On a classic probability space $(\Omega, \mathbb P, \mathcal F)$.
Let $G:\Omega\rightarrow \mathbb R$ a random variable and $P^G$ its law.
Let $\mathcal G_t=\mathcal F_t\vee\sigma (G)$ with $t\in [0, T]$ ($T>0$ fixed).
If we assume that $G$ is independant to $\mathcal F$, why for every $\mathcal G$ adapted processes $Y(G)$ and every $u\in\mathbb R$ we have :
$$ \mathbb E (Y_t(G)\mid\mathcal G_s)=\mathbb E (Y_t(u)\mid\mathcal F_s)_{\{u=G\}}\ ? $$
My answer is the following (and I have a doubt on it I do not know if it is good, I need the approval of someone) :
If $Y(G)$ is $\mathcal G$ adapted hence it is of the form $Y_t(G)(\omega)=y_t(\omega,G(\omega))$ where $y_t$ is $\mathcal F_t\otimes\mathcal B(\mathbb R)$ measurable hence : $$ \mathbb E (Y_t(G)\mid\mathcal G_s)=(\mathbb E (Y_t(.)\mid\mathcal F_s\otimes\mathcal B(\mathbb R)))_{G} $$ Firstly $X(.,.)=\mathbb E (Y_t(.)\mid\mathcal F_s)$ is $\mathcal F_s\otimes\mathcal B(\mathbb R)$ measurable.
For $A\in\mathcal F_s$ and for every $u\in\mathbb R$, $\{u=G\}\in\mathcal B(\mathbb R)$ we have : $$ \mathbb E (Y_t(u)\mathbb I_{A\times\{u=G\}}) = \mathbb E (Y_t(u)\mathbb I_A \mathbb I_{u=G})=\mathbb E^{G} (\mathbb E (Y_t(u)\mathbb I_A) \mathbb I_{u=G}) $$ The last equality follows from $G$ is $\mathcal F$ independant. And : $$ \mathbb E (\mathbb E (Y_t(u)\mid\mathcal F_s)\mathbb I_{A\times\{u=G\}}) = \mathbb E (\mathbb E(Y_t(u)\mid\mathcal F_s)\mathbb I_A \mathbb I_{u=G}))=\mathbb E^{G} (\mathbb E (\mathbb E(Y_t(u)\mid\mathcal F_s)\mathbb I_A) \mathbb I_{u=G}) $$
We use the class monotone argument and : $$ \mathbb E (Y_t(G)\mid\mathcal G_s) = \mathbb E (Y_t(.)\mid\mathcal F_s)_{G} $$ and for every $u$: $$ \mathbb E (Y_t(G)\mid\mathcal G_s) = \mathbb E (Y_t(u)\mid\mathcal F_s)_{u=G} $$ There is no problem if $P^G$ is a probability on $(\Omega,\mathcal F)$ ? There is no problem in my last lines ?