Limit of conditional expected value is the conditional expected value of the limit?

Question

I have the following problem on my Problem Set:

I was able to do itens A) and B), but I'm not sure if my solution of item C) is indeed correct. Here's what I've done:

MCT stands for Monotone Convergence Theorem. I have problems accpeting the second time I used the MCT. Obviously, I was applying the MCT on a sequence that was indeed monotone but I'm not sure that its pointwise limit is $\mathbb{E}[X|\mathcal{G}]$, as required by the theorem. Am I on the right path? Any ideas? Thanks a lot in advance!!

P.S: The "same trick" I talk about is that if $Z = \mathbb{E}(X|\mathcal{G})$, then $\mathbb{E}(Z1_{G}) = \mathbb{E}(X1_{G})$, for any $G \in \mathcal{G}$.

Answer 1

You should mention why $\lim\limits_{n\to\infty} E[X_n|\mathcal{G}]$ is $\mathcal{G}$-measurable but the rest is fine…

But I guess you've written it down too complicated.

It's faster by: Let $Z := \lim\limits_{n\to\infty} E[X_n|\mathcal{G}]$

i) obviously Z is $\mathcal{G}$-measurable

ii) for $G\in\mathcal{G}$ it holds $$E[Z1_G] = E[\lim\limits_{n\to\infty} E[X_n|\mathcal{G}]1_G] = E[\lim\limits_{n\to\infty} E[X_n1_G|\mathcal{G}]] = \lim\limits_{n\to\infty} E[E[X_n1_G|\mathcal{G}]] =\lim\limits_{n\to\infty}E[X_n1_G] = E[\lim\limits_{n\to\infty} X_n1_G] = E[X1_G]$$

But from i) and ii) it follows $Z = E[X|\mathcal{G}]$