I am working on this exercise:
Suppose that $\mathcal{P}$ is a $\pi-$system of subsets of $\Omega$ such that $\Omega\in\mathcal{P}$ and $\mathcal{G}=\sigma(\mathcal{P})\subset\mathcal{F}$. Show that if $X\in L^{1}(\Omega,\mathcal{F},\mathbb{P})$ and $Y\in L^{1}(\Omega,\mathcal{G},\mathbb{P})$ are such that $\mathbb{E}(X;G)=\mathbb{E}(Y;G)$ for all $G\in\mathcal{P}$, then $Y=\mathbb{E}(X|\mathcal{G})$.
I've worked out most parts but I got stuck in the end.
Below is what I've got:
Firstly since $Y\in L^{1}(\Omega, \mathcal{G},\mathbb{P})$, it is automatically $\mathcal{G}-$measurable. It then remains to show that for all $A\in\mathcal{G}$, we have $$\int_{A}Xd\mathbb{P}=\int_{A}Yd\mathbb{P}.$$
Define the collection of sets $\mathcal{L}:=\{A\in\mathcal{F}:\mathbb{E}(X;A)=\mathbb{E}(Y;A)\}$, then we know that $\mathcal{P}\subset\mathcal{L}$ since for all $G\in\mathcal{P}$ by hypothesis we have $\mathbb{E}(X;G)=\mathbb{E}(Y;G)$ so $G\in\mathcal{L}$.
It then suffices to show that $\mathcal{L}$ is a $\lambda-$system, since if $\mathcal{L}$ was a $\lambda-$system, $\pi-\lambda$ theorem and the above result of $\mathcal{P}\subset\mathcal{L}$ would together imply that $\sigma(\mathcal{P})\subset\mathcal{L}$, which would imply that for all $A\in\sigma(\mathcal{P})=\mathcal{G}$, we would have $\mathbb{E}(X;A)=\mathbb{E}(Y;A)$ as desired.
Firstly $\Omega\in\mathcal{P}\subset\mathcal{L}$ is immediate. Now let $A, B\in \mathcal{L}$ and $A\subset B$, then by the definition of $\mathcal{L}$, we know that $$\int_{A}Xd\mathbb{P}=\int_{A}Yd\mathbb{P}\ \text{and}\ \int_{B}Xd\mathbb{P}=\int_{B}Yd\mathbb{P},$$ but $A\subset B$, so we can use the difference of the integral that $$\int_{B\setminus A}Xd\mathbb{P}=\int_{B}Xd\mathbb{P}-\int_{A}Xd\mathbb{P}=\int_{B}Yd\mathbb{P}-\int_{B}Yd\mathbb{P}=\int_{B\setminus A}Yd\mathbb{P},$$ which implies that $B\setminus A\in\mathcal{L}$.
Then, I tried to show $\mathcal{L}$ is cloud under the limit of monotone sequence, i.e. if $A_{i}\in\mathcal{L}$ and $A_{i}\nearrow A$, then $A\in\mathcal{L}$. I was able to prove when $X,Y\geq 0$, but was not able to extend to a general case.
Below is my attempt:
Let $A_{i}\in \mathcal{L}$ be such that $A_{i}\nearrow A$. By hypothesis both $X$ and $Y$ are integrable. Suppose firstly $X, Y\geq 0$, then since $A_{i}\nearrow A$, we must have $X\mathbb{1}_{A_{i}}\nearrow X\mathbb{1}_{A}$ and $Y\mathbb{1}_{A_{i}}\nearrow Y\mathbb{1}_{A},$ it then follows from the monotone convergence theorem that $$\int_{\Omega}X\mathbb{1}_{A_{i}}d\mathbb{P}\nearrow\int_{\Omega}X\mathbb{1}_{A}d\mathbb{P}\ \text{and}\ \int_{\Omega}Y\mathbb{1}_{A_{i}}d\mathbb{P}\nearrow\int_{\Omega}Y\mathbb{1}_{A}d\mathbb{P},$$ but by definition of $\mathcal{L}$, for each $i$, we have $$\int_{\Omega}X\mathbb{1}_{A_{i}}d\mathbb{P}=\int_{A_{i}}Xd\mathbb{P}=\int_{A_{i}}Yd\mathbb{P}=\int_{\Omega}Y\mathbb{1}_{A_{i}}d\mathbb{P},$$ so it follows that $$\int_{A}Xd\mathbb{P}=\int_{\Omega}X\mathbb{1}_{A}d\mathbb{P}=\int_{\Omega}Y\mathbb{1}_{A}d\mathbb{P}=\int_{A}Yd\mathbb{P}.$$
The problem here is that when I tried to extend this result to general case, we normally set $X:=X^{+}-X^{-}$ and $Y:=Y^{+}-Y^{-}$, but then it is not necessary that $\mathbb{E}(X^{+}\mathbb{1}_{A_{i}})=\mathbb{E}(Y^{+}\mathbb{1}_{A_{i}})$ and $\mathbb{E}(X^{-}\mathbb{1}_{A_{i}})=\mathbb{E}(Y^{-}\mathbb{1}_{A_{i}})$, so we cannot go back to the first case and conclude.
It is not necessarily true because the hypothesis is only about $X$ and $Y$, so we have $$\int_{A_{i}}X^{+}-X^{-}d\mathbb{P}=\int_{A_{i}}Y^{+}-Y^{-}d\mathbb{P},$$ from which we cannot conclude the positive parts being equal and the negative parts being equal.
What should I do? Thank you!
Edit 1:
Okay I figured it out, I asked a dumb question…. I will answer my own post.
Best Answer
Okay I figured it out, I asked a dumb question. I don't really need $X\mathbb{1}_{A{i}}\nearrow X\mathbb{1}_{A}$, I only need $X\mathbb{1}_{A{i}}\longrightarrow X\mathbb{1}_{A}$, and then we apply dominated convergence theorem.
Below is the updated proof of the last part:
Let $A_{i}\in \mathcal{L}$ be such that $A_{i}\nearrow A$. By hypothesis both $X$ and $Y$ are integrable. Since $A_{i}\nearrow A$, then we have $X\mathbb{1}_{A_{i}}\longrightarrow X\mathbb{1}_{A}$, and $|X\mathbb{1}_{A_{i}}|\leq |X|$, so it follows from dominated convergence theorem that $$\int_{A_{i}}Xd\mathbb{P}\longrightarrow\int_{A}Xd\mathbb{P},$$ similarly we can conclude that for $Y$, $$\int_{A_{i}}Yd\mathbb{P}\longrightarrow\int_{A}Yd\mathbb{P},$$ but by hypothesis we have $$\int_{A_{i}}Xd\mathbb{P}=\int_{A_{i}}Yd\mathbb{P}\ \text{for all}\ i,$$ and thus it follows that $$\int_{A}Xd\mathbb{P}=\int_{A}Yd\mathbb{P},$$ which implies $A\in\mathcal{L}$.
Therefore, $\mathcal{L}$ is a $\lambda-$system and the proof is then concluded.