The theorem is as follows: If $Q<<P$, then there exists a random variable $\Lambda$, such that $\Lambda\geq 0$, $E_P(\Lambda)=1$, and:
$$Q(A)=E_P(\Lambda I(A))=\int_A\Lambda dP.$$
According to my book from that line it follows that $E_Q(X)=E_P(\Lambda X)$. Can someone explain this implication in an elementary way (I'm quite new to Lebesgue integrals and such)?
From my understanding we have that: $$E_Q(X)=\int_{-\infty}^{\infty}X(\omega)dQ(\omega),$$ and $$E_P(\Lambda X)=\int_{-\infty}^{\infty}\Lambda(\omega)X(\omega)dP(\omega)$$ But how can i connect those expressions? (This is btw why they also call $\Lambda=\frac{dQ}{dP}$, and looking at these two expression it makes sense...So I want to know the connection)