Editing Boole's inequality (section)

=== Proof using induction ===
Boole's inequality may be proved for finite collections of <math>n</math> events using the method of [[Mathematical induction|induction]].{{citation needed|date=February 2025}}

For the <math>n=1</math> case, it follows that

:<math>\mathbb P(A_1) \le \mathbb P(A_1).</math>

For the case <math>n</math>, we have

:<math>{\mathbb P}\left(\bigcup_{i=1}^{n} A_i \right) \le \sum_{i=1}^{n} {\mathbb P}(A_i).</math>

Since <math>\mathbb P(A \cup B) = \mathbb P(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B),</math> and because the union operation is [[associative]], we have

:<math>\mathbb{P}\left(\bigcup_{i=1}^{n+1}A_i\right) = \mathbb{P}\left(\bigcup_{i=1}^n A_i\right) + \mathbb{P}(A_{n+1}) -\mathbb{P}\left(\bigcup_{i=1}^n A_i \cap A_{n+1}\right).</math>

Since

:<math>{\mathbb P}\left(\bigcup_{i=1}^n A_i \cap A_{n+1}\right) \ge 0,</math>

by the [[Probability Axioms#First axiom|first axiom of probability]], we have

:<math>\mathbb{P}\left(\bigcup_{i=1}^{n+1} A_i \right) \le \mathbb{P} \left(\bigcup_{i=1}^n A_i\right) + \mathbb{P}(A_{n+1}),</math>

and therefore

:<math>\mathbb{P}\left(\bigcup_{i=1}^{n+1} A_i \right) \le \sum_{i=1}^{n} \mathbb{P}(A_i) + \mathbb{P}(A_{n+1}) = \sum_{i=1}^{n+1} \mathbb{P}(A_i).</math>