Joining the Conditional $\sigma$-Algebra Does not Change Independence

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In E. Carnal "Markov Properties for Certain Random Fields", there is a lemma 2.2 (i) which states:

If $\mathscr{A}\perp\mathscr{B}\mid\mathscr{C}$, then $\mathscr{A}\vee\mathscr{C}\perp\mathscr{B}\vee\mathscr{C}\mid\mathscr{C}$, where $\mathscr{A}\vee\mathscr{C}=\sigma(\mathscr{A}\cup\mathscr{C})$.

I want to prove this. I notice:

(1) It suffices to prove $\mathscr{A}\vee\mathscr{C}\perp\mathscr{B}\mid\mathscr{C}$.

(2) For any $\Lambda\in\mathscr{K}:=\{A\cap C:A\in\mathscr{A},C\in\mathscr{C}\}\subseteq\mathscr{A}\vee\mathscr{C}$, it holds that $$\mathbb{P}\{\Lambda\cap B\mid\mathscr{C}\}=\mathbb{P}\{\Lambda\mid\mathscr{C}\}\mathbb{P}\{B\mid\mathscr{C}\},\quad\forall B\in\mathscr{B}.$$ (3) Let $\mathscr{H}:=\{H\in\mathscr{A}\vee\mathscr{C}: \mathbb{P}\{H\cap B\mid\mathscr{C}\}=\mathbb{P}\{H\mid\mathscr{C}\}\mathbb{P}\{B\mid\mathscr{C}\},\forall B\in\mathscr{B}\}\supseteq\mathscr{K}$, it suffices to show that $\mathscr{H}$ is a $\sigma$-algebra, in that $\mathscr{H}\supseteq\sigma(\mathscr{K})=\mathscr{A}\vee\mathscr{C}$ $\Rightarrow$ $\mathscr{H}=\mathscr{A}\vee\mathscr{C}$.

(4) To show that $\mathscr{H}$ is a $\sigma$-algebra, I tried to prove $\mathscr{H}$ is both a $\pi$-system and a $\lambda$-system, but I failed on the former one.

Any new insights? Many Thanks!!!

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Sorry, I made a holy mistake.

Instead of proving $\mathscr{H}$ is a $\pi$-system, I shall prove that $\mathscr{K}$ is a $\pi$-system, which is manifest.

Then, by $\pi$-$\lambda$ theorem, the conclusion $\mathscr{H}\supseteq\sigma(\mathscr{K})$ holds immediately.

(I should not prove that $\mathscr{H}$ is a $\sigma$-field, and I was mislead by the form of the conclusion $\mathscr{H}\supseteq\sigma(\mathscr{K})$.)

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$\begin{array}{lcr} \text{Always:} &\mathscr{D}\perp\mathscr{C} &\mid\mathscr{C}\\ \text{Hence:} &(\mathscr{A}\vee\mathscr{B})\perp\mathscr{C} &\mid\mathscr{C}\\ \text{If besides:} &\mathscr{A}\perp\mathscr{B} &\mid\mathscr{C}\\ \text{Then:} &\mathscr{A}\perp\mathscr{B}\perp\mathscr{C} &\mid\mathscr{C}\\ \text{Therefore:} &\mathscr{A}\perp (\mathscr{B}\vee\mathscr{C}) &\mid\mathscr{C} \end{array}$