Here $X_k$s are random variables, not necessarily real-valued. The problem I encounter is $\sigma(X_1, \cdots, X_n)$ might include sets that are not in any of $\sigma(X_1), \cdots, \sigma(X_n)$, which makes it hard to prove $\sigma(X_{n + 1})$ and $\sigma(X_1, \cdots, X_n)$ are independent.
Forgive me if this is a trivial question, the motivation is from a classical example of martingale. $X_k$s are i.i.d. real-valued random variables with zero mean, $S_n = \sum_{k=1}^n X_k$, then $S_n$ is a martingale with respect to the generating filtration. I think I need a positive answer for my question to prove $E[S_{n + 1}|\mathcal{F}_n] = S_n$.
Best Answer
Thanks to the comments, here is an answer to my own question.
First, notice that to prove $X_{n + 1}$ and $\sigma(X_1, \cdots, X_n)$ are independent, it suffices to show $\sigma(X_n)$ and $\mathcal{C}$ are independent, where $\mathcal{C}$ is any $\pi$-system such that $\sigma(\mathcal{C}) \supseteq \sigma(X_1, \cdots, X_n)$. The proof of this claim is not difficult but a bit tedious, refer to this question for details.
$\mathcal{C} = \{\cap_{k=1, \cdots, n}\{X_k \in B_k\}\}$ can be the $\pi$-system that works, where $B_k$s are measurable sets from spaces $X_k$s take value from.
$\mathcal{C}$ is a $\pi$-system since for any $C_1, \cdots, C_m \in \mathcal{C}$ with $C_j = \cap_{k=1, \cdots, n}\{X_k \in B_{jk}\}$, we have $$\cap_{j=1, \cdots, m}C_k = \cap_{k=1, \cdots, n}\{X_k \in \cap_{j = 1, \cdots, m}B_{jk}\} \in \mathcal{C}.$$ Further, $\sigma(\mathcal{C}) \supseteq \sigma(X_1, \cdots, X_n)$ since any $F_l \in \sigma(X_l)$ is in $\mathcal{C}$.
Finally, for $F_{n + 1} \in \sigma(X_{n + 1})$ and $C = \cap_{k=1, \cdots, n}\{X_k \in B_k\} \in \mathcal{C}$, $$P(F_{n + 1} \cap C) = P(F_{n + 1} \cap \cap_{k=1, \cdots, n}\{X_k \in B_k\}) = P(F_{n + 1})\prod_{k=1, \cdots, n}P(\{X_k \in B_k\})$$ since $\{X_k \in B_k\} \in \sigma(X_k)$ for all $k$ and we know $X_k$s are independent.
I wonder if I unnecessarily complicated things, is there an easier proof so that it is actually a "trivial" result?