Definition
Let $X,Y,Z$ be random variables for a common probability space $(\Omega,\mathscr{F},P)$
Then, $X,Y$ are said to be conditionally independent given $Z$ if $P(X\in A, Y\in B|Z)=P(X\in A|Z)P(Y\in B|Z)$ for all borel sets $A,B$.
If $X,Y$ are conditionally independent given $Z$, $P(X\in A|Y,Z)= P(X\in A|Z)$?
This is true if we assume $(X,Y,Z)$ is jointly absolutely continuous. (Because $$p(x|z)=p(x|z)p(y|z)/p(y|z)=p(x,y|z)/p(y|z)=p(x,y|z)p(z)/p(y,z)=p(x,y,z)/p(y,z)=p(x|y,z)$$)
I am curious if this holds for general $X,Y,Z$'s.
This is certainly true.
Let's say that the $\sigma$-algebras $\mathcal{F}$ and $\mathcal{G}$ are independent given $\mathcal{H}$ if
$$ P_{\mathcal{H}}(F \cap G) = P_{\mathcal{H}}(F) P_{\mathcal{H}}(G) $$ for every $F \in \mathcal{F}$ and $G \in \mathcal{G}$.
Then we will show that $$ P_{\mathcal{G} \vee \mathcal{H}}(F) = P_{\mathcal{H}}(F) $$ for every $F \in \mathcal{F}$.
First $P_{\mathcal{H}}(F)$ is certainly $(\mathcal{G} \vee \mathcal{H})$-measurable.
Second, we have, for every $F \in \mathcal{F}$, $G \in \mathcal{G}$ and $H \in \mathcal{H}$,
\begin{align} P(F \cap G \cap H) &= E( P_{\mathcal{H}}(F) P_{\mathcal{H}}(G) I_{H})\\\\ &= E( P_{\mathcal{H}}(F) P_{\mathcal{H}}(G \cap H)) \\\\ &= E( P_{\mathcal{H}}(F) I_{G \cap H}) \end{align} where the first equality comes from the conditional independence of $\mathcal{F}$ and $\mathcal{G}$. Since the equality $P(F \cap G \cap H) = E( P_{\mathcal{H}}(F) I_{G \cap H})$ is true for every set of the form $G \cap H$, it will be true for every set in $\mathcal{G} \vee \mathcal{H}$, by standard arguments.