It is hard for me to see that for any B and for any $A \in \mathscr G$ $$E[I_{B}E[I_AY|\mathscr G ]] = E[I_{B}I_AE[Y|\mathscr G ]] $$
Could someone help me to see it
Note: I would like to understand why this equality hold, I know that since $1_A$ is $\mathscr G$ measurable I could bring it outside the conditional expectation.
Let $(\Omega, \mathcal{F},P)$ be a probability space; $\mathcal{G}$ a sub-$\sigma$-algebra of $\mathcal{F}$; and $Z$ a random variable on $\Omega$ so that $Z\in L^1$. By definition, you might know that $E[Z\mid \mathcal{G}]$ is a random variable on $(\Omega,\mathcal{G})$ (also in $L^1$) satisfying $$E[1_AE[Z\mid\mathcal{G}]]=E[1_AZ]$$ since $A$ is $\mathcal{G}$-measurable. Take the random variable $Z=Y1_B$ on $\Omega$ (with finite expectation), and you are done.
Ok, see that
$$E[1_AE[1_BY\mid\mathcal{G}]]=E[1_A1_BY]=E[1_A1_BE[Y\mid \mathcal{G}]].$$