# Probability measure is countably additive over almost disjoint sets.

Let $A_n, n\ge1,$ be events in a probability space $(\Omega,\mathcal{F}, P)$ that satisfy the property $$P(A_n\cap A_m)=\emptyset, m\not=n$$ Prove $P(\bigcup_{n\ge 1} A_n)=\sum_{n\ge 1}P(A_n).$

I can prove the result for finite unions by using the measure-theoretic version of the Inclusion-Exclusion Principal, but this is useless since $\lim P(\bigcup _{i=1} ^n A_i)$ isn't necessarily equal to $P(\bigcup _{i\ge 1}A_i).$ Does the Inclusion-Exclusion Principle hold for infinite unions? I believe there's an easier way to solve than using that principal, but I am not sure how to do it.

I tried constructing a new sequence $B_n=A_n-\bigcup_{i=1}^{n-1}A_i$, but this doesn't allow me to use the assumption. The sequence $B_n=A_n-\bigcap_{i=1} ^nA_i$ doesn't suffice since it's not a disjoint sequence.

The given condition says that the sets $A_i, A_j$ only intersect in some tiny sets that nobody cares about. So throw it all away. Then show that putting any amount of it back doesn't change anything.
1. Let $N = \bigcup_{n \ne m} (A_n \cap A_m)$. Show that $P(N)=0$.
2. Let $B_n = A_n \setminus N$. Check that $P(B_n) = P(A_n)$. Check that the $B_n$ are disjoint. Let $B = \bigcup_n B_n$ and show that $P(B) = \sum_n P(A_n)$.
3. Show that $B \subset \bigcup_n A_n \subset (B \cup N)$. Conclude that $P(\bigcup_n A_n) = P(B)$.