Suppose two $\sigma$-algebra's $\mathcal{F}_1,\mathcal{F}_2$ are conditionally independent given some $\sigma$-algebra $\mathcal{G}$ i.e. for any $A\in \mathcal{F}_1$, and $B\in\mathcal{F}_2$, we have
$P(A\cap B|\mathcal{G}) = P(A|\mathcal{G})P(B|\mathcal{G})$
is it true that for any $C\in \mathcal{G}$ we have for any $A\in \mathcal{F}_1$, and $B\in\mathcal{F}_2$,
$P(A\cap B|C) = P(A|C)P(B|C)$?
I have been trying to show this and this is what I have
$P(A\cap B|C) = \frac{1}{P(C)} P(A\cap B\cap C) =\frac{1}{P(C)}* \mathbb{E}[\mathbb{E} (1_A *1_B*1_C|\mathcal{G})]$
$ =\frac{1}{P(C)} *\mathbb{E}[\mathbb{E}(1_A*1_C|\mathcal{G})\mathbb{E}(1_B*1_C|\mathcal{G})] $
At this point I feel like I should be able to use the definition of conditional expectation, but I am not sure what to do with the product of conditional expectations.
The reason why you are having difficulty showing this is that the assertion is not true! If it were, then setting $C=\Omega$ we would conclude $$P(A\cap B) = P(A\cap B\mid \Omega) = P(A\mid \Omega) P(B\mid \Omega) = P(A)P(B)$$ for every $A\in\cal F_1$ and every $B\in\cal F_2$, i.e., we would deduce that the sigma-algebras $\cal F_1$ and $\cal F_2$ are unconditionally independent.
What's the intuition behind this negative result? Suppose ${\cal G}=\sigma(Z)$, where $Z$ is a discrete random variable. If $\cal F_1$ and $\cal F_2$ are conditionally independent given $\cal G$, it is straightforward to prove that $$P(A\cap B\mid Z=z)=P(A\mid Z=z)P(B\mid Z=z)$$ for any $z$ and any $A\in\cal F_1$ and $B\in\cal F_2$. However, $\cal G$ contains more events than those of the form $\{Z=z\}$, including events where the asserted claim fails to hold. The problem is that $\cal G$ is too big.