[proofplan] The union $\bigcup A_n$ of possibly overlapping events is rewritten as the union of disjoint increments $B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k$. Countable additivity applies to the disjoint sequence, and the bound follows from the monotonicity $\mathbb{P}(B_n) \le \mathbb{P}(A_n)$. [/proofplan] [step:Disjointify the sequence via $B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k$] Define $B_1 = A_1$ and for $n \ge 2$, \begin{align*} B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k. \end{align*} The sets $B_1, B_2, \ldots$ are pairwise disjoint: for $i < j$, we have $B_i \subset \bigcup_{k=1}^{j-1} A_k$ (since $B_i \subset A_i$ and $i \le j - 1$), while $B_j$ is disjoint from $\bigcup_{k=1}^{j-1} A_k$ by construction. Moreover, $\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty A_n$: the inclusion $\subset$ holds since $B_n \subset A_n$ for each $n$. For the reverse, if $\omega \in \bigcup A_n$, let $m$ be the smallest index with $\omega \in A_m$. Then $\omega \notin \bigcup_{k=1}^{m-1} A_k$, so $\omega \in B_m$. [guided] The events $A_n$ may overlap, so [countable](/page/Countable%20Set) additivity does not apply directly to $\bigcup A_n$. The standard trick is to "disjointify" the sequence: define $B_n$ to consist of those elements that appear in $A_n$ for the first time — that is, they belong to $A_n$ but not to any earlier $A_k$. Formally, $B_1 = A_1$ and $B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k$ for $n \ge 2$. By construction, $B_n$ is disjoint from $\bigcup_{k=1}^{n-1} A_k$. Since $B_i \subset A_i \subset \bigcup_{k=1}^{j-1} A_k$ for $i < j$, the sets $B_i$ and $B_j$ are disjoint. The unions match because each $\omega$ in $\bigcup A_n$ must enter some $A_m$ for the first time (take $m = \min\{n : \omega \in A_n\}$), and then $\omega \in B_m$. [/guided] [/step] [step:Apply countable additivity and bound $\mathbb{P}(B_n) \le \mathbb{P}(A_n)$] Since the $B_n$ are pairwise disjoint with $\bigcup B_n = \bigcup A_n$, countable additivity gives \begin{align*} \mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mathbb{P}(B_n). \end{align*} Since $B_n \subset A_n$ for every $n$, monotonicity of $\mathbb{P}$ gives $\mathbb{P}(B_n) \le \mathbb{P}(A_n)$. Therefore \begin{align*} \mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mathbb{P}(B_n) \le \sum_{n=1}^\infty \mathbb{P}(A_n). \end{align*} [/step]