$ E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y 1_A )$ IF $Y \in \mathcal{L^1}  (\Omega, \mathcal{F}, \Bbb{P})$
and $\mathcal{G_1}  , \mathcal{G_2} ,\mathcal{G_3} $are $\sigma $  fields in $\mathcal{F} $
If we assume that $Y$ is $\mathcal{G_1}$ measurable and $\mathcal{G_3}    $   is independent of $\mathcal{G_1}\bigvee\mathcal{G_2}$.
How can we prove that  $ E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y  1_A )$ for every $A$ formed as $A= B \cap C$ , 
$B \in \mathcal{G_2}  $,$ C \in \mathcal{G_3}  $. 
And  then extend this to  $\mathcal{G_2}\bigvee\mathcal{G_3}$  ( by using Dynkin's $π - λ $ theorem).
I found a similar exercise but I was unable to prove this problem .
 A: $E[E(Y|G_2)1_B1_C]=E[E(Y1_B|G_2)]E[1_C]=E[Y1_B]E[1_C]=E[Y1_A]$
I am unsure from the wording of your question whether you want help with the pi lambda extension. 
In short, the set of A for which the equality holds is easily checked to be a Dynkin system D, sets which are an intersection of elements of $G_2$ and $G_3$ are clearly a pi system P. Hence, the equation is satisfied on $\sigma(P)\subset D$
A: By definition of $A$, $$
E [E (Y\mid\mathcal{G_2}) 1_A ] =  E [E (Y\mid\mathcal{G_2}) 1_B1_C ].$$
Since $E (Y\mid\mathcal{G_2})$ is $\mathcal G_2$-measurable, so is $E (Y\mid\mathcal{G_2}) 1_B$ hence $1_C$ is independent of $Y\mid\mathcal{G_2}) 1_B$ and we get 
$$
E [E (Y\mid\mathcal{G_2}) 1_A ] =  E [E (Y\mid\mathcal{G_2}) 1_B]\mathbb P(C).$$
Using the definition of conditional expectation gives 
$$
E [E (Y\mid\mathcal{G_2}) 1_A ] =  E [Y 1_B]\mathbb P(C).
$$
Conclude using the fact that $1_C$ is independent of $Y1_B$, a $\mathcal G_1\vee\mathcal G_2$-measurable random variable.
To conclude, let $\mathcal G$ be the collection of $\mathcal F$-measurables sets $A$ such that 
$$E [E (Y\mid\mathcal{G_2}) 1_A ] = E (Y  1_A ) a.s..$$
One can show that $\mathcal G$ is a $\lambda$-system containing the sets of the form $G_2\cap G_3$, $G_2\in\mathcal G_2$, $G_3\in\mathcal G_3$.
