I am fairly new to filtrations, discrete stochastic processes etc. I don't understand the last jump this proof makes.
Suppose we have a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and $\mathcal{G}\subset \mathcal{F}$. Let $D>0$ be an $\mathcal{F}$-measurable random variable with $\mathbb{E}^{\mathbb{P}}[D]=1$ and define $\mathbb{Q}$ via $\frac{d\mathbb{Q}}{d\mathbb{P}}=D.$ Let $\xi$ be an $\mathcal{F}$-measurable random variable. Then $$\mathbb{E}^{\mathbb{Q}}[\xi|\mathcal{G}]=\frac{1}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}].$$ $\textit{Proof.}$ By definition $\mathbb{E}^{\mathbb{Q}}[\xi |\mathcal{G}]$ is a $\mathcal{G}$-measurable r.v such that $$\mathbb{E}^{\mathbb{Q}}\left [\mathbb{E}^{\mathbb{Q}}[\xi |\mathcal{G}]\textbf{1}_{A}\right ]=\mathbb{E}^{\mathbb{Q}}[\xi \textbf{1}_A],\;\; \forall A\in \mathcal{G}.$$ We have (by iterative conditioning on expectation) $$\mathbb{E}^{\mathbb{Q}}\left [ \frac{1}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]\textbf{1}_A\right ]=\mathbb{E}^{\mathbb{P}}\left [ \frac{D}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]\textbf{1}_A\right ]$$ $$=\mathbb{E}^{\mathbb{P}}\left [\left .\mathbb{E}^{\mathbb{P}}\left [ \frac{D}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]\textbf{1}_A\right |\mathcal{G}\right ] \right ]=\mathbb{E}^{\mathbb{P}}\left [ \frac{\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]\textbf{1}_A}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]\right ]$$ $$=\mathbb{E}^{\mathbb{P}}[\xi D\textbf{1}_A]=\mathbb{E}^{\mathbb{Q}}[\xi\textbf{1}_A]$$
Now I have a few issues with this proof. I'm not sure why the indicator function is necessary for the expectations in the proof and also why it would disappear at the end when we equate the two equations. Also, generally it does not seem clear to me why if two expectations agree then their arguments do, i.e., why $$\mathbb{E}^{\mathbb{Q}}\left [\mathbb{E}^{\mathbb{Q}}[\xi |\mathcal{G}]\textbf{1}_{A}\right ]=\mathbb{E}^{\mathbb{Q}}\left [ \frac{1}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]\textbf{1}_A\right ]$$ $$\Rightarrow \mathbb{E}^{\mathbb{Q}}[\xi |\mathcal{G}]= \frac{1}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]$$
If anyone could enlighten me I would greatly appreciate it.
The key point here is $\forall A\in\mathcal{G}$.
I recall the definition of the conditional expectation. $\mathbb{E}\left[ X\mid \mathcal{G}\right]$ is the random variable which is $\mathcal{G}$ measurable and that satisfies the property: $$\forall A\in\mathcal{G},\, \mathbb{E}\left[\mathbb{E}\left[X\mid\mathcal{G}\right]\mathbf{1}_A\right]=\mathbb{E}\left[X\mathbb{1}_A\right]$$
Thus, if we denote $$Z=\frac{1}{\mathbb{E}^{\mathbb{P}}[D|\mathcal{G}]}\mathbb{E}^{\mathbb{P}}[\xi D|\mathcal{G}]$$ if we want to prove that $\mathbb{E}^{\mathbb{Q}}[\xi|\mathcal{G}]=Z$, all we have to do is to prove that for any $A\in\mathcal{G}$, we have $$\mathbb{E}^{\mathbb{Q}}\left[\xi\mathbf{1}_A\right]=\mathbb{E}^{\mathbb{Q}}\left[Z\mathbf{1}_A\right]$$ That is what is done in the proof.