We know that for integrable r.v.'s $X$, the defining equation for its conditional expectation $Y=E[X|\mathcal A]$ given some sub-$\sigma$-field $\mathcal A$ is $$E[X\cdot1_A]=E[Y\cdot1_A],\ \forall A\in\mathcal A\quad(*)$$
I was wondering if I could change this to $$E[XZ]=E[YZ],\ \forall\text{ nonnegative, bounded,}\ \mathcal A\text{-measurable r.v. }Z\quad(\#)$$
Now $(\#)\implies(*)$ is obvious. For the other direction. My proof is as follows:
Since $Z$ is $\mathcal A$-measurable, nonnegative and bounded, we can find an increasing sequence of simple functions $\psi_k\geq0$ such that $\psi_k\nearrow Z$ a.s. Moreover, $\psi_k$ has the form $$\psi_k=\sum_{j=1}^{n_k}a_j\cdot1_{A_{kj}},\text{ where}\ A_{kj}\in\mathcal A$$ (existence of $\psi_k$ is due to the proposition on page 274 of Real Analysis by Stein and Shakarchi). Clearly, $E[X\psi_k]=E[Y\psi_k]$ by $(*)$. Let $k\to\infty$ and by dominated convergence we have $E[XZ]=E[YZ]$.
Some questions:
Is my proof correct?
Can I loosen the conditions on $Z$. For example, can it be unbounded but only $L^1$ (maybe a new proof is needed, if the conclusion is true)?
Your proof is correct. The conclusion holds whenever $Z$ is $\mathcal A$ measurable, $E|XZ| <\infty$ and $E|YZ| <\infty$. For this you only have to choose your simple functions $\psi_k$ converging to $Z$ such that $|\psi_k| \leq |Z|$. (DCT is still applicable).