Generate $\sigma$-algebra

Question

Let

$\mathcal{C}:=\{\pi_{t_1,\ldots, t_d}^{-1}(B) : {t_1,\ldots, t_d} \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R}^d)\}$ and

$\mathcal{C}_0:=\{\pi_{t}^{-1}(B) : t\in [0,1]$ and $B \in \mathcal{B}(\mathbb{R})\}$,

where $\pi_t: (C[0,1],\mathcal{B}_{C})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ and $\pi_{t_1,\ldots, t_d}$ the same for $\mathbb{R}^d$. $\mathcal{B}$ denotes always the Borel $\sigma$-field. My question:

Do these two collections generate the same $\sigma$-algebra, ie. $\sigma(\mathcal{C})=\sigma(\mathcal{C}_0)$?

It is clear that $\sigma(\mathcal{C}_0)\subset\sigma(\mathcal{C})$ since $\mathcal{C}_0\subset \mathcal{C}$. Is the opposite direction also true?

Answer 1

Yes. Consider the collection of all Borel sets $B$ in $\mathbb R^{d}$ such that $\pi_{t_1,t_2,...,t_d^{-1}} (B)$ belongs to $\sigma \mathcal (C_0)$. Verify that this class is a sigma algebra and it contains sets of the type $B=B_1 \times B_2 \times ... \times B_d$ where each $B_i$ is Borel in $\mathbb R$. Hence it contains all Borel sets in $\mathbb R^{d}$. It is now clear that $\sigma \mathcal (C) \subset \sigma \mathcal (C_0)$