Let $\mathbb{R}^{\mathbb{T}}$ denote the set of all functions $x:\mathbb{T}\longrightarrow\mathbb{R}$, where $\mathbb{T}$ is just some index sets (the time in the stochastic process).
Then we can define the cylinder set as $$\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B\}\ \text{for some}\ B\in\mathcal{B}(\mathbb{R}^{n}).$$ We can also define the elementary cylinder as $$\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in B_{1}\times\cdots\times B_{n}\}\ \text{for some}\ B_{1},\cdots, B_{n}\in\mathcal{B}(\mathbb{R}).$$
Then, denote the collection of all cylinder sets to be $\mathcal{C}$ and denote the collection of all elementary cylinders to be $\mathcal{E}$. Then I want to show
$\sigma(\mathcal{C})=\sigma(\mathcal{E})$.
I have some attempt but I could not show they coincided:
Here is my attempt:
Firstly, for any $E_{1}\in\mathcal{E}$, it can be written as $$E_{1}=\{x\in\mathbb{R}^{\mathbb{T}}:(x_{t_{1}},\cdots, x_{t_{n}})\in A_{1}\times\cdots\times A_{n}\},$$ for some $A_{1},\cdots, A_{n}\in\mathcal{B}(\mathbb{R})$.
But recall that $\mathcal{B}(\mathbb{R}^{n})=\sigma\Big(\{B_{1}\times\cdots\times B_{n}:B_{1}\in\mathcal{B}(\mathbb{R}),\cdots,B_{n}\in\mathcal{B}(\mathbb{R})\}\Big)$ is the smallest $\sigma-$algebra containing all of such generating sets, and thus it must be that $A_{1}\times\cdots\times A_{n}\in \mathcal{B}(\mathbb{R}^{n})$.
Therefore, $E_{1}$ is also a cylinder set. That is, $E_{1}\in\mathcal{C}$. Thus, $\mathcal{E}\subset\mathcal{C}$.
Note that $\sigma(\mathcal{C})$ is a $\sigma-$algebra and must be $\lambda-$system. Also, it follows from the here: (Only Proof Checking) Show that the collection of all elementary cylinders is a semi-algebra. that $\mathcal{E}$ is a $\pi-$system.
Combining above, it follows from Dynkin's $\pi-\lambda$ Theorem that $\sigma(\mathcal{E})\subset\sigma(\mathcal{C})$.
However, I don't know how to show the other direction.
One book containing a really short argument saying that
Clearly every cylinder belongs to the $\sigma-$algebra generated by elementary cylinders, and therefore $\sigma-$algebra generated by elementary cylinders and all cylinders coincide.
From here: Show that the collection of cylinders form an algebra. we know that $\mathcal{C}$ is an Algebra, and thus $\sigma(\mathcal{C}))=\mathcal{C}$, so if it is indeed true that every cylinder belongs to $\sigma-$algebra generated by elementary cylinders, then we have $\sigma(\mathcal{C})=\mathcal{C}\subset\sigma(\mathcal{E})$.
However, I don't know how to convince myself that every cylinder belongs to the $\sigma-$algebra generated by elementary cylinders…
In addition, it will be really appreciated if someone could have a check my proof for the first inclusion.
Please help! Thank you so much!
Best Answer
Each given cylinder set in $\mathcal C$ is of the form $f^{-1}(B)$ for some $B\in\mathcal B(\mathbb R^n)$, where $n\in\mathbb N$, $t_1,\ldots,t_n\in\mathbb T$ are a finite collection of indices, and the function $f:\mathbb R^{\mathbb T}\to\mathbb R^n$ is defined as $$f(x)\equiv(x_{t_1},\ldots,x_{t_n})\quad\text{for every $x\in\mathbb R^{\mathbb T}$}.$$ You want to show that $f^{-1}(B)\in\sigma(\mathcal E)$, meaning that $\mathcal C\subseteq\sigma(\mathcal E)$, which, in turn, will imply that $\sigma(\mathcal C)\subseteq\sigma(\mathcal E)$. (No need to use Dynkin’s theorem for this last implication—see below my remark on your proof of the reverse inclusion.)
To this end, define the following collection: $$\mathcal B^*\equiv\{B\subseteq\mathbb R^n\,|\,f^{-1}(B)\in\sigma(\mathcal E)\}.$$ It is not difficult to check that $\mathcal B^*$ is a $\sigma$-algebra on $\mathbb R^n$. Also, if $B=B_1\times\cdots\times B_n$ for some $B_1,\ldots,B_n\in\mathcal B(\mathbb R)$, then $f^{-1}(B)$ is an elementary cylinder by definition, so $f^{-1}(B)\in\mathcal E\subseteq\sigma(\mathcal E)$. Therefore, $$\{B_1\times\cdots\times B_n\,|\,B_1,\ldots,B_n\in\mathcal B(\mathbb R)\}\subseteq\mathcal B^*,$$ from which it follows that $$\mathcal B(\mathbb R^n)=\sigma\big(\{B_1\times\cdots\times B_n\,|\,B_1,\ldots,B_n\in\mathcal B(\mathbb R)\}\big)\subseteq\mathcal B^*,$$ since $\mathcal B^*$ is a $\sigma$-algebra. Therefore, if $B\in\mathcal B(\mathbb R^n)$, then $B\in\mathcal B^*$, so that $f^{-1}(B)\in\sigma(\mathcal E)$, which is precisely what you sought to show.
As for the first part of your proof, using Dynkin’s theorem is overkill. The fact that $\mathcal E\subseteq\mathcal C$ directly implies that $\mathcal E\subseteq\sigma(\mathcal C)$. Since $\sigma(\mathcal C)$ is a $\sigma$-algebra that includes $\mathcal E$, the smallest $\sigma$-algebra including $\mathcal E$ must be contained in $\sigma(\mathcal C)$. That is: $\sigma(\mathcal E)\subseteq\sigma(\mathcal C)$.