Conditional-independence equivalences and preservation

Assume (1), fix bounded $g$, and write [F1, F2, F3] $W=\mathbb{E}[g(Z)\mid \mathcal{G}]$. For $C\in \mathcal{G}$ and measurable $A$, [F1] and the defining event-integral property give $\begin{array}{rl}{\displaystyle \mathbb{E}[1_C{1}_{\{Y\in A\}}g(Z)]}& {\displaystyle =\mathbb{E}[1_C\mathbb{E}(1_{\{Y\in A\}}g(Z)\mid \mathcal{G})]}\\ {\displaystyle }& {\displaystyle =\mathbb{E}[1_C\mathbb{E}(1_{\{Y\in A\}}\mid \mathcal{G})W]}\\ {\displaystyle }& {\displaystyle =\mathbb{E}[1_C{1}_{\{Y\in A\}}W],}\end{array}$ where the last equality follows by [F3]. The events $C\cap \{Y\in A\}$ form a pi-system generating $\mathcal{G}\vee \sigma (Y)$. For fixed $g$, the events on which the first and last integrals agree form a lambda-system, so [F2] extends the equality to the whole join. Since $W$ is measurable for that join, it is a version of $\mathbb{E}[g(Z)\mid \mathcal{G}\vee \sigma (Y)]$. This proves (2), including $A=\mathrm{\varnothing }$ and $A$ equal to the whole state space.

Conversely assume (2), put $\mathcal{H}=\mathcal{G}\vee \sigma (Y)$, and take [F1, F3, F4] bounded $f,g$. By [F3], (2), and [F4], $\begin{array}{rl}{\displaystyle \mathbb{E}[f(Y)g(Z)\mid \mathcal{G}]}& {\displaystyle =\mathbb{E}[\mathbb{E}(f(Y)g(Z)\mid \mathcal{H})\mid \mathcal{G}]}\\ {\displaystyle }& {\displaystyle =\mathbb{E}[f(Y)\mathbb{E}(g(Z)\mid \mathcal{H})\mid \mathcal{G}]}\\ {\displaystyle }& {\displaystyle =\mathbb{E}[f(Y)W\mid \mathcal{G}]=\mathbb{E}[f(Y)\mid \mathcal{G}]W\mathrm{.}}\end{array}$ This is [F1], so (1) follows.

If $\varphi$ and $\psi$ are measurable, substitute $f\circ \varphi$ and [F1] $g\circ \psi$ in [F1]; boundedness and measurability are preserved. Hence $\varphi (Y)\perp \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\perp \psi (Z)\mid \mathcal{G}$.

If $V$ is $\mathcal{G}$-measurable, then [step 1.1, step 1.2] $\mathcal{G}\vee \sigma (Y,V)=\mathcal{G}\vee \sigma (Y)$. Thus criterion (2) is unchanged after replacing $Y$ by $(Y,V)$. Symmetry gives the corresponding claim on the $Z$ side. Constant, one-point, and zero-valued $V$ are included.