Editing Boole's inequality (section)

==Proof==

=== 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>

=== Proof without using induction ===
Let events <math>A_1, A_2, A_3, \dots </math>in our [[probability space]] be given. The countable additivity of the measure <math>\mathbb{P}</math> states that if <math>B_1, B_2, B_3, \dots</math> are pairwise disjoint events, then

:<math>\mathbb{P}\left(\bigcup_{i} B_i\right) = \sum_i \mathbb P(B_i).</math>

Set

:<math>B_i := A_i - \bigcup^{i-1}_{j=1} A_j.</math>

Then <math>B_1, B_2, B_3, \dots</math> are pairwise disjoint. We claim that: 

:<math>\bigcup^{\infty}_{i=1} A_i = \bigcup^{\infty}_{i=1} B_i.</math>

One inclusion is clear. Indeed, since <math>B_i \subset A_i</math> for all i, thus <math>\bigcup^{\infty}_{i=1} B_i \subset \bigcup^{\infty}_{i=1} A_i</math>.

For the other inclusion, let <math>x \in \bigcup^{\infty}_{i=1} A_i</math> be given. Write <math>k</math> for the minimum positive [[integer]] such that <math>x \in A_k</math>. Then <math>x \in  A_k - \bigcup^{k-1}_{j=1} A_j = B_k</math>. Thus <math>x \in \bigcup^{\infty}_{i=1} B_i</math>. Therefore <math>\bigcup^{\infty}_{i=1} A_i \subset \bigcup^{\infty}_{i=1} B_i</math>.

Therefore

:<math>\mathbb P\left(\bigcup_iA_i\right) = \mathbb P\left(\bigcup_iB_i\right) = \sum_i \mathbb P (B_i) \leq \sum_i \mathbb P(A_i),</math>

where the last inequality holds because <math>B_i \subset A_i</math> implies that <math>\mathbb P (B_i) \leq \mathbb P(A_i),</math> for all i.