# Show $\sigma$-algebra of cylindrical sets is a certain collection

I am trying to prove that $B\in$ the $\sigma$-algebra generated by the cylindrical sets of the form $\{f: [0,1] \to \mathbb{R} | (f(t_1),...,f(t_n))\in A\}$ for some $n \geq 1$ and Borel subset $A\subset \mathbb{R}^n$, iff $\exists t_1,t_2,...\in [0,1]$, and $C\in \mathcal{F}_0$, where $\mathcal{F}_0$ is the $\sigma$-algebra generated by the cylindrical sets of $\mathbb{R}^{\mathbb{N}}$, s/t $B = \{f\in \{f: [0,1]\to \mathbb{R}\} : (f(t_1), f(t_2), ...)\in C \}$.

I feel lost on this one. Any tips?

Edit: I think I can do this with the good sets principle but I'm having trouble proving that $B = \{f\in \{f: [0,1]\to \mathbb{R}\} : (f(t_1), f(t_2), ...)\in C \}$ is a $\sigma$-algebra. I can show that it's closed under complements but not under countable union or intersection.

Let $(B_n)_{n\geqslant 1}$ be such that $$B_n=\{f, (f(t_{n,k}))_{k=1}^\infty\in C_n\},$$ where $C_n$ is a Borel subset of $\mathbb R^{\mathbb N}$ and $t_{n,k}\in [0,1]$.

Let $\tau\colon\mathbb N\to\mathbb N$ be a bijection. Define $$C:=\{(x_j)_{j\in\mathbb N}, x_j\in C_k\mbox{ if }\pi_1(\tau^{-1}(j))=k\},$$ that is $x_j\in C_k$ if $j$ is reached by an element of the row $k$.

(maybe it is simpler we we deal with the union of two elements: in this case, the corresponding $C$ is $(x_j)_{j\in\mathbb N}, x_{2j}\in C_1,x_{2j+1}\in C_2\}$)

Set \begin{align*} A_1 &:= \{f: [0,1] \to \mathbb{R}; (f(0),f(1)) \in \{0\}\} = \{f:[0,1] \to \mathbb{R}; f(0)=f(1)=0\} \\A_2 &:= \{f: [0,1] \to \mathbb{R}; (f(0),f(1)) \in \{1\}\} = \{f:[0,1] \to \mathbb{R}; f(0)=f(1)=1\}. \end{align*}
Obviously, $A:= A_1 \cup A_2$ is an element of the $\sigma$-algebra generated by the cylindrical sets but there does not exist $C \in \mathcal{F}_0$ such that
$$A = \{f : [0,1] \to \mathbb{R}; (f(0),f(1)) \in C\}.$$
Remark: There is the following characterization of the $\sigma$-algebra generated by the cylindrical sets:
A set $A$ is an element of the $\sigma$-algebra generated by the cylindrical sets if and only if there exists $(t_n)_{n \in \mathbb{N}} \subseteq [0,1]$ such that $$f(t_n)=g(t_n) \, \forall n, f \in A \Rightarrow g \in A.$$