I want to show the following: Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ a sub-$\sigma$-algebra, $Z$ a $\mathcal{G}$-measurable random variable and $Y$ a random variable independent of $\mathcal{G}$. Then for every non-negative measurable function $g$ we have $$ \mathbb{E}[g(Y,Z) \vert \mathcal{G}] = \int_{\mathbb{R}} g(y,Z) dP_{Y}(y). $$
My attempt of the proof: The RHS is $\mathcal{G}$-measurable, so it remains to show that $$ \int_{A} g(Y(\omega),Z(\omega)) d\mathbb{P}(\omega) = \int_{A} \int_{y \in \mathbb{R}} g(y,Z(\omega)) dP_{Y}(y) d\mathbb{P}(\omega) $$ for all $A \in \mathcal{G}$.
Now I would like to write $A = (Y,Z)^{-1}((Y,Z)(A))$ in order to apply the change of variables formula to the LHS, but this is only possible if $(Y,Z)$ is injective, which is not the case in general. How can I remove this issue? Assuming we have $A = (Y,Z)^{-1}((Y,Z)(A))$, the LHS equals (using independence) $$ \int_{y \in Y(A)} \int_{z \in Z(A)} g(y,z) dP_{Y}(y) dP_{Z}(z) \overset{(*)}{=} \int_{y \in Y(A)} \int_{A} g(y,Z(\omega)) dP_{Y}(y) d\mathbb{P}(\omega). $$ In $(*)$ I transformed $z \in Z(A)$ to $\omega \in A$ and I am not sure if this is correct and how to justify it if this is the case. Now the new LHS looks almost like the RHS but I am integrating over $y \in Y(A)$ instead of $y \in \mathbb{R}$, so what went wrong in my sloppy calculations?