Proving $\omega\mapsto B(\cdot, \omega)$ is measurable.

Question

Let $(B_t)_{t \geq 0}$ be a Brownian motion on the probability space $(\Omega, \mathcal{A},P)$, and $\pi_t$ the canonical projection at time $t$, and $C(\mathbb{R}^+_0,\mathbb{R}^d)$ is the set of continuous functions from $\mathbb{R}^+_0$ to $\mathbb{R}^d$.

How do we prove that $\omega\mapsto B(\cdot, \omega)$ is $\mathcal{A}/\mathcal{R_C}$-measurable, where $$\mathcal{R_C}=\sigma(\pi_t:t\in[0, \infty[) \cap \{w: w \in C(\mathbb{R}^+_0,\mathbb{R}^d) \land w(0)=\mathbf{0}\}$$ ?

Answer 1

Let $\mathcal{C}:=\{x\in C(\mathbb{R}^+_0,\mathbb{R}^d):x(0)=0\}$ and $\Omega':=\{\omega\in \Omega:t\mapsto B_t(\omega) \in \mathcal{C}\}$. Then the BM is continuous and starts at $0$ for all $\omega\in \Omega'$. Define $\varphi:\Omega'\to\mathcal{C}$ by $$ (\varphi(\omega))(t)=B_t(\omega). $$

Then $\varphi$ is an $\mathcal{A}'/\mathcal{R_C}$ measurable map, where $\mathcal{A}'=\mathcal{A}\mid_{\Omega'}$, because for any cylinder set $I_n=\{x\in \mathcal{C}:(\pi_{t_1}(x),\ldots,\pi_{t_n}(x))\in D\}$ with $D\in \mathcal{B}(\mathbb{R}^{d\times n})$, $$ \varphi^{-1}(I_n)=\{\omega\in \Omega':(B_{t_1}(\omega),\ldots,B_{t_n}(\omega))\in D\}\in \mathcal{A}', $$ and $\mathcal{R_C}$ is generated by the cylinder sets.

Answer 2

It follows from the fact that $B_t$ is a random variable for each $t$ (i.e., $B_t\in\mathcal A$ for each $t$).

Let $f:\Omega\rightarrow C([0,\infty),\mathbb R^d),\ \omega\mapsto B(\cdot,\omega)$. The question is whether $f$ is $\mathcal A/\mathcal R_{\mathcal C}$-measurable. Since $\mathcal R_{\mathcal C}$ is generated by cylinder sets, it suffices to ask whether $$ \{\omega:f_t(\omega)\in H\}\in\mathcal A\quad\text{for all }t\in[0,\infty)\text{ and }H\in\mathcal B(\mathbb R^d). $$ But $f_t(\omega)$ is none other than $B_t(\omega)$, and so $\{\omega:f_t(\omega)\in H\}=B_t^{-1}(H)\in\mathcal A$.