Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?

According to 1 2, the third Kolmogorov axiom is

for disjoint sets $(A_n)_{n \in \mathbb{N}}$

$P(\cup_n A_n) = \sum_n P(A_n)$

Is that really disjoint rather than pairwise disjoint?

If we have events $A, B, C$ s.t.

$A \cap B = \emptyset$

$A \cap C = \emptyset$

$B \cap C \neq \emptyset$

$P(B \cap C) > 0$,

then A, B and C are disjoint but not pairwise disjoint...I think? (*)

I don't think it follows that $P(A \cup B \cup C) = P(A) + P(B) + P(C)$.

I think $P(A \cup B \cup C) = P(A) + P(B) + P(C \setminus B)$ ?

(*) From what I remember in advanced probability class:

$\{A_n\}_n$'s are disjoint if $\cap_n A_n = \emptyset$

$\{A_n\}_n$'s are pairwise disjoint if $A_i \cap A_j = \emptyset$ for distinct indices i,j

From Larsen and Marx (book used in my elementary probability class):

I find this strange. If 'disjoint' and 'pairwise disjoint' are equivalent (ie disjoint does not mean what I said above), why even say that $A_i \cap A_j = \emptyset$ for distinct indices i,j? Why not just say disjoint?

On the other hand, disjointness is used to justify the $P(\cup_n A_n) = \sum_n P(A_n)$ statements later on. Seems kind of inconsistent.

Solution 1:

If the sets $A_i$ are pairwise disjoint then any intersection incorporating at least two different $A_i$ is empty, and conversely: If any intersection incorporating at least two different $A_i$ is empty then the $A_i$ are in particular pairwise disjoint. Therefore it is sufficient to call them "disjoint".

It's another thing with "independent" in probability theory: If the events $A_i$ are pairwise independent then they need not be "mutually independent", but "mutually independent" events are of course also pairwise independent.