Conditional independence given a sigma-algebra

Assume the axiom of choice so that the library's conditional-expectation classes are available. Let $Y$ and $Z$ be random elements and let $\mathcal{G}\subseteq \mathcal{F}$ be a sigma-algebra. We say that $Y$ and $Z$ are conditionally independent given $\mathcal{G}$, and write $Y\perp \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\perp Z\mid \mathcal{G}$, if for every pair of bounded measurable real functions $f,g$, $\mathbb{E}[f(Y)g(Z)\mid \mathcal{G}]=\mathbb{E}[f(Y)\mid \mathcal{G}]\mathbb{E}[g(Z)\mid \mathcal{G}]\text{a.s.}$ This is an equality of almost-everywhere classes and hence does not depend on representatives.

For sigma-algebras ${\mathcal{H}}_1,{\mathcal{H}}_2\subseteq \mathcal{F}$, the notation ${\mathcal{H}}_1\perp \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\perp {\mathcal{H}}_2\mid \mathcal{G}$ means the same identity for every bounded ${\mathcal{H}}_1$-measurable $U$ and bounded ${\mathcal{H}}_2$-measurable $V$.

The definition is symmetric. If $\mathcal{G}=\{\mathrm{\varnothing },\mathrm{\Omega }\}$ modulo null sets, it reduces to ordinary independence. If one side is $\mathcal{G}$-measurable, conditional independence is automatic because that factor is already known when conditioning. The choices of the zero function or the constant-one function cause no exceptional case.