Let $X$ be a random variale such that $E(X \mid G)$ has the same distribution as $X$, $E|X|<\infty$. Prove that $$\text{sgn}(X) = \text{sgn}(E(X \mid G)).$$
The solution is given here where I don't understand a step:
Let $Y=E(X \mid G)$. Then for any Borel measurable $A\subseteq R$,
$$E(X\textbf{1}_{\{X\in A\}}) = E(Y\textbf{1}_{\{Y\in A\}}) = E(X\textbf{1}_{\{Y\in A\}})$$
I understand that the first equality is true because $X$ and $Y$ have the same distribution but for the second equality, the reason given is "since $Y\in G\implies \{Y\in A\}\in G$" but I don't understand how that implies the second equality. Thanks and appreciate a hint.
Since $Y= \mathbb{E}(X \mid G)$ we have
$$\int_B Y \, d\mathbb{P} = \int_B X \, d\mathbb{P}$$
for all $B \in G$. As $Y$ is $G$-measurable, it holds that $\{Y \in A\} \in G$ for any Borel set $A$, and therefore we get
$$\int_{\{Y \in A\}} Y \, d\mathbb{P} = \int_{\{Y \in A\}} X \, d\mathbb{P},$$
i.e.
$$\mathbb{E}(Y 1_{\{Y \in A\}}) = \mathbb{E}(X 1_{\{Y \in A\}}).$$