TITLE: interpretation of independence events
QUESTION [5 upvotes]: $\{A_i, i \in \mathbb{N} \}$  are defined to be independent, if $P(\cap_{k=1}^{n} A_{i_k}) = \prod_{k=1}^{n} P(A_{i_k}) $ for any finite subset of $\{A_i, i \in \mathbb{N} \}$.

We know $P(\cup_{i=1}^{\infty} A_i)
    = \sum_{i=1}^{\infty} P(A_i) $ iff $\{A_i , i \in \mathbb{N}\}$ are
disjoint, which is independent of
the probability measure and purely
depends on the relation between the
sets. I was wondering if it is
possible to similarly
characterize/interpret $\{A_i , i
    \in \mathbb{N}\}$  being independent
purely from relation between sets,
and make it independent of the
probability measure as much as
possible if completely is impossible?
Is the definition of $\{A_i, i \in
    \mathbb{N} \}$  being independent
equivalent to $P(\cap_{i=1}^{\infty}
    A_{i}) = \prod_{i=1}^{\infty}
    P(A_{i}) $. What is the purpose of
considering any finite subset
instead?
Is generalization of independence
from probability space to general
measure space meaningful? 
The only interpretations of
independence I know are: measure can be exchanged with product/intersection on independent sets,
and intuitively, independent events occur
independently of each other. Are
there other interpretation,
especially in the general measure
space setting?

Thanks and regards!

REPLY [2 votes]: Not sure your point 2 was addressed, so let me state that defining independence as suggested would lead to a trivial notion, quite different from independence as one wants it. 
To wit, any collection of sets $(A_i)_{i\ge1}$, finite or infinite, could be made part of a larger collection $(A_i)_{i\ge0}$ such that the condition stated in 2 holds: simply add $A_0=\emptyset$. One would be led to say that a sequence is independent while one of its subsequences is not. 
So, the problem has nothing to do with infinite sequences: to define the independence of $A$, $B$ and $C$ by the only condition that $P(A\cap B\cap C)=P(A)P(B)P(C)$ (thus forgetting the supplementary conditions that $P(A\cap B)=P(A)P(B)$, $P(B\cap C)=P(B)P(C)$ and $P(A\cap C)=P(A)P(C)$) already leads to a notion too weak to model any kind of independence, since the supplementary conditions I just wrote can fail.