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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!!!

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  1. 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.

  2. 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.

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  • $\begingroup$ So it this $(\Omega,\mathcal{C},\mathbb{P})$ is NOT a probability space, right? I had the above question was because I thoughtwe always want a measurable space. Regarding the second question: hence Wiener measure only "select" those path with Brownian property, and it kind of makes sense, since it's not defined on the whole continuous function space anyway. Am I understanding it correctly? Thank you for your quick and helpful answer! $\endgroup$ Commented Mar 30 at 2:26
  • $\begingroup$ How can we show those paths that don't satisfy Brownain property has Wiener measure $0$? And a follow-up question, when we say we equip a space with a sigma algebra, we don't necessarily mean the sigma algebra on the aforementioned space? $\endgroup$ Commented Mar 30 at 2:40
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    $\begingroup$ @RandomHarry No, $(\Omega, \mathcal C, \mathbb{P})$ is a probability space. The fact that $\Omega$ is not a measurable subset of some other $\sigma$-algebra has nothing to do with anything. Wiener measure is defined on $\mathcal C$, and is defined to be the law of Brownian motion. So if a property holds a.s. for Brownian motion, the paths which don't have that property have Wiener measure $0$ by definition of Wiener measure. $\endgroup$ Commented Mar 30 at 2:49
  • $\begingroup$ Yes, then my question would be why $\Omega\in \mathcal{C}$? I thought it is not the case, since $\mathcal{C}$ is the same as the product sigma algebra defined on ${\mathbb{R}}^{\mathbb{R}_+}$, and $\Omega$ is not in it (argument is provided in the link in my post), or am I wrong? $\endgroup$ Commented Mar 30 at 2:53
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    $\begingroup$ @RandomHarry In your post, $\mathcal C$ is the smallest $\sigma$-algebra on $\Omega$ such that the coordinate mappings are all measurable, so $\Omega \in \mathcal C$ by the definition of $\sigma$-algebra. There is no reason to expect it to be the same as $\mathcal{B}(\mathbb{R}^{\mathbb{R}})$. $\endgroup$ Commented Mar 30 at 2:57

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