Conditional independence of events $A,B$ given a nonnegligible event $C$ is defined as $\mathbb{P}(A|B,C)=\mathbb{P}(A|C)$. This is equivalent to $\mathbb{P}(A,B|C)=\mathbb{P}(A|C)\mathbb{P}(B|C)$. When we move onto random variables, we use this second characterisation: we say that random variables $X,Y$ are independent conditionally on random variable $W$ if for every $A\in\sigma(X)$ and $B\in \sigma(Y)$, $\mathbb{P}(A,B|\sigma(W))=\mathbb{P}(A|\sigma(W))\mathbb{P}(B|\sigma(W))$ a.s.
My question is: does the first characterisation still hold? I.e. is it true that this is equivalent to $\mathbb{P}(A|\sigma(Y,W))=\mathbb{P}(A|\sigma(W))$ for all $A\in\sigma(X)$? How do we prove/disprove it? Here I denoted $\sigma(Y,W)=\sigma(\sigma(Y)\cup\sigma(W))$. Thanks for any help.
My attempt: perhaps it suffices (but by what standard argument?) to show this for generators of the sigma-algebras involved. Denote with $B$ a generic generator of $\sigma(Y)$ and $C$ a generator of $\sigma(W)$. Assuming that $B\cap C$ describes a generic generator of $\sigma(Y,W)$, the result is true by the characterisation given for events. Thus the problem seems to boil down to whether $\sigma(Y,W)$ is truly generated by sets $B\cap C$ for $B$ a generator of $\sigma(Y)$ and $C$ a generator of $\sigma(W)$. I know how to show that in general $\sigma(Y,W)$ is generated by all the possible intersections of the sets in $\sigma(Y)$ and $\sigma(W)$, but I am not sure if we can consider JUST the intersections of the generators of the two and still get $\sigma(Y,W)$.
The characterisation is true. In fact, random variables are not needed.
Theorem. Let $\mathscr{A},$ $\mathscr{B}$ and $\mathscr{F}$ be three sigma fields. A necessary and sufficient condition for $$ P(A \cap B \mid \mathscr{F}) = P(A \mid \mathscr{F}) P(B \mid \mathscr{F}) \qquad \forall A \in \mathscr{A}, B \in \mathscr{B} $$ is that $$ P(A \mid \mathscr{B} \vee \mathscr{F}) = P(A \mid \mathscr{F}). $$ (As it is somewhat common, $\mathscr{B} \vee \mathscr{F} = \sigma(\mathscr{B} \cup \mathscr{F})$.)
Proof of sufficiency. Consider $A$ and $B$ as in the statement. The definition and $\mathscr{F}$-measurability imply, $$ P(A \mid \mathscr{F}) P(B \mid \mathscr{F}) = E(\mathbf{1}_B E(\mathbf{1}_A \mid \mathscr{F}) \mid \mathscr{F}); $$ the hypothesis allows replacing $E(\mathbf{1}_A \mid \mathscr{F})$ by $E(\mathbf{1}_A \mid \mathscr{B} \vee \mathscr{F}),$ then $\mathbf{1}_B$ can enter this expectation by $(\mathscr{B} \vee \mathscr{F})$-measurability, so we get $$ P(A \mid \mathscr{F}) P(B \mid \mathscr{F}) = E(E(\mathbf{1}_A \mathbf{1}_B \mid \mathscr{B} \vee \mathscr{F}) \mid \mathscr{F}) = E(\mathbf{1}_A \mathbf{1}_B \mid \mathscr{F}) = P(A \cap B \mid \mathscr{F}), $$ the penultimate equality is the tower property of conditional expectation. Thus, sufficiency has been proved.
Proof of necessity. We are required to prove $$ \begin{equation}\tag{*} P(A \cap H) = \int\limits_H P(A \mid \mathscr{F})\ dP \end{equation} $$ for all $H \in \mathscr{B} \vee \mathscr{F}.$ Note that both sides in the equation above define measures on $H.$
The following can be proved as a basic exercise on sigma fields, $$ \mathscr{B} \vee \mathscr{F} = \sigma\left( \left\{ \bigcup_{\alpha \in \Gamma} B_\alpha \cap F_\alpha \middle| B_\alpha \in \mathscr{B}, F_\alpha \in \mathscr{F} \right\} \right), $$ the right hand side is the field of all finite unions of sets of the form $B \cap F.$ Now, monotone class theorem shows that suffices to prove (*) when $H = B \cap F.$ Then $$ P(A \cap B \cap F) = \int\limits_F P(A \cap B \mid \mathscr{F})\ dP = \int\limits_F P(A \mid \mathscr{F})P(B \mid \mathscr{F})\ dP. $$ Now, as above, $$ P(A \mid \mathscr{F}) P(B \mid \mathscr{F}) = E(\mathbf{1}_B E(\mathbf{1}_A \mid \mathscr{F}) \mid \mathscr{F}), $$ and integrating over a set in $\mathscr{F}$ will cancel out the last conditioning (by definition), $$ P(A \cap B \cap F) = \int\limits_F \mathbf{1}_B E(\mathbf{1}_A \mid \mathscr{F})\ dP = \int\limits_{F \cap B} P(A \mid \mathscr{F})\ dP, $$ completing the proof. QED