I am trying to understand the proofs of the properties of conditional expectation.
I first start with the definition of conditional expectation:
let $X$ be an integrable r.v. on the probability space $(\Omega,\mathcal F,\mathbb P)$ and $\mathcal G\subset \mathcal F$ a sigma-algebra.
Then a r.v. $Y=\mathbb E(X|\mathcal G)$, $\mathcal G$-measurable function for which holds $\mathbb E(XI_A)=\mathbb E(YI_A)$ for each $A\in \mathcal G$ is called conditional expectation of X given $\mathcal G$.
Now, if I want to prove the "pull out what is known" I have to prove: $\mathbb E(XY|\mathcal G)=Y\mathbb E(X|\mathcal G)$ if Y is $\mathcal G$-measurable
How to prove this? Do I have to show that $\mathbb E(XY|\mathcal G)$ is $\mathcal G$-measurable and that $\mathbb E(XYI_A)=\mathbb E(\mathbb E(XY|\mathcal G)I_A) $?
I have no clue, I see everywhere on the proofs I find that "clearly $ Y\mathbb E(X|\mathcal G)$ is $\mathcal G$-measurable", why?
If I use the definition of conditional expectation I may say that $\mathbb E(X|\mathcal G)$ is $\mathcal G$-measurable, and that Y is same by the assumptions, but I don't know what happens to their product.
The second point I don't understand is that we can prove the equality if we show $\mathbb E(Y\mathbb E(X|\mathcal G)I_A)=\mathbb E(XYI_A)$. The steps after this I can understand but I don't know why I have to show this.
To me these two quantities can be re-written as $\mathbb E (\mathbb E(XY|\mathcal G)I_A)=\mathbb E(YXI_A)$
I tried to ask the prof and he said that we need to show that the property is a conditional expectation by checking the measurability and then the equation on top for each A, but for me it's very confusing.
You'll want to do it in four parts: prove it for constant functions, simple functions, positive functions, then all functions. I'm pretty sure you'll also need the dominated convergence theorem. Many examples of this 4-part style proof can surely be found in whatever textbook you're using (if you're not using one, let me know and I'll link you one).
This should've been a comment but I didn't have enough rep to make one.. sorry about that. Hopefully it helps.