I've been thinking about an exercise of conditional expectations and I think I have almost solve it, but I would be more comfortable if someone could confirm to me that my reasoning is correct. The problem is the following:
Let $X$ be a random vector taking values in $(\mathbb{R},\cal{B})$ and $\varphi(x)$ a random variable on $(\mathbb{R},\cal{B})$. Let $Y$ a random variable on $(\Omega,\sigma)$ and $E|Y|, E|\varphi(X)|,E|Y\varphi(X)|<\infty$. I want to prove that $$E(\varphi(X)Y|X)=\varphi(X)E(X|Y)$$ This is my attemp: First, we must see that $\varphi(X)E(Y|X)$ is measurable in $\sigma(X)$, but it is obviously given the definition of $E(Y|X)$ and being $\varphi$ measurable. So the problem will be solved if we prove that
$$\int_D\varphi(X)E(Y|X)dP=\int_D\varphi(X)YdP \hspace{0.4cm} \forall D\in \sigma(X)$$
Considering the properties of the integral and the construction of random variables from indicators, it will (almost) proven if we can see the last equality for $\varphi=I_B$. So it's enough to see that
$$\int_DI_{X\in B}E(Y|X)dP=\int_DI_{X\in B}YdP \hspace{0.4cm} \forall D\in \sigma(X) \hspace{0.4cm} \forall B\in \cal{B}$$
But the sets $\{X \in B \}$ are the same as the ones in $ \sigma(X) $, which is closed under intersections. (I just fix in the case $X \in B$ because in the opposite the indicator function is 0). By definition of $E(Y|X)$,
$$\int_{D'}E(Y|X)dP=\int_{D'}YdP \hspace{0.4cm} \forall D'\in \sigma(X)$$
So the previous equality is true, just taking $D'=D \cap \{X\in B\}$.
You proved the result when $\varphi$ is the characteristic function of a Borel subset. When $\varphi$ is bounded, the random variable $\varphi(X)$ can be uniformly approximated by linear combinations of functions of the form $\sum_{i=1}^nc_i\chi(X\in B_i)$, hence the result is true when $\varphi$ is bounded.
When $\varphi$ is non-negative but not necessarily bounded, define $\varphi_n:=\varphi\cdot\chi\{\varphi \leqslant n\}$ and use the previous case to get the wanted result.
In general, decompose $\varphi$ into positive and negative part.