Let's suppose we have a probability space $(\Omega,\mathcal{F},P)$ and a sub sigma algebra $\mathcal{G} \subset \mathcal{F}$. Let's say we have $X$ that is $\mathcal{G}$-measurable and $Y$ that is $\mathcal{G}$-independent. If we have a bounded measurable function $\phi$, then the following should apply:
$$E[\phi(X,Y)|\mathcal{G}]= E[\phi(x, \cdot)]|_{x=X}$$
The point is that I haven't found a simple example of the above formula, and I want to check if that means what I think. Now, I have to use this to prove that processes with independent increments are Markov processes, so let's take that. Let $\phi(X_{t+h})= 2 X_{t+h} -1$. Then $\phi(X_{t+h})=\psi(X_t,X_{t+h}-X_t)= 2(X_{t+h} -X_t) +2X_t -1$. If we put $Y=X_{t+h}- X_t$, then we'd have $\psi(X_t,Y)=2Y +2X_t -1$. So, if I interpreted things correctly, we'd have $G(x)=E[\psi(x,\cdot)]=E[Y] +2x -1$ as a function of $x$. So for instance $G(3)=E[Y]+5$. Then:
$$E[\phi(X_{t+h})| \mathcal{G}]=E[\psi(x,\cdot)]|_{x=X_t} = E[Y] + 2 X_t -1$$
Is this correct?
This proposition says that if you let $f:x\mapsto E(\phi(x,Y))$, then $E(\phi(X,Y)|\mathcal G) = f(X)$.
Yes, what you've done in the example is correct.