[proofplan] The proof is just the defining identity for [conditional probability](/page/Conditional%20Probability), rearranged by multiplication. When the conditioning event has positive probability, the relevant quotient is well-defined. Applying this once with conditioning event $B$ and once with conditioning event $A$ gives the two formulas. [/proofplan] [step:Rewrite the definition of $\mathbb P(A \mid B)$ when $\mathbb P(B) > 0$] Assume $\mathbb P(B) > 0$. Since $A, B \in \mathcal F$ and $\mathcal F$ is closed under finite intersections, $A \cap B \in \mathcal F$. By the definition of conditional probability, \begin{align*} \mathbb P(A \mid B) = \frac{\mathbb P(A \cap B)}{\mathbb P(B)}. \end{align*} Multiplying both sides by the positive real number $\mathbb P(B)$ gives \begin{align*} \mathbb P(A \mid B)\mathbb P(B) = \mathbb P(A \cap B). \end{align*} Thus \begin{align*} \mathbb P(A \cap B) = \mathbb P(A \mid B)\mathbb P(B). \end{align*} [guided] Assume $\mathbb P(B) > 0$. The positivity condition is exactly what makes conditioning on $B$ meaningful, because the definition of $\mathbb P(A \mid B)$ divides by $\mathbb P(B)$. First, since $A, B \in \mathcal F$ and $\mathcal F$ is a $\sigma$-algebra, it is closed under finite intersections. Hence $A \cap B \in \mathcal F$, so $\mathbb P(A \cap B)$ is defined. The conditional probability of $A$ given $B$ is defined by \begin{align*} \mathbb P(A \mid B) = \frac{\mathbb P(A \cap B)}{\mathbb P(B)}. \end{align*} Because $\mathbb P(B) > 0$, we may multiply both sides by $\mathbb P(B)$ without changing the equality: \begin{align*} \mathbb P(A \mid B)\mathbb P(B) = \mathbb P(A \cap B). \end{align*} Reversing the two sides of this equality gives \begin{align*} \mathbb P(A \cap B) = \mathbb P(A \mid B)\mathbb P(B). \end{align*} [/guided] [/step] [step:Rewrite the definition of $\mathbb P(B \mid A)$ when $\mathbb P(A) > 0$] Assume $\mathbb P(A) > 0$. Since $A \cap B = B \cap A$, the definition of conditional probability gives \begin{align*} \mathbb P(B \mid A) = \frac{\mathbb P(B \cap A)}{\mathbb P(A)} = \frac{\mathbb P(A \cap B)}{\mathbb P(A)}. \end{align*} Multiplying both sides by the positive real number $\mathbb P(A)$ gives \begin{align*} \mathbb P(B \mid A)\mathbb P(A) = \mathbb P(A \cap B). \end{align*} Therefore \begin{align*} \mathbb P(A \cap B) = \mathbb P(B \mid A)\mathbb P(A). \end{align*} [/step]