[proofplan] We decompose $A$ into the countable disjoint union $A = \bigcup_n (A \cap B_n)$ using the partition $(B_n)$, apply countable additivity to the disjoint pieces, and rewrite each term $\mathbb{P}(A \cap B_n)$ using the definition of conditional probability. [/proofplan] [step:Decompose $A$ using the partition $(B_n)$] Since $(B_n)_{n \ge 1}$ is a partition of $\Omega$, every $\omega \in A$ belongs to exactly one $B_n$, so \begin{align*} A = A \cap \Omega = A \cap \bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty (A \cap B_n). \end{align*} The sets $A \cap B_n$ are pairwise disjoint: for $i \ne j$, $(A \cap B_i) \cap (A \cap B_j) = A \cap (B_i \cap B_j) = A \cap \varnothing = \varnothing$, since the $B_n$ are pairwise disjoint. [/step] [step:Apply countable additivity and rewrite using conditional probability] By [countable](/page/Countable%20Set) additivity applied to the pairwise disjoint sequence $(A \cap B_n)_{n \ge 1}$, \begin{align*} \mathbb{P}(A) = \sum_{n=1}^\infty \mathbb{P}(A \cap B_n). \end{align*} Since $\mathbb{P}(B_n) > 0$ for each $n$, the definition of conditional probability gives $\mathbb{P}(A \cap B_n) = \mathbb{P}(A \mid B_n) \, \mathbb{P}(B_n)$. Substituting: \begin{align*} \mathbb{P}(A) = \sum_{n=1}^\infty \mathbb{P}(A \mid B_n) \, \mathbb{P}(B_n). \end{align*} [/step]