proof of : $ E^Q \left[ Y|F(s) \right] = \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$ To prove this Lemma : 
$$ E^Q \left[ Y|F(s) \right] = \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$
with $P$ and $Q$ two equivalent probability measures and $Z(t)$ is the expectation of the Radon Nikodym derivative $Z(t) = E^P \left[\frac{dQ}{dP}|F(t) \right]$, and $Y$ is $F(t)-$ measurable.
I found that in Shreve's Stochastic Calculus for Finance II, the result is concluded only by checking the partial averaging property :
$$ \int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A E^Q \left[ Y|F(s) \right] dP$$
My question is why is it sufficient to check the partial averaging property to conclude the result? Why don't we just do without taking the expecation of both sides? Thank you
 A: You have a question to this link I gave you in your other question…


*

*why don't you ask there if you have a question?

*why don't you link to the answer you have a question to?


Nevertheless let's answer your question: 
You are wrong, in "Shreve's Stochastic Calculus for Finance II" they don't check $$\int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A E^Q \left[ Y|F(s) \right] dP$$ but they check if $$\int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A Y dQ$$ holds for $A \in F(s)$. Or if you write it with expectations instead of the integral form $$E_Q\left[1_A\frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]\right] = E_Q\left[1_AY\right]$$
And this is because this had necessarily to hold if $$\frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$ should be the conditional expectation of $Y$ w.r.t to $F(s)$ besides the $F(s)$ - mesurability... this those is obvious.
Maybe it's clearer to you if we set $$X := \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right]$$
Then $X$ is the $Q$-conditional expectation of $Y$ w.r.t. $F(s)$ iff 


*

*$X$ is $F(s)$ mb

*$E_Q[X1_A] = E_Q[Y1_A]$ for all $A \in F(s)$


And this is exactly what is checked by:
$$E_Q[X1_A] = \int_A X dQ = \int_A \frac{1}{Z(s)}E^P\left[ Y Z(t) | F(s) \right] dQ = \int_A Y dQ = E_Q[Y1_A]$$
