[proofplan] We show that conditional probability $\mathbb{P}(\cdot \mid B)$ is itself a probability measure on $(\Omega, \mathcal{F})$. Countable additivity of $\mathbb{P}(\cdot \mid B)$ then follows directly from countable additivity of $\mathbb{P}$. [/proofplan] [step:Express the conditional probability using the definition and distribute the intersection] Let $(A_n)_{n \ge 1}$ be pairwise disjoint events. By definition of conditional probability, \begin{align*} \mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n \;\Big|\; B\right) = \frac{\mathbb{P}\!\left(\left(\bigcup_{n=1}^\infty A_n\right) \cap B\right)}{\mathbb{P}(B)}. \end{align*} Distributing the intersection over the union, \begin{align*} \left(\bigcup_{n=1}^\infty A_n\right) \cap B = \bigcup_{n=1}^\infty (A_n \cap B). \end{align*} [/step] [step:Verify disjointness of $A_n \cap B$ and apply countable additivity of $\mathbb{P}$] Since the $A_n$ are pairwise disjoint, the sets $A_n \cap B$ are also pairwise disjoint: if $i \ne j$, then $(A_i \cap B) \cap (A_j \cap B) = (A_i \cap A_j) \cap B = \varnothing \cap B = \varnothing$. By [countable](/page/Countable%20Set) additivity of $\mathbb{P}$, \begin{align*} \mathbb{P}\!\left(\bigcup_{n=1}^\infty (A_n \cap B)\right) = \sum_{n=1}^\infty \mathbb{P}(A_n \cap B). \end{align*} [/step] [step:Divide by $\mathbb{P}(B)$ to obtain the result] Substituting back and dividing by $\mathbb{P}(B) > 0$: \begin{align*} \mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n \;\Big|\; B\right) = \frac{\sum_{n=1}^\infty \mathbb{P}(A_n \cap B)}{\mathbb{P}(B)} = \sum_{n=1}^\infty \frac{\mathbb{P}(A_n \cap B)}{\mathbb{P}(B)} = \sum_{n=1}^\infty \mathbb{P}(A_n \mid B), \end{align*} where interchanging the sum and the division by the constant $\mathbb{P}(B)$ is valid because the sum has non-negative terms. [/step]