Question:
Suppose we have an independent family of random variables $$ \{X_{ij} \mid i \in \{1,\dots,m\}, j \in \mathbb{N} \} $$ on the same probability space $(\Omega, \mathcal{A}, \mathbb{P})$.
How can we show that $X_1 := \sum_{j=1}^\infty X_{1j}, \; \dots \;, X_m := \sum_{j=1}^\infty X_{mj}$ are independent?
Thoughts:
I can show that this holds for $X_1^{(k_1)} := \sum_{j=1}^{k_1} X_{1j}, \; \dots \;, X_m^{(k_m)} := \sum_{j=1}^{k_m} X_{mj}$ for arbitrary $k_1, \dots, k_m \in \mathbb{N}$.
But I don't know how to make the transition to the infinite case.
Define $$\mu_{i,j}=\sigma((X_{i,k})_{1 \leq k \leq j}).$$
Define $$\mu_i=\bigcup_{j \geq 1}{\mu_{i,j}}.$$
$\mu_i$ is stable under finite intersection so the monotone class it generates is $$\sigma_i=\sigma((X_{i,j})_{j \geq 1}).$$
Note that all $X_{i,j}$ and X_i are $\sigma_i$-measurable.
We prove that all $\sigma_i$ are independent.
Define $$\mathscr{I}=\{(A_i) \in \prod_i{\sigma_i},\,P(\forall i,\, X_i \in A_i)=\prod_i{P(X_i \in A_i)}\}.$$
Since everything is finite, $\mathscr{I}$ contains the product of $\mu_i$.
We show by induction over $0 \leq i \leq m$ that $$\mathscr{I} \supset \sigma_1 \times \ldots \times \sigma_i \times \mu_{i+1} \times \ldots \times \mu_m.$$
It’s obvious for $i=0$.
Let $1 \leq i \leq m$ be such that the statement holds for $i-1$.
Let $A_1, \ldots, A_{i-1}, A_{i+1}, \ldots A_m$ with $A_x \in \sigma_x$ (if $x < i$) and $A_x \in \mu_x$ (if $x >i$).
The set of all $B \in \sigma_i$ such that $(A_1, \ldots, A_{i-1},B,A_{i+1},\ldots,A_m) \in \mathscr{I}$ is a monotone class containing $\mu_i$ by induction hypothesis. By the above this set must contain $\sigma_i$.
As a consequence, $\mathscr{I}=\prod_i{\sigma_i}$, thus the $\sigma_i$ are independent. Since $X_i$ is $\sigma_i$-measurable, the $X_i$ are independent.