Looking for a quick check on this "proof", since I think the book's one is too hand wavy (or I'm just not mature enough yet):
Theorem: If $P$ is a probability function, then $$ P ( \cup_{i=1}^\infty A_i) \leq \sum_{i = 1}^\infty P(A_i) $$
Proof:
Not sure about how to do this one:
$$ \cup_{i=1}^\infty (A_i / \cup_{j=1}^{i-1} A_j) = \cup_{i=1}^\infty A_i $$
Lets assume that we have proved that.
Let $A_i^* = A_i / \cup_{j=1}^{i-1} A_j.$
Lets prove the statement of the theorem now:
Let $i$ and $k$ be any arbitrary integers such that $i > k$. The opposite case is symmetric. Prove $A_i^* \cap A_k^* = \emptyset$.
$$ A_i^* \cap A_k^* $$
$$ A_i \cap (\cap_{j=1}^{i-1} A_j^C) \cap A_k \cap (\cap_{l=1}^{k-1} A_l^C) $$
Using associative law, idempotent law and the assumption that $i > k$,
$$ A_i \cap A_k \cap (\cap_{j=1}^k A_j^C). $$
Assume that
$$ A_i \cap A_k \cap (\cap_{j=1}^k A_j^C) \neq \emptyset.$$
Let $x \in A_i \cap A_k \cap (\cap_{j=1}^{k-1} A_j^C)$. Then $x \in A_i, x \in A_k, x \in A_1, x \in A_2, ..., x \in A_{i-1}$.
But then $x \in A_k$ and $x \in A_k^C$, so $A_k = \emptyset$. But then $A_i \cap A_k \cap (\cap_{j=1}^k A_j) = \emptyset$, which is a contradiction.
So then the $A^{*}$'s are disjoint.
So $P(\cup_{i=1}^\infty A_i*) = \sum_{i=1}^\infty P(A_i)$. Since $A_i^* \subset A_i$ by construction, $P(A_i^*) \leq P(A_i)$ for any $i$. So the theorem is true.