Proof of the law of total probability

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let $(B_i)_{i \in I}$ be the given family, with $I = \set{1, \dots, N}$ or $I = \set{1, 2, \dots}$.
2. So the $B_i$ are pairwise disjoint events with $\bigcup_{i \in I} B_i = \Omega$ and $P(B_i) > 0$. They form a partition of $\Omega$. Let $A$ be an event.
3. For each integer $i \ge 1$ let $E_i = A \cap B_i$ if $i \in I$, and $E_i = \varnothing$ if $i \notin I$. Each $E_i$ is in $\mathcal{F}$, because $\mathcal{F}$ is a σ-algebra.
4. The $E_i$ are pairwise disjoint. For $i \ne j$ in $I$ we have $E_i \cap E_j \subseteq B_i \cap B_j = \varnothing$, and the empty sets meet nothing.
5. Their union is $\bigcup_{i \in I} (A \cap B_i) = A \cap \bigcup_{i \in I} B_i = A \cap \Omega = A$.
6. By the complement rule, $P(\varnothing) = 1 - P(\Omega) = 0$.
7. By the countable additivity of the probability measure $P$ and step 6, $P(A) = \sum_{i \in I} P(A \cap B_i)$.
8. For $i \in I$ we have $P(B_i) > 0$. By the definition of conditional probability, $P(A \cap B_i) = P(A \mid B_i) \, P(B_i)$.
9. Substituting step 8 into step 7 gives $P(A) = \sum_{i} P(A \mid B_i) \, P(B_i)$.