A family $\mathcal A=(A_i:i\in I)$ of events is called independent, if for all finite subsets $J$ of $I$,
$$
\mathbb P\left(\bigcap_{i\in J}A_i\right)=\prod_{i\in J}\mathbb P(A_i).
$$
I'm wondering why we need $J$ to be a finite subset? Infinite products are defined after all, and we also have the following:
$$
\mathbb P\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty \mathbb P(A_i),
$$
where $A_1,A_2,\dots$ are disjoint events in $\mathcal F$. So I don't see why the definition requires $J$ to be finite; where would it go wrong otherwise?
Best Answer
In short, the two definitions are equivalent, and the one with finite sets is more convenient.
Claim. A family $\mathcal A=(A_i:i\in I)$ of events is independent, if and only if for all finite or countably infinite subsets $J$ of $I$, $$ \mathbb P\left(\bigcap_{i\in J}A_i\right)=\prod_{i\in J}\mathbb P(A_i). $$
Proof. Clearly this implies independence, since finite sets are 'finite or countably infinite'.
Conversely, for any countably infinite set $J$, let $\{J_n\}_{n=1}^{\infty}$ be an increasing sequence of finite sets such that $J=\bigcup_{n=1}^{\infty}J_n$. Then by continuity of measure, $$ \mathbb P\left(\bigcap_{i\in J}A_i\right)=\lim_{n\to\infty}\mathbb P\left(\bigcap_{i\in J_n}A_i\right)=\lim_{n\to\infty}\prod_{i\in J_n}\mathbb P(A_i)=\prod_{i\in J}\mathbb P(A_i). $$
So one may define independence using either finite or countably infinite sets. In practice, it is more convenient to use the definition with finite sets to verify that a given family of random variables is independent. Thus people stick with the first definition, since using countably infinite sets doesn't gain anything more.