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.

  • $\begingroup$ oh alright then. I guess there is an error in the book I'm working on claiming that the family of all cylinder sets was an algebra. Of course if it's only that they generate a $\sigma$-algebra then it's fine. Thanks! $\endgroup$ – user26320 Mar 15 '17 at 22:19
  • $\begingroup$ Yes, I think there should be an error. I give you a reference from the book I recently studied for a Stochastic Processes course. On chapter 4 of "Brownian Motion", R. L. Schilling, L. Partzsch we read: "Since the family of all cylinder sets is a $\cap$-stable generator of the trace $\sigma$-algebra $C^{0} \cap \mathcal{B}^{I}(\mathbb{R}^{d})$...". Where $\mathcal{B}^{I}(\mathbb{R}^{d})=\sigma \{ \pi_{t}^{-1} (B): B \in \mathcal{B}(\mathbb{R}^d), t \in I\}$, $\pi_{t} : (\mathbb{R}^d)^{I} \rightarrow \mathbb{R}^d$, $w \mapsto w(t)$ and $C^{0}$ is the space of vector valued conitnuous functions $\endgroup$ – GaC Mar 16 '17 at 12:17
  • $\begingroup$ Of course, he considers cylinders sets of continuous functions, but the point about closure wrt to unions and intersection stability is the same, as you can read in the citation. $\endgroup$ – GaC Mar 16 '17 at 12:20

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