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.
 A: 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.


*

*Let $N = \bigcup_{n \ne m} (A_n \cap A_m)$.  Show that $P(N)=0$. 

*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 $P(B) = \sum_n P(A_n)$.

*Show that $B \subset \bigcup_n A_n \subset (B \cup N)$.  Conclude that $P(\bigcup_n A_n) = P(B)$.
A: Yes, Inclusion-Exclusion Principle holds for infinite unions.
Since you can show that $P(\bigcup_{i=1}^{n}A_i)=\sum_{i=1}^{n}P(A_i)$, then let's continue on here.
Since $\bigcup_{i=1}^{n}A_i \subseteq \bigcup_{i=1}^{\infty}A_i$, then we have
$$\sum_{i=1}^{n}P(A_i)=P(\bigcup_{i=1}^{n}A_i)\leq P(\bigcup_{i=1}^{\infty}A_i)$$ for all n.
Therefore, we have
$$\sum_{i=1}^{\infty}P(A_i)\leq P(\bigcup_{i=1}^{\infty}A_i)\leq \sum_{i=1}^{\infty}P(A_i)$$ where the second inequality is given by sub-additivity property. Thus, we got the equality.
Alternatively, you can apply continuity of probability, so
$$\lim_{n \rightarrow \infty} P(\bigcup_{i=1}^{n}A_i)=P\Big(\lim_{n \rightarrow \infty}(\bigcup_{i=1}^{n}A_i)\Big)=P(\bigcup_{i=1}^{\infty}A_i)$$
$$\lim_{n \rightarrow \infty}\sum_{i=1}^{n}P(A_i)=\sum_{i=1}^{\infty}P(A_i)$$
