$X$, $Y$, $Z$ are 3 random variables.
Let $g(\cdot)$ denote a conditional expectation such that $g(X,Z)=E(Y|X,Z)$
Need to show $E(Y|X, g(X,Z))$. I think $E(Y|X, g(X,Z))=E(Y|X,Z)$ and my rationale is given below:
\begin{align} E(Y|X, g(X,Z))&=E(E(Y|X,g(X,Z),Z)|X,g(X,Z)) \\ &=E(E(Y|X,Z)|X,g(X,Z)) \qquad \because X \, \text{and} \, Z \, \text{determine} \, g(X,Z) \\ &=E(g(X,Z)|X,g(X,Z)) \\ &=g(X,Z)=E(Y|X,Z) \end{align}
Any issue with my rationale? thanks,
I think you are right. Since I am more comfortable using a notation with $\sigma$-algebras let's introduce $$ {\cal A}:=\sigma\Big(X,g(X,Z)\Big)\,,\quad {\cal B}:=\sigma(X,Z) $$ which clearly satisfy ${\cal A}\subseteq {\cal B}$ so that we can use the tower property:
\begin{align}\tag{1} E[\,Y\,|\,{\cal A}\,]=E\Big[\,E[\,Y\,|\,{\cal B}\,]\,\Big|\,{\cal A}\,\Big]\,. \end{align} By the Doob-Dynkin factorization lemma: $E[\,Y\,|\,{\cal B}\,]=g(X,Z)$ so that the RHS of (1) becomes $$ E\Big[\,g(X,Z)\,\Big|\,{\cal A}\,\Big]=g(X,Z). $$ This holds because $g(X,Z)$ is trivially ${\cal A}$-measurable. In other words, we have shown that $$\tag{2} \boxed{E[\,Y\,|\,{\cal A}\,]=g(X,Z)\,.} $$
