Conditional expectation and the Monotone Class Theorem

Question

I'm trapped by the following problem 1.5 from Exercises in Probability A Guided Tour from Measure Theory to Random Processes, via Conditioning

In probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $\mathcal{G}_{}\subset \mathcal{F}$ and $Y\in \mathcal{G}$, If $\mathop{{}\mathbb{E}}_{\mathcal{G}}g(X)=g(Y)$ for any bounded positive $g$, show that $X = Y \text{ a.s. }$.

Where $\mathbb{E}_{\mathcal{G}}$ is expectation conditioning in $\mathcal{G}$

The solution is extend the identity to $$ \mathop{{}\mathbb{E}}_{\mathcal{G}} G(X,Y)=G(Y,Y) $$ by monotone class theorem, then taking $G(x,y)=\mathbf{1}_{x \neq y}$. But I'm confused how to get this from MCT.

My attempt is to prove all $G$ forms a monotone class $\mathcal{H}$, then I have to prove $\mathbf{1}_{A}(x,y)$ belong to this class for any borel $A\in \mathcal{B}(\mathbb{R}^2)$. The section $\mathbf{1}_{A}(X,y)$ is bounded and positive thus belong to $\mathcal{H}$, but how to proceed to say $\mathbf{1}_{A}$ also belong to $\mathcal{H}$?

Answer 1

I think you should use the $\pi-\lambda$ theorem instead of the Monotone Class Theorem.

$ \mathop{{}\mathbb{E}}_{\mathcal{G}} G(X,Y)=G(Y,Y) $ holds for all non-negative measurable $G$. To see this first consider the case $G=I_{A\times B}$ where $A, B \in \mathcal B (\mathbb R)$. In this case this equality follows by just multiplying the given equation by $I_B(Y)$. Now the class of all $E$ for which $ \mathop{{}\mathbb{E}}_{\mathcal{G}} I_E(X,Y)=I_E(Y,Y) $ is a $\lambda$ system and it contains the $\pi$ system of sets of the form $A\times B$. Hence, the result holds for all $E$ in the product $\sigma-$ algebra. Now go to simple functions and then non-negative mesurable funcstions, as usual.

Answer 2

$$P(X\in A\cap Y\in B)=\int_B dP_Y \int I_A(x) dP_{X/Y}= \int_B I_A(y)dP_Y=P_Y(A\cap B)$$ So $P(X\neq Y)=0$

They tell me to clarify the answer, but it's all so obvious that I don't know where to start. I will focus on the points that could be darker:
First.
$\int I_A(x) dP_{X/Y}=I_A(y)$ comes from choosing $g(x)=I_A(x)$ in: $\,\mathop{{}\mathbb{E}}_{\mathcal{G}}g(X)=g(Y)$
Second:
$P(X\in A\cap Y\in B)=0$ if $A$ and $B$ are disjoint, which implies that $P(X\neq Y)=0$.