Denote $\mathbb{R}^{\mathbb{T}}$ to be the set of functions $x:\mathbb{T}\longrightarrow\mathbb{R}$, where $\mathbb{T}$ is an indexing set (the "time" in the stochastic process).
Now, define the cylinder sets as $$C(t_{1},\cdots, t_{n}, B):=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B\}\ \text{for some}\ B\in\mathcal{B}(\mathbb{R}^{n}).$$
Then the cylindrical $\sigma$-algebra is defined as $$\mathcal{B}(\mathbb{R}^{\mathbb{T}})=\sigma(\text{cylinders}).$$
Now, I want to show that:
Define $\mathcal{F}_{T}:=\sigma\Big(\{C(t_{1},\cdots, t_{n}, B):t_{1}\cdots, t_{n}\in T\}\Big)$ for $T\subset\mathbb{T}$. Prove that $$\mathcal{B}(\mathbb{R}^{\mathbb{T}})=\bigcup_{\text{countable}\ T\subset\mathbb{T}}\mathcal{F}_{T},$$ where the union is taking over all countable subset $T$ of $\mathbb{T}$.
I had some attempt as follows:
Denote $\mathcal{C}$ to be the collection of all cylinder sets. Let $A$ be a cylinder set, then it can be written as $A=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B\}$ for some $B\in\mathcal{B}(\mathbb{R}^{n})$. Then $t_{1},\cdots, t_{n}$ must belong to some index subset $T_{1}$ of $\mathbb{T}$, and thus $$A\subset \{C(t_{1},\cdots, t_{n}, B):t_{1},\cdots, t_{n}\in T_{1}\}.$$
Thus, if we define $\mathcal{C}_{T_{1}}$ to be the collection of all the set of the form as the RHS of the above inclusion, we then have $$\mathcal{C}\subset\mathcal{C}_{T_{1}}\subset\mathcal{F}_{T_{1}}.$$
But $\mathcal{B}(\mathbb{R}^{n}):=\sigma(\mathcal{C})$ is the smallest $\sigma-$algebra containing $\mathcal{C}$, and thus $$\mathcal{B}(\mathbb{R}^{n})\subset\mathcal{F}_{T_{1}}\subset\bigcup_{\text{countable T}\subset\mathbb{T}}\mathcal{F}_{T}.$$
However, I have no idea about how to show the inverse inclusion.
Also, is my proof for $\subset$ correct? I am really bad at measure theory, so I don't really have any confidence at all…
This question is related to:
What is the sigma algebra of cylindrical sets?
Cylindrical sigma algebra answers countable questions only.,
but there were not any complete proof there, and the notions of cylindrical $\sigma-$algebra were not really the same.
Thank you so much!
Edit 1:
As comments pointed out, I should prove $\bigcup_{T}\mathcal{F}_{T}$ is a $\sigma-$algebra. Inspired by saz, I generated a proof about $\sigma-$algebra, and seems proved $\subset$. However, I still don't know how to show $\supset$.
With a little bit notation abuse, denote the RHS of the desired equality to be $\mathcal{F}$.
Lemma:$\mathcal{F}$ is a $\sigma-$algebra.
Proof of lemma:
Indeed, since $\mathcal{F}_{T}$ is defined to be a $\sigma-$algebra for any countable subset $T\subset\mathbb{T}$, it must contain $\varnothing$, and thus $\varnothing\in\mathcal{F}$.
Secondly, if $E\in \mathcal{F}$, then $E\in\mathcal{F}_{T}$ for some $\mathcal{F}_{T}$, but it is a $\sigma-$algebra, so $E^{c}$ is in that set and thus $E^{c}\in\mathcal{F}$.
Finally, let $\{E_{j}\}_{j=1}^{\infty}$ be a countable collection of sets that are in $\mathcal{F}$, then $E_{j}\in\mathcal{F}_{T_{j}}$ for some countable $T_{j}\subset\mathbb{T}$. Consider the set defined by $T^{*}:=\bigcup_{j=1}^{\infty}T_{j},$ it is again a countable subset of $\mathbb{T}$ because it is a countable union of countable sets. Also, by construction, we must have for each $j$, $$\{C(t_{1},\cdots, t_{n}, B):t_{1}\cdots, t_{n}\in T_{j}\}\subset\{C(t_{1},\cdots, t_{n}, B): t_{1},\cdots, t_{n}\in T^{*}\}\subset\mathcal{F}_{T^{*}},$$ but $\mathcal{F}_{T_{j}}$ is the smallest $\sigma-$algebra containing the LHS, and thus $\mathcal{F}_{T_{j}}\subset\mathcal{F}_{T^{*}}$ for each $j$.
Therefore, $E_{j}\in\mathcal{F}_{T*}$ for each $j$. Hence, $\bigcup_{j=1}^{\infty}E_{j}\subset\mathcal{F}_{T^{*}}\subset\mathcal{F}.$
Proof of $\subset$:
Now, denote $\mathcal{C}$ to be the collection of all cylinder sets and let $A\in\mathcal{C}$. Then A can be written as $$A=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B\}\ \text{for some}\ B\in\mathcal{B}(\mathbb{R}^{n}).$$ But $t_{1},\cdots, t_{n}$ must belong to some countable index subset $T$ of $\mathbb{T}$ (the most convenient way is to define $T:=\{t_{1},\cdots, t_{n}\}$). Therefore, $A\in\mathcal{F}$, so $\mathcal{C}\subset\mathcal{F}$.
However, we have showed that $\mathcal{F}$ is a $\sigma-$algebra, and we know that $\mathcal{B}(\mathbb{R}^{\mathbb{T}})=\sigma(\mathcal{C})$ is the smallest $\sigma-$algebra containing $\mathcal{C}$ and thus $\mathcal{B}(\mathbb{R}^{\mathbb{T}})\subset\mathcal{F}$.
I'd like to express my great appreciate to saz who really spent lots of time on my dumb questions. Thank you so much saz :)!
Edit 2: Proof of saz's remark:
As I am really bad at measure theory, I am going to prove saz's remark as an exercise:
I claim that $\mathcal{F}_{S}\subset\mathcal{F}_{T}$ for any two (not necessarily countable) $S,T\subset\mathbb{T}$ with $S\subset T$.
Indeed, every set in the collection $\mathcal{A}_{1}:=\{C(t_{1},\cdots, t_{n}, B), t_{1}\cdots, t_{n}\in S, B\in\mathcal{B}(\mathbb{R}^{n})\}$ must belong the collection $\mathcal{A}_{2}:=\{C(t_{1},\cdots, t_{n}, B), t_{1}\cdots, t_{n}\in T, B\in\mathcal{B}(\mathbb{R}^{n})\}$, since $S\subset T$. This implies that $$\mathcal{A}_{1}\subset\mathcal{A}_{2}\subset\sigma(\mathcal{A}_{2})=\mathcal{F}_{T},$$ but $\mathcal{F}_{S}=\sigma(\mathcal{A}_{1})$ is the smallest the $\sigma-$algebra containing $\mathcal{A}_{1}$, and thus $\mathcal{F}_{S}\subset\mathcal{F}_{T}$.
It definitely follows immediately from saz's comments about the comparison of generating set.
I am really grateful of the help from saz, thank you so so so much!
Best Answer
Your proof of "$\subseteq$" looks okay now. Isn't the converse actually trivial? By definition,
$$\mathcal{F}_T = \sigma(C(t_1,\ldots,t_n,B); t_1,\ldots,t_n \in T, B \in \mathcal{B}(\mathbb{R}^n)),$$
and so
$$\mathcal{F}_T \subseteq \sigma(C(t_1,\ldots,t_n,B); t_1,\ldots,t_n \in \mathbb{T}, B \in \mathcal{B}(\mathbb{R}^n)) \stackrel{\text{def}}{=} \mathcal{B}(\mathbb{R}^T).$$
Hence, $\mathcal{F}_T \subseteq \mathcal{B}(\mathbb{R}^{\mathbb{T}})$ for each $T \subseteq \mathbb{T}$, and this gives $\bigcup_{T \, \text{ctble}} \mathcal{F}_T \subseteq \mathcal{B}(\mathbb{R}^{\mathbb{T}})$.
Remark: What we used here (and what you also used in the proof of closedness of $\mathcal{F}$ under countable unions) is that $\mathcal{F}_S \subseteq \mathcal{F}_T$ for any two $S,T \subseteq \mathbb{T}$ with $S \subseteq T$ (no matter whether the sets $S,T$ are countable or not).