$ measurable, and hence the pushforward measure $\mu = \mathbb{P} \circ \psi^{-1}$ defines a probability measure on $(C_0, \mathcal{B}(C_0))$.

This measure $\mu$ is the Wiener measure. It is then straightforward to check that on cylindrical sets in $\text{Cyl}(C_0)$, i.e. sets of the form

$$ \Gamma = \{ f \in C_0 \mid f(t_1) \in C_1, \dots, f(t_n) \in C_n \} \qquad C_i \in \mathcal{B}(\mathbb{R}^d) $$

we have that

$$ \mu(\Gamma) = \mu (\pi_{t_1} \in C_1, \dots, \pi_{t_n} \in C_n) = \mathbb{P}(B(t_1) \in C_1, \dots, B(t_n) \in C_n) $$

where $\pi_t: C_0 \to \mathbb{R}^d$ is the evaluation functional $\pi_t: f \mapsto f(t)$. One can also check that $\text{Cyl}(C_0)$ is closed under finite intersections and generates $\mathcal{B}(C_0)$. From this, we can conclude that $\mu$ is uniquely determined by its values on $\text{Cyl}(C_0)$.

Everything up to this point is clear to me, but my confusion is now as follows. The authors then conclude that the process $(\pi_t)_{t \geq 0}$ is a Brownian motion on $(C_0, \mathcal{B}(C_0), \mu)$. By construction, the process $(\pi_t)_{t \geq 0}$ has $\pi_0 = 0$ and continuous sample paths. But how can I see that, e.g., $\pi_t - \pi_s \sim \mathcal{N}(0, (t-s)I)$? This is intuitively clear to me, but I'm not sure how to argue this more formally.


1 Answer 1


You can argue directly like this. Let $f \in C_0$, let $t> s$ fixed. Define the functional $f\mapsto f_t-f_s$. This functional is continuous. Indeed, suppose $f^n\to f$ in the local uniform topology. Then, let $N\geq \lceil t\rceil$, we get $$|(f^n_t-f^n_s)-(f_t-f_s)|\leq |f_t^n-f_t|+|f_s^n-f_s|\leq 2\sup_{u\leq N}|f_u^n-f_u|\to 0$$ Therefore $f\mapsto f_t-f_s$ is a $\mathscr{B}(C_0)/\mathscr{B}(\mathbb{R}^d)$-measurable function, as it is continuous. Denote with $\phi(f)=f_t-f_s$ such function. We get for $A \in \mathscr{B}(\mathbb{R}^d)$ (I show all passages for clarity) $$\begin{aligned}\mu(\{f:f_t-f_s \in A\})&=\mu(\{f:\phi(f)\in A \})\\ &=P(B^{-1}(\{f:\phi(f)\in A\}))\\ &=P(B^{-1}(\phi^{-1}(A)))\\ &=P((\phi\circ B)^{-1}(A))\\ &=P(\{\omega: (\phi\circ B)(\omega)\in A\})\\ &=P(\{\omega: B(t,\omega)-B(s,\omega)\in A\})\\ &=\int_A\frac{1}{(2\pi(t-s))^{d/2}}e^{-\frac{|x|^2}{2(t-s)}}\lambda^d(dx)\end{aligned}$$

  • $\begingroup$ Okay, this is exactly what I had thought as well. But does this not follow immediately from the fact that $\mu$ is the pushforward of $\mathbb{P}$ along $\psi$? Where do we use the discussion regarding the cylindrical sets, if at all? $\endgroup$ Sep 7, 2023 at 22:10
  • $\begingroup$ @GregorSamsa one of the characterisations of BM is through the finite dimensional distributions. See the section in Schilling's book in Chapter 2. $\endgroup$
    – Snoop
    Sep 7, 2023 at 22:24

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