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 $\mathcal{H}$?
Best Answer
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.