Conditional expectation given a sigma algebra Let $Z$ be a random variable defined on $(\Omega,\mathcal{F},p)$, $E(Z)=1$ and $Z\geq0$ $a.s$. 

Let's take the probability measure $q$ induced by $Z$ as $q(A)=E(I_{A}Z)$ for any $A \in \mathcal{F}$ now define $U=E(Z|\mathcal{G})$ where $\mathcal{G}$ is a sub $\sigma$-algebra of $\mathcal{F}$

Prove that $E_q(X|\mathcal{G})=\frac{E(XZ|\mathcal{G})}{U}$ for any bounded $X$.

$\mathbf{Solution}$  

By definition $E_q(X|\mathcal{G})$ is such that $E_q(E_q(X|\mathcal{G})S)=E_{q}(XS)$ for any $S$ bounded and $\mathcal{G}$ measurable.
This is the same as
$E(E_q(X|\mathcal{G})SZ)=E(XSZ)$.

Now by taking $XZ$ as a random variable one has $E(XSZ)=E(E(XZ|\mathcal{G})S)$

$E(E_q(X|\mathcal{G})SZ)=E(E(XZ|\mathcal{G})S)$

But now I am missing the part of $U$, anyone knows what to do from here on?
Let us first show that $Y := E_q(X|\mathcal G)$ is a.s. $p$-bounded. Let $|X|\le C$. Acually, $Y$ is defined via $\int YZS\,dp=\int XZS\,dp$, so $Y$ can be anything where $Z=0$. Let us put $Y = C/2$ on $\{Z=0\}$. Then
$$
C\int_{Y\ge 2C}Z\,dp\ge\int_{Y\ge 2C}XZ\,dp = \int_{Y\ge 2C}YZ\,dp\ge 2C\int_{Y\ge 2C}Z\,dp.
$$
Hence, $\int_{Y\ge 2C}Z\,dp= 0$, which implies $Z=0$ on $\{Y\ge 2C\}$. By our definition of $Y$ on $\{Z=0\}$, this is impossible, so that $p(Y\ge 2C)=0$ follows. Similarly, $p(Y\le -2C) = 0$.

Now, for all bounded $S$ we have that $E_q(YS) = E_q(XS)$, that is, $E((YZ)S) = E((XZ)S)$. The  latter means that $E(XZ|\mathcal G) = ZY$. Hence, since $Y$ is bounded,
$$
E(E(XZ|\mathcal G)S) = E(Z(YS)) = E(E(Z|\mathcal G)YS).
$$
This implies
$$
E(XZ|\mathcal G) = E(E(XZ|\mathcal G)|\mathcal G) = E(Z|\mathcal G)Y,
$$
which we had to prove.
Let us first show that $Y := E_q(X|\mathcal G)$ is a.s. $p$-bounded. Let $|X|\le C$. Acually, $Y$ is defined via $\int YZS\,dp=\int XZS\,dp$, so $Y$ can be anything where $Z=0$. Let us put $Y = C/2$ on $\{Z=0\}$. Then
$$
C\int_{Y\ge 2C}Z\,dp\ge\int_{Y\ge 2C}XZ\,dp = \int_{Y\ge 2C}YZ\,dp\ge 2C\int_{Y\ge 2C}Z\,dp.
$$
Hence, $\int_{Y\ge 2C}Z\,dp= 0$, which implies $Z=0$ on $\{Y\ge 2C\}$. By our definition of $Y$ on $\{Z=0\}$, this is impossible, so that $p(Y\ge 2C)=0$ follows. Similarly, $p(Y\le -2C) = 0$.

Now, for all bounded $S$ we have that $E_q(YS) = E_q(XS)$, that is, $E((YZ)S) = E((XZ)S)$. The  latter means that $E(XZ|\mathcal G) = ZY$. Hence, since $Y$ is bounded,
$$
E(E(XZ|\mathcal G)S) = E(Z(YS)) = E(E(Z|\mathcal G)YS).
$$
This implies
$$
E(XZ|\mathcal G) = E(E(XZ|\mathcal G)|\mathcal G) = E(Z|\mathcal G)Y,
$$
which we had to prove.