In E. Carnal "Markov Properties for Certain Random Fields", there is a lemma 2.2 (ii) which states:
If $\mathscr{A}\perp\mathscr{B}\mid\mathscr{C}$, $\mathscr{D}\subseteq\mathscr{A}\cup\mathscr{B}$ where $\cup$ is the union of set families, then $\mathscr{A}\perp\mathscr{B}\mid\mathscr{C}\vee\sigma(\mathscr{D})$.
I want to prove this. Any insights? Many Thanks!!!
I've tried in this way: For any $A\in\mathscr{A}$, $B\in\mathscr{B}$, $\Lambda\in\mathscr{C}\vee\sigma(\mathscr{D})$ or $$\Lambda\in\mathscr{K}:=\left\{C\cap D: C\in\mathscr{C}, D\in\sigma(\mathscr{D})\right\},$$ does it somehow hold that \begin{align*} (*) \int_{\Lambda}\mathbb{E}\left\{\mathbb{1}_A\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\,\mathrm{d}\mathbb{P} = \int_{\Lambda}\mathbb{E}\left\{\mathbb{1}_A\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\mathbb{E}\left\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\,\mathrm{d}\mathbb{P}? \end{align*}
I've also noticed that
(1) $\sigma(\mathscr{D})\subseteq\mathscr{A}\vee\mathscr{B}$;
(2) \begin{align*} \text{LHS of (*)} &=\int_{\Lambda}\mathbb{1}_A\mathbb{1}_B\,\mathrm{d}\mathbb{P}\\ &=\int_{C}\mathbb{1}_D\mathbb{1}_A\mathbb{1}_B\,\mathrm{d}\mathbb{P}\\ &=\cdots; \end{align*}
(3) \begin{align*} \text{RHS of (*)} &=\int_{\Lambda}\mathbb{E}\left\{\mathbb{1}_A\mathbb{E}\left\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\,\mathrm{d}\mathbb{P}\\ &=\int_{\Lambda}\mathbb{1}_A\mathbb{E}\left\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\,\mathrm{d}\mathbb{P}\\ &=\int_{C}\mathbb{E}\left\{\mathbb{1}_D\mathbb{1}_A\mathbb{E}\left\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\right\}\mid\mathscr{C}\right\}\,\mathrm{d}\mathbb{P}; \end{align*} Well, I can hardly apply the condition $\mathscr{A}\perp\mathscr{B}\mid\mathscr{C}$, and can hardly convert between $\mathscr{C}$ and $\mathscr{C}\vee\sigma(\mathscr{D})$.
I also tried in this way, which I think is fairly close to the whole proof. Suppose $C\cap D$ is any atom of $\mathscr{C}\vee\sigma(\mathscr{D})$, where $C$ is any atom of $\mathscr{C}$, and $D$ is any atom of $\sigma(\mathscr{D})$. Since $\mathscr{D}\subseteq\mathscr{A}\cup\mathscr{B}$ rather than $\mathscr{A}\vee\mathscr{B}$, we can show that $D$ is of the form $D=A_0\cap B_0$ where $A_0\in\mathscr{A}$ and $B_0\in\mathscr{B}$. Therefore, on any atom $C\cap D$ of $\mathscr{C}\vee\sigma(\mathscr{D})$, it holds that, $\forall A\in\mathscr{A}$, $\forall B\in\mathscr{B}$, \begin{align*} \mathbb{P}(A\cap B\cap C\cap D)\mathbb{P}(C\cap D)\mathbb{P}^2(C) &=\mathbb{P}((A\cap A_0)\cap(B\cap B_0)\cap C)\mathbb{P}(C)\mathbb{P}(A_0\cap B_0\cap C)\mathbb{P}(C)\\ &=\mathbb{P}(A\cap A_0\cap C)\mathbb{P}(B\cap B_0\cap C)\mathbb{P}(A_0\cap C)\mathbb{P}(B_0\cap C)\\ &=\mathbb{P}(A\cap A_0\cap C)\mathbb{P}(B_0\cap C)\mathbb{P}(B\cap B_0\cap C)\mathbb{P}(A_0\cap C)\\ &=\mathbb{P}(A\cap A_0\cap B_0\cap C)\mathbb{P}(C)\mathbb{P}(B\cap B_0\cap A_0\cap C)\mathbb{P}(C)\\ &=\mathbb{P}(A\cap C\cap D)\mathbb{P}(B\cap C\cap D)\mathbb{P}^2(C), \end{align*} where the second and the second to the last equality use the condition $\mathscr{A}\perp\!\!\!\perp\mathscr{B}\mid\mathscr{C}$. Thus, it has been showed that $\mathbb{E}\{\mathbb{1}_A\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\}$ and $\mathbb{E}\{\mathbb{1}_A\mid\mathscr{C}\vee\sigma(\mathscr{D})\}\mathbb{E}\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\}$ do match on every atom of $\mathscr{C}\vee\sigma(\mathscr{D})$.
In this approach, I think there are still several issues:
(1) Do the matchings on all atoms suggest the equality $$\mathbb{E}\{\mathbb{1}_A\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\}=\mathbb{E}\{\mathbb{1}_A\mid\mathscr{C}\vee\sigma(\mathscr{D})\}\mathbb{E}\{\mathbb{1}_B\mid\mathscr{C}\vee\sigma(\mathscr{D})\};$$
(2) Can the approach still hold when the probability of some atoms, say $C$ and $D$, are zero?
To prove this lemma, we need to show that for any events $A$ in $\mathscr{A}$ and $B$ in $\mathscr{B}$, we have $P(A \cap B | \mathscr{C} \vee \sigma(\mathscr{D})) = P(A|\mathscr{C}\vee\sigma(\mathscr{D}))P(B|\mathscr{C}\vee\sigma(\mathscr{D}))$.
First, note that since $\mathscr{A} \perp \mathscr{B}|\mathscr{C}$, we have $P(A \cap B|\mathscr{C}) = P(A|\mathscr{C})P(B|\mathscr{C})$.
Now, consider the event $A \cap B$. This event can be written as $(A\cap B\cap \mathscr{C})\cup(A\cap B\cap \mathscr{C}^c)$. Since $\mathscr{A}\perp\mathscr{B}|\mathscr{C}$, we have $A\perp B|\mathscr{C}^c$, which means that $P(A\cap B|\mathscr{C}^c)=P(A|\mathscr{C}^c)P(B|\mathscr{C}^c)$. Therefore,
\begin{align*} P(A \cap B | \mathscr{C} \vee \sigma(\mathscr{D})) &= P((A \cap B \cap \mathscr{C}) \cup (A \cap B \cap \mathscr{C}^c) | \mathscr{C} \vee \sigma(\mathscr{D}))\\ &= P(A \cap B \cap \mathscr{C} | \mathscr{C} \vee \sigma(\mathscr{D})) + P(A \cap B \cap \mathscr{C}^c | \mathscr{C} \vee \sigma(\mathscr{D}))\\ &= P(A|\mathscr{C} \vee \sigma(\mathscr{D}))P(B|\mathscr{C} \vee \sigma(\mathscr{D})) + P(A|\mathscr{C}^c \vee \sigma(\mathscr{D}))P(B|\mathscr{C}^c \vee \sigma(\mathscr{D}))\\ &= P(A|\mathscr{C} \vee \sigma(\mathscr{D}))P(B|\mathscr{C} \vee \sigma(\mathscr{D})) \end{align*}
where the second equality follows from the law of total probability and the fact that $A\perp B|\mathscr{C}^c$, and the third equality follows from the definition of conditional probability and the fact that $A$ and $B$ are both in $\mathscr{D}$ or their complements are in $\mathscr{D}$.
Therefore, we have shown that $P(A \cap B | \mathscr{C} \vee \sigma(\mathscr{D})) = P(A|\mathscr{C} \vee \sigma(\mathscr{D}))P(B|\mathscr{C} \vee \sigma(\mathscr{D}))$, which completes the proof.