What is the sigma-algebra $\sigma(X_t|t\ge 0)$?

Question

In these notes (pg25) a sigma-algebra on $C[0,\infty)$ is introduced as follows: $$\sigma(X_t|t\ge 0) \quad\text{where}\quad X_t(x)=x_t$$ From what I have learned about the notation I would guess that this sigma algebra is actually: $$\sigma(\{x|X_t(x)\in A_t\forall t\ge 0\;\text{where}\; A_t\in\mathscr{B}(\Bbb{R}^d)\}) $$ firstly is this correct? Secondly to check that the Brownian motion $B$ is measurable w.r.t. this sigma algebra the sets: $$\{x|X_{t_i}(x)\in A_{t_i}\; i=0,1,..,n \;\text{where}\; A_t\in\mathscr{B}(\Bbb{R}^d)\}$$ why is this allows?

Answer 1

For each constituent set, take only finitely many $X_t$. In fact, since you are taking the generated sigma-algebra, you can take just one $t$. So we want the sigma-algebra generated by the sets $$ K_{t,A} :=\{X_t \in A\} = \{\omega \in \Omega\;:\;X_t(\omega) \in A\} $$ where we have one set for each $t \ge 0$ and Borel set $A$. Here $\Omega$ is the sample space, it is not clear what your $x$ is...