Independence of infinitely many random variables I was reading this wiki article about probabilistic independence, it says

An infinite family of σ-algebras is said to be independent if all its finite subfamilies are independent.

So when we say a set of countably infinitely many random variables is mutually independent, what we really mean is any of its finite subset is mutually independent. My question is, what's preventing us from dropping the modifier "finite" here? Does it make sense if one would want to extend the definition to include infinite subsets/subfamilies? I was told it definitely doesn't, if you assume the axiom of choice. How come?
 A: For a countable family of $\sigma$-algebras, the definitions are equivalent.  More precisely, let $(\Omega, \mathcal F, \mathbb{P})$ be a probability space and $(\tau_i)_{i \in \mathbb{N}}$ be a family of sub $\sigma$-algebras of $\mathcal F$.  Then every finite sub-family is independent iff we have $\mathbb{P}\left(\bigcap_{i=1}^\infty A_i\right) = \prod_{i=1}^\infty \mathbb{P}(A_i)$ for every collection of sets $(A_i)_{i \in \mathbb{N}}$ with $A_i \in \tau_i$ for all $i$.  This property clearly implies independence of any finite sub-family (take $A_i = \Omega$ if $\tau_i$ is not in the sub-family), and the reverse implication follows directly from the monotone convergence theorem for sets.
In the case that we have an uncountable family of $\sigma$-algebras $(\tau_i)_{i \in \mathcal I}$, it is not obvious how to define $\prod_{i \in \mathcal I} \mathbb{P}(A_i)$ when $\mathcal I$ is uncountable.  However, since $\mathbb{P}(A) \in [0,1]$ for all $A$, a reasonable definition might be \begin{align*}
\prod_{i \in \mathcal I} \mathbb{P}(A_i) &:= \inf_{\substack{F \subset \mathcal I \\ F \text{ finite}}} \prod_{i \in F} \mathbb{P}(A_i).
\end{align*}
In this case, the property $\mathbb{P}\left(\bigcap_{i\in \mathcal I} A_i\right) = \prod_{i\in \mathcal I} \mathbb{P}(A_i)$ for every collection of sets $A_i \in \tau_i$ is not equivalent to every finite sub-family of $(\tau_i)$ being independent.  For starters, since $\mathcal I$ is uncountable, there is no guarantee that $\bigcap_{i\in \mathcal I} A_i$ is even measurable.  Even if we tried to change the property to "$\mathbb{P}\left(\bigcap_{i\in \mathcal I} A_i\right) = \prod_{i\in \mathcal I} \mathbb{P}(A_i)$ for every collection of sets $A_i \in \tau_i$ such that $\bigcap_{i \in \mathcal I} A_i$ is measurable," it still would not be equivalent.
For example, take $\Omega = [0,1]$ with $\mathbb{P}$ being Lebesgue measure.  Let $\mathcal I := [0,1]$ and $\tau_x := \{\emptyset,\Omega,\{x\},(\Omega \setminus \{x\})\}$ for all $x \in \mathcal I$.  Since $\mathbb{P}(A_x) \in \{0,1\}$ for every $A_x \in \tau_x$, any finite sub-family of $(\tau_x)_{x \in \mathcal I}$ is independent.  However, if we take $A_x := \Omega \setminus \{x\}$, then $\prod_{x \in \mathcal I} \mathbb{P}(A_i) = 1$ but $\bigcap_{x \in \mathcal I} A_x = \emptyset$ so $\mathbb{P}\left(\bigcap_{x\in \mathcal I} A_x\right) = 0 \ne \prod_{x \in \mathcal I} \mathbb{P}(A_i)$.
