[proofplan] Each property follows from the axioms of a probability measure: $\mathbb{P}(\Omega) = 1$ and countable additivity over disjoint events. Parts (i) and (ii) use the decomposition $\Omega = A \cup A^c$; part (iii) uses the decomposition $B = A \cup (B \setminus A)$; part (iv) uses $A \cup B = A \cup (B \setminus A)$ together with part (iii). [/proofplan] [step:Decompose $\Omega = \Omega \cup \varnothing$ to prove $\mathbb{P}(\varnothing) = 0$] Since $\Omega = \Omega \cup \varnothing$ is a disjoint union and $\varnothing \in \mathcal{F}$, [countable](/page/Countable%20Set) additivity gives $\mathbb{P}(\Omega) = \mathbb{P}(\Omega) + \mathbb{P}(\varnothing)$. Since $\mathbb{P}(\Omega) = 1 < \infty$, subtracting yields $\mathbb{P}(\varnothing) = 0$. [/step] [step:Use $\Omega = A \cup A^c$ to prove the complement rule] The sets $A$ and $A^c$ are disjoint with $A \cup A^c = \Omega$. By countable additivity (applied to the two-term sequence $A, A^c, \varnothing, \varnothing, \ldots$), \begin{align*} 1 = \mathbb{P}(\Omega) = \mathbb{P}(A) + \mathbb{P}(A^c). \end{align*} Rearranging gives $\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$. [/step] [step:Decompose $B = A \cup (B \setminus A)$ to prove monotonicity] Suppose $A \subset B$. Then $B = A \cup (B \setminus A)$ is a disjoint union. By countable additivity, \begin{align*} \mathbb{P}(B) = \mathbb{P}(A) + \mathbb{P}(B \setminus A). \end{align*} Since $\mathbb{P}(B \setminus A) \ge 0$ (probability measures are non-negative by definition), we conclude $\mathbb{P}(B) \ge \mathbb{P}(A)$. [/step] [step:Write $A \cup B$ as a disjoint union to prove inclusion-exclusion] We decompose $A \cup B = A \cup (B \setminus A)$, which is a disjoint union. By countable additivity, \begin{align*} \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B \setminus A). \end{align*} It remains to express $\mathbb{P}(B \setminus A)$ in terms of $\mathbb{P}(A \cap B)$ and $\mathbb{P}(B)$. Since $B = (A \cap B) \cup (B \setminus A)$ is a disjoint union, countable additivity gives \begin{align*} \mathbb{P}(B) = \mathbb{P}(A \cap B) + \mathbb{P}(B \setminus A), \end{align*} so $\mathbb{P}(B \setminus A) = \mathbb{P}(B) - \mathbb{P}(A \cap B)$. Substituting, \begin{align*} \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B). \end{align*} [guided] The key idea is to break $A \cup B$ into pieces that are disjoint, so that countable additivity applies. The natural decomposition is $A \cup B = A \cup (B \setminus A)$, where $A$ and $B \setminus A$ are disjoint by construction (any element of $B \setminus A$ does not belong to $A$). Applying countable additivity to this disjoint union: \begin{align*} \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B \setminus A). \end{align*} We now need to relate $\mathbb{P}(B \setminus A)$ to the quantities appearing in the desired formula. The set $B$ itself splits as $B = (A \cap B) \cup (B \setminus A)$, and these two pieces are disjoint (an element of $B$ is either in $A$ or not). Countable additivity gives \begin{align*} \mathbb{P}(B) = \mathbb{P}(A \cap B) + \mathbb{P}(B \setminus A), \end{align*} which rearranges to $\mathbb{P}(B \setminus A) = \mathbb{P}(B) - \mathbb{P}(A \cap B)$. Substituting into the first equation: \begin{align*} \mathbb{P}(A \cup B) &= \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B). \end{align*} [/guided] [/step]