Question
This question is now concerning stochastic processes. Let $(X_t)_{t\geq0}$ be defined on the probability space $(\Omega,\mathcal{F},P)$ with $\mathcal{F}_t=\sigma(X_s:s\leq t)$. Assume that for the restircted measures it holds $P_{|\mathcal{F}_t}\ll Q_{|\mathcal{F}_t}$.
Then $$f(\mathcal{F}_t)=\frac{dP_{|\mathcal{F_t}}}{dQ_{|\mathcal{F}_t}} \tag1$$ is $\mathcal{F}_t$ measurable by radon-nikodym and thus is a.s. a function of $(X_s)_{s\leq t}$, say $g((X_s)_{s\leq t})$.
Can we conclude that the following holds, considering the push forward measures $P^{(/X_s)(\omega))_{s\leq t}}$?
$$f(\mathcal{F}_t)(\omega)=\frac{dP_{|\mathcal{F_t}}}{dQ_{|\mathcal{F}_t}}(\omega)=\frac{dP^{(X_s(\omega))_{s\leq t}}}{dQ^{(X_s(\omega))_{s\leq t}}}=g((X_s(\omega)_{s\leq t}) \tag2$$
What can we say about this situation? What does this mean?
Are the pushforward measures induced by the sample path $(X_s)_{s\leq t}$? And can we think of the Radon-Nikodym-Density of $P\ll Q$ on $\mathcal{F}_t$ as the Likelihood-function of the joint-density of $(X_s)_{s\leq t}$?
Or do we say that (1) can be expressed almost surely as a function of the sample path $(X_s)_{s\leq t}$?
What I know from more elementary probability theory
Let $X$ be a random variable mapping to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $\mathcal{F}^{X}=X^{-1}(\mathcal{B}(\mathbb{R}))$. Assume there is a measure $Q$, such that $P_{|\mathcal{F}^{X}}\ll Q_{|\mathcal{F}^{X}}$. We define the Radon-Nikodym density as
$$f(\mathcal{F}^{X})=\frac{dP_{|\mathcal{F}^{X}}}{dQ_{|\mathcal{F}^{X}}}$$
If this is case for the pushforward measures it holds $P^{X}\ll Q^{X}$, where $Q^{X}(B)=Q(A)$ and $A:=X^{-1}(B)\in \mathcal{F}^{X}$. We can define a Radon-Nikodym density as
$$g(X)=\frac{dP^{X}}{dQ^{X}}$$
Now the transformation theorem states, that
$$f(\mathcal{F}^{X})(\omega)=g(X(\omega))\quad Q-a.s.$$