Is it feasible to prove this property of conditional expectation without too many lemmas? I'm reading a proposition about conditional expectation operator in the lecture note:



My lecture note does not provide the proof. I would like to ask it's possible to prove it from below definitions and theorems without too many intermediate results. If it is the case, I will give it a try. If the authors do not present the proof because it relies on too many auxiliary results, I think proving it by myself is almost an impossible mission.
Thank you so much for your help!







 A: I've just figured a proof of which I posted here. It would be great if someone helps me verify it. Thanks!

$\textbf{My attempt}$
Because $\mathbb E[X]$ is a constant function, it's clearly $\mathcal G$-measurable. It remains to show that $\forall \Lambda  \in \mathcal G: \mathbb E[\mathbb E[X] \mathbb{1}_\Lambda] = \mathbb E[X \mathbb{1}_\Lambda]$ or equivalently $\forall \Lambda  \in \mathcal G: \mathbb E[X] \mathbb P[\Lambda] = \mathbb E[X \mathbb{1}_\Lambda]$.
First, we need the following lemma.

$\textbf{Lemma} \quad$ For a non-negative random variable $X$, there is an increasing sequence $(X_n)$ of non-negative simple functions such that $(X_n)$ converges to $X$. In particular, $$\forall (n,i) \in \mathbb N \times \{0,\ldots,2^n\}: X_n = \sum_{i=0}^{2^n} b_i \mathbb{1}_{B_i}$$ where $b_i = in2^{-n}$ and $B_i =\{\omega \in \Omega \mid  in2^{-n} \le X(\omega) < (i+1)n2^{-n}\}$.

Assume that $X \ge 0$. By lemma, there is an increasing sequence $(X_n)$ of nonnegative simple functions that converges uniformly to $X$. Then we have, for any $\Lambda \in \mathcal G$ $$\mathbb E[X_n \mathbb{1}_\Lambda ] = \mathbb E \left [ \left (\sum_{i=0}^{2^n} b_i \mathbb{1}_{B_i} \right ) \mathbb{1}_\Lambda \right ] = \sum_{i=0}^{2^n} b_i \mathbb E \left [ \mathbb{1}_{B_i}  \mathbb{1}_\Lambda \right ]= \sum_{i=0}^{2^n} b_i \mathbb E \left [   \mathbb{1}_{B_i \cap \Lambda} \right ] = \sum_{i=0}^{2^n} b_i \mathbb P[B_i \cap \Lambda ]$$
Because $B_i = X^{-1} \big ([in2^{-n}, (i+1)n2^{-n}) \big)$, $B_i$ is $\sigma(X)$-measurable and thus $B_i \in \mathcal F_X$. Moreover, $X$ and $\mathcal{G}$ are independent. So $\mathbb P[B_i \cap \Lambda ] = \mathbb P[B_i] \mathbb P[ \Lambda ]$ and thus $\mathbb E[X_n \mathbb{1}_\Lambda ] = \mathbb P[ \Lambda ] \sum_{i=0}^{2^n} b_i \mathbb P[B_i]  = \mathbb P[ \Lambda ] \sum_{i=0}^{2^n} b_i \mathbb E[ \mathbb{1}_{B_i}]= \mathbb P[ \Lambda ] \mathbb E[ \sum_{i=0}^{2^n} b_i  \mathbb{1}_{B_i}] = \mathbb P[ \Lambda ] \mathbb E[X_n]$. To sum up, we have $$\mathbb E[X_n \mathbb{1}_\Lambda ] = \mathbb P[ \Lambda ]\mathbb E[ X_n]$$
Taking the limit on both sides and applying $\textit{monotone convergence theorem}$, we get $\mathbb E[X \mathbb{1}_\Lambda ] = \mathbb P[ \Lambda ] \mathbb E[X]$.
Come back to the general case, we write $X = X^+ -X^-$. Then the linearity of conditional expectation operator completes the proof.
