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 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 on 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 all finite n,
$ \blacksquare $
