If $(E,\mathcal E)$ is a measurable space and $(X_t)_{t≥0}$ is a $E$-valued process, then $σ((X_{t_1},…,X_{t_n}))=σ(X_{t_1},…,X_{t_n})$ Let $(E,\mathcal E)$ be a measurable space and $(X_t)_{t\ge 0}$ be a $(E,\mathcal E)$-valued stochastic process.

How can we show that $$\mathcal A_I:=\sigma\left(\left(X_{t_1},\ldots,X_{t_n}\right)\right)=\sigma(X_t,t\in I)=:\mathcal B_I$$ for all $I=\left\{t_1,\ldots,t_n\right\}\in\mathcal I:=\left\{I\subseteq[0,\infty):I\text{ is finite}\right\}$?

Let $$\pi_i:E^n\to E\;,\;\;\;x\mapsto x_i$$ for $i\in\left\{1,\ldots,n\right\}$. Let's assume $(E,\mathcal E)=(\mathbb R,\mathcal B(\mathbb R))$. Since $\pi_i$ is continuous, $$X_{t_i}=\pi_i\left(\left(X_{t_1},\ldots,X_{t_n}\right)\right)$$ is $\mathcal A_I$-measurable for all $i\in\left\{1,\ldots,n\right\}$. Thus, $$\mathcal B_I\subseteq\mathcal A_I\;.$$ Let $(e_1,\ldots,e_n)$ be the standard basis of $E^n$ and $$\rho_i:E\to E^n\;,\;\;\;x\mapsto xe_i$$ for $i\in\left\{1,\ldots,n\right\}$. Since $\rho_i$ is continuous, $$\left(X_{t_1},\ldots,X_{t_n}\right)=\sum_{i=1}^n\rho_i\left(X_{t_i}\right)$$ $\left(X_{t_1},\ldots,X_{t_n}\right)$ is $\mathcal B_I$-measurable for all $i\in\left\{1,\ldots,n\right\}$. Thus, $$\mathcal A_I\subseteq\mathcal B_I$$ and we are done.

Can we prove the statement without the assumption $(E,\mathcal E)=(\mathbb R,\mathcal B(\mathbb R))$?

 A: Let's prove the following general statement:

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces and denote by $\mathcal{B}^n$ the product-$\sigma$-algebra on $Y^n$. Then a mapping $f: (X,\mathcal{A}) \to (Y^n,\mathcal{B}^n)$ is measurable, if and only if, the projections $\pi_j(f): (X;\mathcal{A}) \to (Y,\mathcal{B})$ are measurable for all $j=1,\ldots,n$.

Proof:
"$\Rightarrow$": Fix $j \in \{1,\ldots,n\}$. Since the product $\sigma$-algebra $\mathcal{B}^n$ contains sets of the form $$Y \times \ldots \times Y \times B \times Y \times \ldots \times Y$$ for $B \in \mathcal{B}$,  we find that $$\pi_j^{-1}(B) = Y \times \ldots \times Y \times B \times Y \times \ldots \times Y \in \mathcal{B}^n$$ for any $B \in \mathcal{B}$, i.e. $\pi_j: (Y^n,\mathcal{B}^n) \to (Y,\mathcal{B})$ is measurable. This implies that $\pi_j(f)$ is measurable as composition of measurable functions.
"$\Leftarrow$": Recall that $$\mathcal{G}^n := \left\{ B_1 \times \ldots \times B_n; B_j \in \mathcal{B} \right\}$$ is a generator of $\mathcal{B}^n$. Therefore, it suffices to show that $f^{-1}(G) \in \mathcal{A}$ for any $G$ of the form $$G = \times_{j=1}^n B_j.$$ Since $$\times_{j=1}^n B_j = \bigcap_{j=1}^n \pi_j^{-1}(B_j)$$ we get $$f^{-1}(G) = \bigcap_{j=1}^n f^{-1}(\pi_j^{-1}(B_j))  = \bigcap_{j=1}^n (\pi_j(f))^{-1}(B_j) \in \mathcal{A}.$$

Applying the above result to the random vector $(X_{t_1},\ldots,X_{t_n})$ we find that $(X_{t_1},\ldots,X_{t_n}): (\Omega,\mathcal{A}) \to (E^n,\mathcal{E}^n)$ is measurable if, and only if $X_{t_j}: (\Omega,\mathcal{A}) \to (E,\mathcal{E})$ is measurable for all $j=1,\ldots,n$.
