> [!NOTE] Theorem (Boole's Inequality) > Let $(\Omega,\mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. For all $n\in \mathbb{N}_{\geq 2},$ if $A_{1},A_{2},\dots,A_{n}\in \mathcal{F},$ then the [[Probability Measure|probability measure]] $\mathbb{P}$ satisfies $\mathbb{P}\left(\bigcup_{i=1}^nA_i\right)\leqslant\sum_{i=1}^n\mathbb{P}(A_{i})\tag{1}$ **Proof**: For $n=2,$ by [[Probability Measure is Strongly Additive]] $\mathbb{P}(A_{1} \cup A_{2}) = \mathbb{P}(A_{1})+\mathbb{P}(A_{2})-\mathbb{P}(A_{1} \cap A_{2}) \leq \mathbb{P}(A_{1})+\mathbb{P}(A_{2})$since probabilities are non-negative. Suppose $(1)$ is true for all $j\leq n.$ Let $A_{1}A_{2},\dots,A_{n+1}\in \mathcal{F}.$ Then $\begin{aligned} \mathbb{P}\left(\bigcup_{j=1}^{n+1}A_{j}\right)& =\mathbb{P}\left(\left(\bigcup_{j=1}^nA_j\right)\cup A_{n+1}\right) \\ &=\mathbb{P}\left(\bigcup_{j=1}^nA_j\right)+\mathbb{P}\left(A_{n+1}\right)-\mathbb{P}\left(\left(\bigcup_{j=1}^nA_j\right)\cap A_{n+1}\right) \\ &\leqslant\mathbb{P}\left(\bigcup_{j=1}^nA_j\right)+\mathbb{P}(A_{n+1}) \\ &\leqslant\sum_{j=1}^n\mathbb{P}(A_j)+\mathbb{P}(A_{n+1})=\sum_{j=1}^{n+1}\mathbb{P}(A_j).& \end{aligned}$Therefore, the statement is true by [[Induction Principle|induction]]. Proof: Follows from [[Bounds for Probability of Union (Bonferroni Inequalities)]].