I have confusion when I went through the canonical construction of 1-dimensional Brownian Motion. Here we take $\Omega:=C(\mathbb{R}_+,\mathbb{R})$, and we equip $\Omega$ with the smallest sigma algebra $\mathcal{C}$ such that all coordinate mappings are measurable, and $\mathbb{P}$ to be the Wiener measure.
Question 1: by definition of sigma algebra, we should have $\Omega\in\mathcal{C}$ however, from this link: Formally show that the set of continuous functions is not measurable I doubt this is the case (i.e $\Omega\not\in\mathcal{C}$) Comment: I think $\mathcal{C}$ here is the "prouct measure" in the above link, or please correct me if I am wrong.
Question 2: In the construction, we then set for every $t$, $B_{t}(\omega)=\omega(t)$ for $\forall \omega\in\Omega$. However here $\omega\in\mathcal{C}$ is just a random continuous function, which doesn't necessarily to have Brownian property, e.g not differentiable everywhere, recurrent at level $0$, etc. Why can we still make such construction? Or is it because of the Wiener measure we equipped on the space? Can you please still elaborate the reason behind it? Thank you a lot!!!
Yes, $\Omega$ is the set of continuous functions from $[0,\infty)$ to $\mathbb{R}$. The fact that $\Omega$ is not a measurable set of some other $\sigma$-algebra is not at all relevant.
It's true that there are some elements $\omega \in \Omega$ that don't satisfy properties that Brownian motion does almost surely, such as being non-differentiable. However, those elements have probability $0$ under the Weiner measure. It's the same way we can define a uniform random variable on $\Omega = [0,1]$ with Lebesgue measure by $X(\omega) = \omega$, even though some $\omega$s are rational and a uniform random variable is almost surely irrational.