If $P(\cup_{i\in\mathbb{N}}A_{i}) = \sum_{i\in\mathbb{N}} P[A_{i}]$, prove that A_{i} are almost disjoint.

Question

I started with the inclusion exclusion principle. Let

$B_{1} = A_{1}, B_{2}=A_{2} - A_{1}, \ldots , B_{n} = A_{n} - (\cup_{i=1}^{n-1}A_{i})$

So, $A_{n} = B_{n} \cup (\cap_{i=1}^{n-1}A_{i})$.

We know that $P(\cup A_{i}) = P(\cup B_{i}) = \sum P(B_{i}) = \sum P(A_{i}).$

Also, $P(A_{n}) \leq P(B_{n}) + P(\cap_{i=1}^{n-1}A_{i})$ and $P(A_{n}) \geq P(B_{n})$.

But I am not sure what to do after this. Also, the question asks for countable events and I am doing this for finite events. Any help will be appreciated.

Answer 1

Ιf existed $A_j,A_i$ such that $P(A_i\cap A_j)>0$ then

$P(A_i \cup A_j)=P(A_i)+P(A_j)-P(A_j\cap A_i)<P(A_i)+P(A_j)$

So $$P(\bigcup_kA_k) < P(\bigcup_{k\neq i,j}A_k )+P(A_i)+P(A_j) \leq \sum_{k \neq i,j}p(A_k) +P(A_j)+P(A_i)=\sum_kP(A_k)$$

which is a contradiction