[proofplan] We encode the successive intersections by the events $B_k = A_1 \cap \cdots \cap A_k$. The positivity hypothesis ensures that each [conditional probability](/page/Conditional%20Probability) $\mathbb{P}(A_k \mid B_{k-1})$ is defined. By the definition of conditional probability, each factor is a quotient $\mathbb{P}(B_k)/\mathbb{P}(B_{k-1})$, and multiplying these quotients makes all intermediate probabilities cancel. [/proofplan] [step:Define the successive intersections and verify the conditional probabilities are defined] For each $1 \le k \le n$, let \begin{align*} B_k := \bigcap_{j=1}^{k} A_j. \end{align*} Since $A_j \in \mathcal{F}$ for every $1 \le j \le n$ and $\mathcal{F}$ is closed under finite intersections, each $B_k$ belongs to $\mathcal{F}$. By hypothesis, $\mathbb{P}(B_{k-1}) > 0$ for every $2 \le k \le n$, so each conditional probability $\mathbb{P}(A_k \mid B_{k-1})$ is defined. [/step] [step:Rewrite each conditional probability as a quotient of successive intersections] For every $2 \le k \le n$, the definition of conditional probability gives \begin{align*} \mathbb{P}(A_k \mid B_{k-1}) = \frac{\mathbb{P}(A_k \cap B_{k-1})}{\mathbb{P}(B_{k-1})}. \end{align*} Because \begin{align*} A_k \cap B_{k-1} = A_k \cap \bigcap_{j=1}^{k-1} A_j = \bigcap_{j=1}^{k} A_j = B_k, \end{align*} we obtain \begin{align*} \mathbb{P}(A_k \mid B_{k-1}) = \frac{\mathbb{P}(B_k)}{\mathbb{P}(B_{k-1})}. \end{align*} [guided] The purpose of introducing $B_k$ is to make the repeated intersections manageable. For a fixed integer $k$ with $2 \le k \le n$, the event $B_{k-1}$ is \begin{align*} B_{k-1} = \bigcap_{j=1}^{k-1} A_j. \end{align*} The hypothesis says $\mathbb{P}(B_{k-1}) > 0$, so the conditional probability $\mathbb{P}(A_k \mid B_{k-1})$ is defined by \begin{align*} \mathbb{P}(A_k \mid B_{k-1}) = \frac{\mathbb{P}(A_k \cap B_{k-1})}{\mathbb{P}(B_{k-1})}. \end{align*} Now compute the event in the numerator. Since intersection is associative and commutative, \begin{align*} A_k \cap B_{k-1} = A_k \cap \bigcap_{j=1}^{k-1} A_j = \bigcap_{j=1}^{k} A_j = B_k. \end{align*} Therefore each conditional factor is exactly the ratio of two consecutive intersection probabilities: \begin{align*} \mathbb{P}(A_k \mid B_{k-1}) = \frac{\mathbb{P}(B_k)}{\mathbb{P}(B_{k-1})}. \end{align*} This quotient form is the key point, because multiplying consecutive quotients will cancel the intermediate terms. [/guided] [/step] [step:Multiply the quotient identities and telescope the product] If $n=1$, then $B_1=A_1$ and the asserted identity reads $\mathbb{P}(B_1)=\mathbb{P}(A_1)$, so the result holds. Assume now that $n \ge 2$. Since $B_1=A_1$, multiplying the quotient identities from the previous step gives \begin{align*} \mathbb{P}(A_1)\prod_{k=2}^{n}\mathbb{P}(A_k \mid B_{k-1}) &= \mathbb{P}(B_1)\prod_{k=2}^{n}\frac{\mathbb{P}(B_k)}{\mathbb{P}(B_{k-1})} \\ &= \mathbb{P}(B_1) \frac{\mathbb{P}(B_2)}{\mathbb{P}(B_1)} \frac{\mathbb{P}(B_3)}{\mathbb{P}(B_2)} \cdots \frac{\mathbb{P}(B_n)}{\mathbb{P}(B_{n-1})} \\ &= \mathbb{P}(B_n). \end{align*} Thus \begin{align*} \mathbb{P}(B_n) = \mathbb{P}(A_1)\prod_{k=2}^{n}\mathbb{P}(A_k \mid B_{k-1}). \end{align*} Substituting $B_k=\bigcap_{j=1}^{k}A_j$ gives the displayed chain rule formula for $A_1,\dots,A_n$. [/step]