I am trying to prove this statement using the monotone class theorem for functions :
$\mathbb{E}Y1_A = \mathbb{E}X1_A, \forall A \in G \iff \mathbb{E}YZ = \mathbb{E}XZ, \forall Z \in L^\infty(G)$
I think I managed to prove $\Rightarrow$ :
Since $L^\infty(G)$ is a vector space of functions from $\Omega \rightarrow \mathbb{R}$ containing the constant functions and if $f_n \subset L^\infty(G)$ is a non-decreasing bounded function series from $L^\infty(G)$, then $lim_nf_n \in L^\infty(G)$ and let $C \subset L^\infty(G)$ be the set of all $1_A$ such that $A\in G$, then $C$ is stable under multiplication and we can apply the monontone class theorem :
$L^0_b(\sigma(C)) \subset L^\infty(G)$ and hence for $Z:= 1_A$ for a certain $A \in G$, we have that since is $\sigma(C)-$mesureable, then $Z$ is $L^\infty(G)-$mesureable and we have that $\mathbb{E}Y1_A = \mathbb{E}Y1_A, \forall A \in G \Rightarrow \mathbb{E}YZ = \mathbb{E}YZ, \forall Z \in L^\infty(G)$
But for the other statement to prove, I am stuck. How can I get from $L^\infty(G)$ functions to indicator functions ?
Any idea ?