conditional expectation conjugate exponents If I know that $X\in L^p(\Omega,F,P)$ and $Y\in L^q(\Omega,G,P), \ G\subset F$, $\frac{1}{p}+\frac{1}{q}=1$, $F$ is $\sigma$ algebra on probability space $\Omega$, $G$ is sub $\sigma$ algebra.
How can I show that $E(XY  \mid G)=Y \cdot E(X \mid G)$?
So far I can do it for $p=1,q=\infty$  but I don't know how proceed from here
thanks 

Hints:


*

*Show that it follows from Hölder's inequality that $X \cdot Y \in L^1$.

*Consider $$X_n(\omega) := (-n) \vee X(\omega) \wedge n := \begin{cases} X(\omega), & \text{if} \, |X(\omega)| \leq n, \\ n, & \text{if} \, |X(\omega)|>n \end{cases}$$ and $Y_n := (-n) \vee Y \wedge n$. Then $X_n \in L^1$ and $Y_n \in L^{\infty}$ for all $n \in \mathbb{N}$. Using the result for the case $p=1$ and $q=\infty$, show that $$\mathbb{E}(X_n Y_n \mid G) = Y_n \mathbb{E}(X_n \mid G). \tag{1}$$

*Let $n \to \infty$ in $(1)$. To this end, use the dominated convergence theorem and the fact that $|X_n Y_n| \leq |X Y|$. 

