[proofplan] We fix an event in one $\pi$-system and show that the collection of events in the other $\sigma$-algebra for which the product formula holds is a d-system. The [Dynkin $\pi$-system lemma](/theorems/505) upgrades agreement on the $\pi$-system to agreement on the full $\sigma$-algebra. Two applications of this argument -- one for each coordinate -- complete the proof. [/proofplan] [step:Fix $A_1 \in \mathcal{A}_1$ and extend the product formula to all of $\sigma(\mathcal{A}_2)$] Fix $A_1 \in \mathcal{A}_1$. Define \begin{align*} \mathcal{D}_1 = \{ B \in \sigma(\mathcal{A}_2) : \mathbb{P}(A_1 \cap B) = \mathbb{P}(A_1)\,\mathbb{P}(B) \}. \end{align*} [claim:D One Is A D System Containing A Two] $\mathcal{D}_1$ is a d-system on $\Omega$ containing $\mathcal{A}_2$. [/claim] [proof] By hypothesis, $\mathcal{A}_2 \subseteq \mathcal{D}_1$. We verify the d-system axioms. First, $\mathbb{P}(A_1 \cap \Omega) = \mathbb{P}(A_1) = \mathbb{P}(A_1)\,\mathbb{P}(\Omega)$, so $\Omega \in \mathcal{D}_1$. Second, if $B, C \in \mathcal{D}_1$ with $B \subseteq C$, then \begin{align*} \mathbb{P}(A_1 \cap (C \setminus B)) = \mathbb{P}(A_1 \cap C) - \mathbb{P}(A_1 \cap B) = \mathbb{P}(A_1)(\mathbb{P}(C) - \mathbb{P}(B)) = \mathbb{P}(A_1)\,\mathbb{P}(C \setminus B). \end{align*} Third, if $(B_n)$ is increasing in $\mathcal{D}_1$, by continuity from below, \begin{align*} \mathbb{P}\!\left(A_1 \cap \bigcup_n B_n\right) = \lim_n \mathbb{P}(A_1 \cap B_n) = \lim_n \mathbb{P}(A_1)\,\mathbb{P}(B_n) = \mathbb{P}(A_1)\,\mathbb{P}\!\left(\bigcup_n B_n\right). \end{align*} [/proof] By the [Dynkin $\pi$-system lemma](/theorems/505), $\sigma(\mathcal{A}_2) \subseteq \mathcal{D}_1$. This means: for every $A_1 \in \mathcal{A}_1$ and $B_2 \in \sigma(\mathcal{A}_2)$, \begin{align*} \mathbb{P}(A_1 \cap B_2) = \mathbb{P}(A_1)\,\mathbb{P}(B_2). \end{align*} [/step] [step:Fix $B_2 \in \sigma(\mathcal{A}_2)$ and extend the product formula to all of $\sigma(\mathcal{A}_1)$] Fix $B_2 \in \sigma(\mathcal{A}_2)$ and define \begin{align*} \mathcal{D}_2 = \{ B \in \sigma(\mathcal{A}_1) : \mathbb{P}(B \cap B_2) = \mathbb{P}(B)\,\mathbb{P}(B_2) \}. \end{align*} [claim:D Two Is A D System Containing A One] $\mathcal{D}_2$ is a d-system on $\Omega$ containing $\mathcal{A}_1$. [/claim] [proof] The previous step established $\mathbb{P}(A_1 \cap B_2) = \mathbb{P}(A_1)\,\mathbb{P}(B_2)$ for all $A_1 \in \mathcal{A}_1$, so $\mathcal{A}_1 \subseteq \mathcal{D}_2$. The d-system verification is identical to the previous claim with roles exchanged. [/proof] By the [Dynkin $\pi$-system lemma](/theorems/505), $\sigma(\mathcal{A}_1) \subseteq \mathcal{D}_2$. [/step] [step:Conclude independence of the generated $\sigma$-algebras] For every $B_1 \in \sigma(\mathcal{A}_1)$ and every $B_2 \in \sigma(\mathcal{A}_2)$, \begin{align*} \mathbb{P}(B_1 \cap B_2) = \mathbb{P}(B_1)\,\mathbb{P}(B_2). \end{align*} This is precisely the definition of independence of $\sigma(\mathcal{A}_1)$ and $\sigma(\mathcal{A}_2)$. [/step]