Conditional expectation with Radon Nikodym derivative is in $L^1(\Omega, \mathbb F, \mathbb Q)$ Let $\mathbb {P, Q}$ be equivalent probability measures on $(\Omega, \mathbb F)$,  $X \in L^1(\Omega, \mathbb F, \mathbb Q)$, ${d\mathbb Q}\over {d \mathbb P}$ be the Radon Nikodym derivative and $ \mathbb G \subset \mathbb F$ a subsigma-algebra.
I know that ${{d\mathbb Q}\over {d \mathbb P}} \in L^1(\Omega, \mathbb F, \mathbb P)$ by de Radon Nikodym theorem, hence $\mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]\in L^1(\Omega, \mathbb F, \mathbb P)$ and is positive with probability $1$
I was able to prove that given the hypothesis, the product $X{{d\mathbb Q}\over {d \mathbb P}}\in L^1(\Omega, \mathbb F, \mathbb P)$ and hence 
 $\mathbb E_{\mathbb P}[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]\in L^1(\Omega, \mathbb F, \mathbb P)$
My question: given that $X \in L^1(\Omega, \mathbb F, \mathbb Q)$ how can I show that 
$${1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G] \in L^1(\Omega, \mathbb F, \mathbb Q)$$
and 
$$ \mathbb E_{\mathbb Q}[X| \mathbb G]={1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]$$
My issue is regarding the first question since the last one is just working out with the properties of conditional expectation. I would really appreciate any hints or suggestions. 
 A: So, you want to show that 
$$\mathbb{E}_\mathbb{Q} \left|{1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}\left[X{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G\right]\right| < \infty.$$
By the conditional triangle inequality, it is enough to show
$$\mathbb{E}_\mathbb{Q} \left[{1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}\left[|X|{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G\right]\right] < \infty.$$
But by definition of $\frac{d\mathbb{Q}}{d\mathbb{P}}$, the latter is the same as showing
$$\mathbb{E}_\mathbb{P} \underbrace{\left[{1 \over \mathbb E_{\mathbb P}[{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G]}\mathbb E_{\mathbb P}\left[|X|{{d\mathbb Q}\over {d \mathbb P}}| \mathbb G\right] \frac{d\mathbb{Q}}{d\mathbb{P}}\right]}_{\text{call this $Y$}} < \infty.$$
Now note that $E_{\mathbb{P}}[Y] = E_\mathbb{P}[E_{\mathbb{P}}[Y | \mathbb{G}]]$ and use some properties of conditional expectation.
