# Construction of the Classical Wiener Measure

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.

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}
• 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? Sep 7, 2023 at 22:10