Proof of the multiplication rule

Proof

We argue by induction on the number of events in the intersection.

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A_1, \dots, A_n \in \mathcal{F}$ with $n \ge 2$ and $P(A_1 \cap \dots \cap A_{n-1}) > 0$.
2. For $1 \le k \le n$ let $C_k = A_1 \cap \dots \cap A_k$. Each $C_k$ is in $\mathcal{F}$, because $\mathcal{F}$ is a σ-algebra.
3. Let $1 \le k \le n - 1$. Then $C_{n-1} \subseteq C_k$. By monotonicity of probability, $P(C_k) \ge P(C_{n-1}) > 0$.
4. So for $1 \le k \le n - 1$ the conditional probability $P(A_{k+1} \mid C_k) = P(A_{k+1} \cap C_k) / P(C_k)$ is defined.
5. Since $A_{k+1} \cap C_k = C_{k+1}$, step 4 gives $P(C_{k+1}) = P(C_k) \, P(A_{k+1} \mid C_k)$ for $1 \le k \le n - 1$.
6. We claim that $P(C_m) = P(A_1) \prod_{k=1}^{m-1} P(A_{k+1} \mid C_k)$ for $1 \le m \le n$. For $m = 1$ the product is empty and the claim reads $P(C_1) = P(A_1)$, which holds since $C_1 = A_1$.
7. Suppose the claim holds for some $m$ with $1 \le m \le n - 1$. Step 5 with $k = m$ multiplies $P(C_m)$ by $P(A_{m+1} \mid C_m)$, so the claim holds for $m + 1$.
8. The case $m = n$ of the claim reads $P(A_1 \cap \dots \cap A_n) = P(A_1) \, P(A_2 \mid A_1) \cdots P(A_n \mid A_1 \cap \dots \cap A_{n-1})$.