# Problem:

Let $$(\Omega, \mathcal{F}, P)$$ be a probability measure space. Let $$Q$$ be another (finite) measure on $$(\Omega, \mathcal{F})$$. Let $$Q \ll P$$ so by the Radon-Nikodym derivative there exists a $$Y \in \mathcal{L}^{1}(\Omega, \mathcal{F}, P)$$ such that $$\frac{dQ}{dP} = Y$$ with $$Q(A) = \int_A Y~dP$$. What is $$E(Y)$$ ?

# My attempt:

Intuitively, we should have that $$E(Y) = E(\frac{dQ}{dP}) = \int_{\Omega} \frac{dQ}{dP}~dP$$ so the $$dP$$s should "cancel out" so that we get $$Q(\Omega)$$ as the final answer. This makes sense: if $$Q$$ is finite, then $$Y \in \mathcal{L}^{1}$$.

I know from Williams 1991, chapter 14, that $$Q(F) = \int_F Y~dP, \forall F \in \mathcal{F}$$, and I think this makes it very likely that my guess is correct, but I couldn't figure out how to use this to prove it. Any help would be appreciated.

Just put $$F=\Omega$$; you can do this since $$\Omega$$ is an element of the sigma-algebra by definition. Then you have $$Q(\Omega)=\int_{\Omega} YdP=E(Y)$$.