Proving that $X$ and $Y$ are conditionally independent given $Z=z$ Here is the exact wording of the problem whose solution I would like verified.
Suppose there are functions (of sets) $f_z$ and $g_z$ such that for all sets $A,B$ we have $$P(X \in A,Y \in B\mid Z=z)=f_z(A)g_z(B)$$
Show that $X$ and $Y$ are conditionally independent given $Z=z$. Here is my argument. First we'll show that $$f_z(\Omega_X)g_z(\Omega_Y)=1$$ This is because
$$f_z(\Omega_X)g_z(\Omega_Y)=P(X\in \Omega_X ,Y \in \Omega_Y\mid Z=z)=1$$
Moreover, notice $$P(X \in A\mid Z=z)=P(X \in A,Y \in \Omega_Y|Z=z)=f_z(A)g_z(\Omega_Y)$$ $$P(Y \in B\mid Z=z)=P(X \in \Omega_X,Y \in B\mid Z=z)=f_z(\Omega_X)g_z(B)$$ Hence $$P(X \in A\mid Z=z)P(Y \in B\mid Z=z)=\big[f_z(A)g_z(\Omega_Y)\big]\big[f_z(\Omega_X)g_z(B)\big]=f_z(A)g_z(B)=P(X \in A,Y \in B\mid Z=z)$$
How does this look?
 A: That looks great to me, although there is nothing specific about conditional independence here since the same argument applies if you remove any mention of $Z$. In what context did you see this question?
A: Yes, that is a good look.
You needed to show, for all values $z$ in the support for $Z$, and subsets $\mathcal A$ and $\mathcal B$ of the supports for $X$ and $Y$ respectively,$^\star$ that:
$$\mathsf P(X{\in}\mathcal A, Y{\in}\mathcal B\mid Z{=}z)=\mathsf P(X{\in}\mathcal A\mid Z{=}z)\,\mathsf P(Y{\in}\mathcal  B\mid Z{=}z)$$
And, indeed you did so.

Letting $\mathcal X, \mathcal Y,$ and $\mathcal Z$ be the supports for $X,Y,$ and $Z$ respectively,$^\star$ then, for all $\mathcal A\subseteq\mathcal X$, $\mathcal B\subseteq\mathcal Y$, and $z\in\mathcal Z$ ...
$$\begin{align}&\quad\mathsf P(X{\in}\mathcal A\mid Z{=}z)\cdot\mathsf P(Y{\in}\mathcal B\mid Z{=}z)\\[1ex]&=\mathsf P(X{\in}\mathcal A, Y{\in}\mathcal Y\mid Z{=}z)\cdot\mathsf P(X{\in}\mathcal X, Y{\in}\mathcal B\mid Z{=}z)&&\text{LoTP}\\[1ex]&=f_{z}(\mathcal A)\cdot g_{z}(\mathcal Y)\cdot f_{z}(\mathcal X)\cdot g_{z}(\mathcal B)&&\text{definition of }f_z\cdot g_z\\[1ex]&=f_z(\mathcal A)\cdot\mathsf P(X{\in}\mathcal X,Y{\in} \mathcal Y\mid Z{=}z)\cdot g_z(\mathcal B)&&\text{definition of }f_z\cdot g_z\\[1ex]&=f_z(\mathcal A)\cdot 1\cdot g_z(\mathcal B)&&\text{Maximum probability}\\[1ex]&=\mathsf P(X{\in}\mathcal A, Y{\in}\mathcal B\mid Z{=}z)&&\text{definition of }f_z\cdot g_z\end{align}$$

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$\star~\tiny\text{(the supports for their marginal probability functions, to be pedantic.)}$
