# Radon-Nikodim Derivative at time 0

I have a very basic question about filtrations and Radon-Nikodym derivatives. I am reading the Andersen-Piterbarg, more in particular Eq. (1.12). They define the process $$\zeta(t) = E^P_t[\frac{dQ}{dP}]$$, where $$Q\sim P$$ are equivalent measures. Now, their claim is that obviously $$\zeta(0) = 1$$. Now, I see that the whole sample space $$\Omega$$ belongs to $$\mathcal{F}_0$$, which thus implies $$\zeta(0) = 1$$ (using the definition of expected value). But why it doesn't hold for every $$t$$? I mean, aren't the $$\mathcal{F}_t$$ also sigma-algebras, and thus contain $$\Omega$$, which would imply, by the definition of conditional expectation, $$\int_{\Omega} \zeta(t, \omega) dP(\omega) = \int_{\Omega} E^P_t[\frac{dQ}{dP}](\omega)dP(\omega) \stackrel{def \:\&\: \Omega \in \mathcal{F}_t}{=} \int_{\Omega} \frac{dQ}{dP}(\omega)dP(\omega) = \int_{\Omega} dQ(\omega) = 1.$$ What am I doing wrong here? Does $$\Omega$$ not belong to $$\mathcal{F}_t$$? Thanks in advance!

At time $$t=0$$, you get \begin{align*} \zeta(0)=E^P_0\left[\frac{\mathrm{d}Q}{\mathrm{d}P}\right]=E^P\left[\frac{\mathrm{d}Q}{\mathrm{d}P}\right]=\int_\Omega \frac{\mathrm{d}Q}{\mathrm{d}P}\mathrm{d}P=\int_\Omega \mathrm{d}Q = Q(\Omega)=1, \end{align*} because $$Q$$ is a probability measure.
But at a general time point $$t$$, you cannot write $$E_t^P[X]=\int_\Omega X \mathrm{d}P$$. That integral is the definition of the unconditional expectation! In fact, it only works at time $$t=0$$ if you assume that the filtration begins with the trivial $$\sigma$$-algebra $$\{\emptyset,\Omega\}$$.