For $0\leq t_1<t_1<...<t_n \leq 1$, $n\in \mathbb{N}$ we define $$\pi_{t_1,...t_n}:(C[0,1],\mathbb{B}(C[0,1]))\to (\mathbb{R}^n,\mathbb{B}(\mathbb{R}^n)), \quad x\to (x(t_1),...,x(t_n)).$$ $\mathbb{B}(C[0, 1]) := \sigma(G \subseteq C[0, 1] : G \text{ is open})$
How can I show that $\pi_{t_1,...t_n}$ is measurable?
My thinking:
I know that it is enough to show measurability on the generating set $\mathbb{R^n}$. I can write $$\pi_{t_1,...t_n}^{-1}(\mathbb{R}^n)\subseteq \mathbb{B}(C[0,1]).$$ This is equivalent as $$\{x\in C[0,1]|\pi_{t_1,...t_n}(x)\in\mathbb{R}^n\}=\{x\in C[0,1]|(x(t_1),...,x(t_n))\in \mathbb{R}^n\}.$$
I would say that this set is, by definition of $x$, a subset of $C[0,1]$, therefore also a subset $\mathbb{B}(C[0,1]).$
Is this the right way of thinking and more importantly, am I even understanding it correctly? Am I missing something?
Consider the space $C[0,1]$ with the sup norm: $d(f,g):=\sup_{x\in[0,1]}|f(x)-g(x)|$ and the topology on $C[0,1]$ induced by that norm - $\mathcal{O}(C[0,1])$. The $\sigma$-algebra under question is $\mathcal{B}\equiv\sigma(\mathcal{O}(C[0,1]))$. Then every continuous function on $[0,1]$ is measurable w.r.t. $\mathcal{B}$ ($\because$ such functions belong to $\mathcal{O}(C[0,1])$). Therefore, it suffices to show that $\pi_{t_j}$, $j=1,\ldots,n$ is continuous.