easy question about conditional expectations

Question

I got a quite easy exercise I just don't get.

Let P be a probability measure, $\frac{dQ}{dP}=Z$, $Z>0$ a.s. and $E[Z]=1$, hence Q is an equivalent proba-measure to P. Then I shall prove that for a sub-sigma field G we have: $E_{Q}[X|G](\omega)=\frac{E_{P}[XZ|G](\omega)}{E_{P}[Z|G](\omega)}=:h(\omega)$, what seems easy as pie, but when using the definition, I need to show that $E[X(\omega)*\mathbb{1}_A(\omega)]=E[h(\omega)*\mathbb{1}_A(\omega)]$ for all $A\in G$ what makes me confused. these double integrals and the fraction of integrals...

Thx for you help!!

Answer 1

Let $A\in G$. We have \begin{align*} \int_A \frac{E_P[XZ\mid G]}{E_P[Z\mid G]}\, dQ &= \int_A \frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,Z \, dP\\ &= \int_A E_P\left[\frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,Z\Biggm | G \right]\, dP\\ &= \int_A \frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,E_P[Z\mid G] \, dP & \text{the fraction is $G$-measurable}\\ &= \int_A E_P[XZ\mid G]\,dP\\ &= \int_A XZ\, dP\\ &= \int_A X\, dQ. \end{align*} Hence $$ E_Q[X \mid G] = \frac{E_P[XZ\mid G]}{E_P[Z\mid G]} $$ as wished.