> [!NOTE] Theorem > Let $(\Omega, \mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. Then the [[Probability Measure|probability measure]] $\mathbb{P}$ is strongly additive: that is for all $A,B\in\mathcal{F},$ $\mathbb{P}(A\cup B) = \mathbb{P}(A)+ \mathbb{P}(B)-\mathbb{P}(A\cap B)$ **Proof**: By [[Set Difference and Intersection From Partition]], $A$ is the union of the disjoint sets $A-B$ and $A \cap B$ and similarly, $B$ is the union of the disjoint sets $B-A$ and $A\cap B.$ Thus since $\mathbb{P}$ is finitely additive, $\begin{align} \mathbb{P}(A) & =\mathbb{P}(A-B) + \mathbb{P}(A \cap B) \\ \mathbb{P}(B) & = \mathbb{P}(B-A) + \mathbb{P}(A \cap B ) \end{align}$Also by [[Set Difference is Disjoint with Reverse]], $(A-B)\cap (B-A)=\emptyset$Hence $\begin{align} \mathbb{P}(A)+\mathbb{P}(B)&=\mathbb{P}(A-B)+2 \mathbb{P}(A \cap B) + \mathbb{P}(B-A) \\ & = \mathbb{P}((A-B)\cup (B-A)\cup (A \cap B)) + \mathbb{P}(A \cap B) \\ &= \mathbb{P}(A \cup B) + \mathbb{P}(A\cap B). \end{align}$