Proof of Bayes' theorem

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A, B \in \mathcal{F}$ with $P(A) > 0$ and $P(B) > 0$.
2. Since $P(B) > 0$, the definition of conditional probability gives $P(A \mid B) = P(A \cap B) / P(B)$.
3. Since $P(A) > 0$, the same definition gives $P(B \mid A) = P(B \cap A) / P(A)$.
4. Multiplying step 3 by $P(A)$ and using $B \cap A = A \cap B$ gives $P(A \cap B) = P(B \mid A) \, P(A)$.
5. Substituting step 4 into step 2 gives $P(A \mid B) = \dfrac{P(B \mid A) \, P(A)}{P(B)}$.