How to use from Bayes' Conditional Expectation to show $E_{P}[\frac{dP}{dQ}\vert \mathcal{G}]>0$ $Q-$a.s. Let $ Q \ll P$ be two measure on a measurable space $(\Omega, \mathcal{F})$. How to use Bayes' Conditional Expectation to show $E_{P}[\frac{dQ}{dP}\vert \mathcal{G}]>0$ $Q-$a.s. 
By the conditional Bayes Formula, we know that for any sub-$\sigma$-algebra $\mathcal{G}\subseteq \mathcal{F}:$
$E_{Q}[X\vert \mathcal{G}]E_{P}[\frac{dQ}{dP}\vert \mathcal{G}]=E_{P}[X\frac{dQ}{dP}\vert \mathcal{G}]$ $P-$a.s.
My Idea thus far: 
Assume there exists $A\in \mathcal G$ where $Q(A)>0$ so that $E_{P}[\frac{dQ}{dP}\vert \mathcal{G}]=0$ on $A$. Then $P(A)>0$ and by Bayes' Formula: 
$E_{P}[X\frac{dQ}{dP}\vert \mathcal{G}]=0$ on $A\Rightarrow \int_{A}X\frac{dQ}{dP}dP=0$ but $0=\int_{A}X\frac{dQ}{dP}dP=E_{Q}[X1_{A}]$
Where do I go from here?
 A: Note that $E_{P}[\frac{dQ}{dP}\vert \mathcal{G}]$ is non-negative Q-a.s and itself can be seen as the Radon–Nikodym derivative of a new measure $\tilde{Q}$ defined by $$\frac{d\tilde{Q}}{dP} :=E_{P}\left[\frac{dQ}{dP}\bigg\vert \mathcal{G}\right]$$
because of $$\frac{d\tilde{Q}}{dP} \ge 0\quad P-a.s$$ and $$E_P\left[\frac{d\tilde{Q}}{dP}\right] = 1$$
Then for each $A\in \mathcal{G}$ it holds $Q(A) = \tilde{Q}(A)$ what follows directly from the definition and the tower property of the conditional expectation:
$$\begin{align*} Q(A) = E_Q[1_A] &= E_P\left[\frac{dQ}{dP}1_A\right] \\ &= E_P\left[1_A E_P\left[\frac{dQ}{dP} \bigg| \mathcal{G}\right]\right] \\ &= E_P\left[1_A\frac{d\tilde{Q}}{dP}\right] \\ &= E_\tilde{Q}[1_A]= \tilde{Q}(A)\end{align*}$$
Now it holds $$A := \left\{E_{P}\left[\frac{dQ}{dP}\bigg\vert \mathcal{G}\right] = 0\right\} =  \left\{\frac{d\tilde{Q}}{dP} = 0\right\} \in \mathcal{G}$$ and so it follows: $$\begin{align*}Q(A) = \tilde{Q}(A) &= E_\tilde{Q}\left[1_A\right] \\ &=E_P\left[\frac{d\tilde{Q}}{dP}1_A\right] \\ &= E_P[0] = 0\end{align*}$$ hence $$E_{P}\left[\frac{dQ}{dP}\bigg\vert \mathcal{G}\right] > 0 \quad Q-a.s.$$
