In chapter 7 of Durrett's Probability theory book official pdf link here, he presents a construct of Brownian motion. One step of his argument is unclear to me.
On page 356, he states
We tidy things up by moving our probability measure to $(C, \mathcal C)$, where $C = \{ \text{continuous}~\omega : [0, \infty) \to \mathbb R\}$ and $\mathcal C$ is the $\sigma$-field generated by the coordinate maps $t \to \omega(t)$. To do this, we observe that the map $\psi$ that takes a uniformly continuous point in $\Omega_q$ to its unique continuous extension in $C$ is measurable,...
Note that $\Omega_q$ are functions from the dyadic rationals to the reals.
- I'm not sure why this map $\psi$ exists (I think it is a real-analysis theorem? Something about the dyadic rationals being dense in the reals we can extend it to continuous functions?)
- Why is this function measurable (from $(\Omega_q, \mathcal F_q)$ to $(C, \mathcal C))$? It is not clear to me how any set of continuous functions in the sigma-algebra $E \subseteq \mathcal C$, means that $\psi^{-1}(E) \in \mathcal F_q$, where $\mathcal F_q$ is the sigma-algebra generated by $\Omega_q$.
For 1, note that a uniformly continous function is Cauchy-continuous and thus can be uniquely extended to the Cauchy completion of its domain; in this case, $\omega$ can be extended to $\mathbb{R}^+=[0,\infty)$. For more details see Wikipedia. For 2, you can check as a routine exercise that $$ \psi^{-1}(\{B_t<a\})=\psi^{-1}(\{\bar{\omega}\in C(\mathbb{R}^+)|\bar{\omega}(t)<a\}) \\=\bigcup_{k\in\mathbb{N}} \bigcup_{n\in\mathbb{N}} \bigcap_{m \geq n} \{\omega \in C(\mathbb{Q}_2^+)|\omega(q_m)<a-\frac{1}{k}\} $$ where $(q_n)$ is any sequence of dyadic rationals converging to $t$. Since sets of the form $\{B_t<a\}$ generate $\mathcal{F}_q$, we have measurability.