Let $A_{1,2,\cdots,n}\subset \Omega$ be independent events, and suppose $\{i_1,i_2,\cdots,i_k\},\{j_1,j_2,\cdots, j_s\}$ are disjoint subsets of $\{1,2,\cdots ,n\}$, is there a straightforward and computationally easy way to prove that $\sigma(A_{i_1},\cdots,A _{i_k})$ and $\sigma(A_{j_1},\cdots,A _{j_s})$ are independent fields ($\sigma$-algebras in other words; or in our case even simpler, algebras actually)? Here of course $\sigma(\cdot)$ denotes the $\sigma$-algebra generated by the bracketed elements.
The motivation is to remove computational pain in a lot of cases. Example: given $A,B,C,D$ independent, with the result above I can directly claim that things like $A\cup B, C\cap D^c$ are independent without having to take pains to verify them.
There is one proof that requires knowledge of independent r.v.: it is based on the theorem which states that if $X_{1,2,\cdots n}$ are independent r.v., and $\{i_1,i_2,\cdots,i_k\},\{j_1,j_2,\cdots, j_s\}$ are disjoint subsets of $\{1,2,\cdots ,n\}$, then for continuous functions $f,g$ we have $f(X_{i_1},\cdots,X _{i_k})$ and $g(A_{j_1},\cdots,A _{j_s})$ are independent r.v. The result follows if we let $X_i:=1_{A_i}$, since independence between events is equivalent to independence between corresponding characteristic functions. However, I believe there are other ways that don't involve r.v. at all. Could you enlighten me? Best regards!
The independence of your $\sigma$-fields follows from
Theorem. Let $(A_{i,j})$ be an array of independent events. Let $\mathcal{F}_i$ be the $\sigma$-field generate by the events in the $i^{th}$ row of this array. Then the $\mathcal{F}_i$ are independent.
The proof can be found in Billingsley (Corollary 2 of Theorem 4.2). It uses just the $\pi$-$\lambda$ theorem and elementary facts about independent events, not random variables.