Consider $(C([0,1],\mathbb{R}^d),\mathcal{F},P)$ the Canonical Probability Space for Brownian Motion, where $\mathcal{F}:=\mathcal{B}(C([0,1],\mathbb{R}^d))$, $P$ is the Wiener measure, and $C([0,1],\mathbb{R}^d)$ is equipped with the topology induced by the topology of uniform convergence on compact sets.
Let $\mathcal{C}$ be the collection of finite-dimensional cylinder sets of the form $C^{A_1,...,A_n}_{t_1,...,t_n}=\lbrace \omega\in C([0,1],\mathbb{R}^d) : \omega(t_1)\in A_1,...,\omega(t_n)\in A_n \rbrace,n\geq 1, A_1,...,A_n\in\mathcal{B}(\mathbb{R}^d),t_1,...,t_n\in[0,1]$.
Problem 2.4.2 in Karatzas and Sheve gives $\mathcal{F}=\sigma(\mathcal{C})$. We also have the coordinate mapping process $W_t(\omega):=\omega(t)$ on $C([0,1],\mathbb{R}^d)$ is a Brownian motion.
My question is the following. Defining $\mathcal{F}_t:=\sigma(W_s,0\leq s\leq t)$, is $\mathcal{F}_1=\mathcal{F}$? My logic is as follows:
Given $C^{A_1,...,A_n}_{t_1,...,t_n}\in \mathcal{C}$ defined above, we have $W_{t_i}^{-1}(A_i)=\lbrace\omega\in C([0,1],\mathbb{R}^d) : \omega(t_i)\in A_i\rbrace=C^{A_i}_{t_i}\in\sigma(W_{t_i})$ Then $C^{A_1,...,A_n}_{t_1,...,t_n}=\cap_{i=1}^nC^{A_i}_{t_i}\in \sigma(W_s,0\leq s\leq 1). $
So $\mathcal{C}\subset \sigma(W_s,0\leq s\leq 1)\Rightarrow \sigma(\mathcal{C}) =\mathcal{F}\subset \sigma(\sigma(W_s,0\leq s\leq 1)) = \sigma(W_s,0\leq s\leq 1) = \mathcal{F}_1.$ Clearly $\mathcal{F}_1\subset \mathcal{F}$, so we are done.
This doesn't feel right to me. In particular, I've seen papers taking the augmentation of $\mathcal{F}_1$, which seems redundant if this was the case.