The statement:
Take two probability measures $\mathbb{P}$ and $\mathbb{Q}$ on $(\Omega, \mathcal{F})$, such that $\mathbb{Q}\ll\mathbb{P}$ with $$d\mathbb{Q}=\Lambda d \mathbb{P}.$$ Let $\mathcal{G}$ be a sub-$\sigma$-algebra on $\mathcal{F}$. Then for any $\mathcal{F}$-random variable $X$ we have $$\mathbb{E}^{\mathbb{Q}}\left[ X | \mathcal{G}\right]=\frac{\mathbb{E} \left[ X \Lambda | \mathcal{G}\right]}{\mathbb{E} \left[ \Lambda | \mathcal{G}\right] }.$$
Part of the proof I understand:
Assume $X \geq 0$. We note that by the definition of conditional expectation, for all $G \in \mathcal{G}$: $$\int_G \mathbb{E} \left[ X \Lambda | \mathcal{G}\right]d \mathbb{P} =\int_G X \Lambda d\mathbb{P} =\int_G X d\mathbb{Q} =\int_G \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] d \mathbb{Q}=\int_G \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \Lambda d \mathbb{P}.$$ Now we use the definition of conditional expectation to take another conditional expectation with respect to $\mathcal{G}$. Since $G \in \mathcal{G}$: $$\int_G \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \Lambda d \mathbb{P}= \int_G \mathbb{E} \left[ \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \Lambda | \mathcal{G} \right] d \mathbb{P}.$$ But $ \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right]$ is $\mathcal{G}$-measurable, and so $$\int_G \mathbb{E} \left[ \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \Lambda | \mathcal{G} \right] d \mathbb{P}=\int_G \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \mathbb{E} \left[ \Lambda | \mathcal{G} \right] d \mathbb{P}.$$ Thus, since in particular $\Omega \in \mathcal{G}$, we get $$\int_{\Omega} \mathbb{E} \left[ X \Lambda | \mathcal{G}\right]d \mathbb{P} = \int_{\Omega} \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \mathbb{E} \left[ \Lambda | \mathcal{G} \right] d \mathbb{P}.$$
??? Part of the proof:
Since $X \geq 0$ (and $\Lambda > 0$) this means that $\mathbb{P}$-a.s. and hence $\mathbb{Q}$-a.s. we have $$\mathbb{E}^{\mathbb{Q}}\left[ X | \mathcal{G}\right]=\frac{\mathbb{E} \left[ X \Lambda | \mathcal{G}\right]}{\mathbb{E} \left[ \Lambda | \mathcal{G}\right] }.$$
Could somebody give me a hint about what happened in that last part of the proof. How can we just omit those expectations? I was thinking it's maybe because the random variable $\mathbb{E} \left[ X \Lambda | \mathcal{G}\right] - \mathbb{E}^{\mathbb{Q}} \left[X | \mathcal{G} \right] \mathbb{E} \left[ \Lambda | \mathcal{G} \right]$ is non-negative, then we could omit expectations if I'm correct, but I don't know how to prove that it is.
Thanks for any help on this. I appreciate it.