Theorem
Let A$ _1 $,A$ _2 $,...,A$ _n $ be n pairwise disjoint events in event space. Then
Proof
By Kolmogorov axioms, for pairwise disjoint events A$ _1 $ and A$ _2 $ in event space F,
Induction Hypothesis: Let A$ _1 $,A$ _2 $,...,A$ _n $ be pairwise disjoint events in F, i.e.
Then,
Now, I want to show that for pairwise disjoint events A$ _i $ i = 1,2,...,n+1
Let A$ _{n+1} $ ∈ F such that A$ _{n+1} $ is pairwise disjoint from events A$ _i $, i = 1,2,...,n.
Let
Note that B ∩ A$ _{n+1} $ = ø because if x ∈ A$ _{n+1} $, then x ∉ A$ _i $ ∀i = 1,2,...,n ⇒ x ∉ B. On the other hand if x ∈ B, then x ∈ A$ _i $ for exactly one i since A$ _i $'s are pairwise disjoint. A$ _{n+1} $ and A$ _i $ are also disjoint so x ∉ A$ _{n+1} $
We mentioned earlier that the probability of the union of two disjoint events is the sum of the probabilities of those events. Hence,
Thus by induction, we have that for a finite collection of pairwise disjoint events A$ _1 $ through A$ _n $ in F,
$ \blacksquare $
