Proof of Boole's inequality by disjoint pieces

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A_1, A_2, \dots$ be events.
2. If the family is finite, say $A_1, \dots, A_N$, set $A_i = \varnothing$ for $i > N$. This changes neither the union nor the sum, because $P(\varnothing) = 1 - P(\Omega) = 0$ by the complement rule. So we may assume an infinite sequence.
3. Let $C_1 = A_1$ and $C_n = A_n \setminus (A_1 \cup \dots \cup A_{n-1})$ for $n \ge 2$. Each $C_n$ is in $\mathcal{F}$, because $\mathcal{F}$ is a σ-algebra.
4. By construction $C_n \subseteq A_n$ for each $n$.
5. The sets $C_n$ are pairwise disjoint. Let $m < n$. Then $C_m \subseteq A_m \subseteq A_1 \cup \dots \cup A_{n-1}$, and $C_n$ has no point in this union.
6. We claim $\bigcup_n C_n = \bigcup_n A_n$. The inclusion $\subseteq$ follows from step 4.
7. For $\supseteq$, let $\omega \in \bigcup_n A_n$ and let $n$ be the least index with $\omega \in A_n$. Then $\omega \notin A_1 \cup \dots \cup A_{n-1}$, so $\omega \in C_n$.
8. By the countable additivity of the probability measure $P$, $P\left(\bigcup_n A_n\right) = P\left(\bigcup_n C_n\right) = \sum_n P(C_n)$.
9. By step 4 and monotonicity of probability, $P(C_n) \le P(A_n)$ for each $n$.
10. So each partial sum satisfies $\sum_{n=1}^{N} P(C_n) \le \sum_{n=1}^{N} P(A_n) \le \sum_{n=1}^{\infty} P(A_n)$. The last sum lies in $[0, \infty]$, because its terms are nonnegative.
11. Letting $N \to \infty$ in step 10 and using step 8 gives $P\left(\bigcup_n A_n\right) \le \sum_n P(A_n)$.