What is the sigma algebra of cylindrical sets? This is a basic question but still, let $C$ be the space of real-valued continuous functions $f$ on $[0,t]$. Then a cylindrical subset of $C$ is defined as a set of the form
$$
S = \{\, f\in C; \,(f(t_1),\dots\,f(t_n))\in B\}
$$
where $B\in \mathcal{B}^n$ and $0<t_1<\dots<t_n<t$. So, take
$$
S_1 = \{\, f\in C;\, f(t_1) \in B_1 \}
$$
$$
S_2 = \{\, g\in C;\, g(t_2) \in B_2 \}
$$
How is the union of these two sets a cylindrical subset of C as defined above? The union of the sets of functions $f$ and $g$ such that $f(t_1)$ is in some interval and $g(t_2)$ is in some other interval, isn't it the set of functions $h$ such that either $h(t_1)$ or $h(t_2)$ belong to the said intervals.
Or is it that the sum of the sets $S_1$ and $S_2$ is defined as the cylindrical set that corresponds to the Borel set $(B_1\times\mathbb{R})\, \cup \,(\mathbb{R}\times B_2)$?
(Apologies for any lack of rigor)

Be careful: the (or maybe "a") family of cylinder sets need not to be a priori closed under unions! You gave the right definition of cylinder set, but note that  the family:
$$ \{ C_{t_1 \dots t_n} (B) : B \in (\mathcal{B})^n\, \ n \in \mathbb{N}, \ t_1 \dots t_n \in [0,T] \} $$
is just a $\pi$-system (i.e, it is closed under finite intersections), not an algebra or even a $\sigma$-algebra. If you want a "closed-under union" property you must consider the $\sigma$-algebra generated by all cylinder sets. Here's an idea: let $\mathcal{F}_{t_1 \dots t_n}$ be the $\sigma$-algebra generated by the above family and define:
$$ \overline{\mathcal{F}} = \bigcup_{n=1}^{+\infty} \bigcup_{t_i \in [0,T], i \le n} \mathcal{F_{t_1 \dots t_n}}. $$
Then $ \overline{\mathcal{F}}$ is actually an algebra of cylinder sets.
