Radon-Nikodym Derivative for paths of stochastic process

Question

I am currently trying to understand this paper https://arxiv.org/pdf/1505.07612.pdf and have problems understanding lemma 3.1. We have a stochastic Process given by \begin{equation} dX_t=rh_t(X) 1_{ \{\theta \leq t\} }dt+\sigma \sqrt{h_t(X)}dW_t \end{equation} where $\theta$ is a Random Variable and $W$ a Brownian motion. Let \begin{equation} \begin{split} \mu_{u,t}(A)&=P\left(X|_{[0,t]}\in A|\theta = u \right)& \quad \mu_{t}(A)&=P\left(X|_{[0,t]}\in A\right) \end{split} \end{equation} for $A\in \mathcal{B}\left( C\left([0,t],\mathbb{R}\right)\right)$. Then the lemma states that the Radon-Nikodym Derivative of $\mu_{u,t}$ w.r.t $\mu_{t,t}$ is given by \begin{equation} \frac{d\mu_{u,t}}{d\mu_{t,t}}=\exp\left(\int_{u}^{t}\frac{r}{\sigma^2}dX_s-\frac{1}{2}\int_{u}^{t}\frac{r^2}{\sigma^2}h_s\left(X\right)ds\right) \end{equation} The lemma therefore says: \begin{equation} \mu_{u,t}(A)=\int_{A}\frac{d\mu_{u,t}}{d\mu_{t,t}}(z)d\mu_{t,t}(z) \end{equation} Right ? My Problem with this is that $\frac{d\mu_{u,t}}{d\mu_{t,t}}$ is a random variable i.e. $\frac{d\mu_{u,t}}{d\mu_{t,t}}:\Omega \rightarrow \mathbb{R}$. But $\mu_{u,t}$ and $\mu_{t,t}$ are measures on $\mathcal{B}\left( C\left([0,t],\mathbb{R}\right)\right)$. So they are acting on different spaces. How does this then make sense ? I have already read the proof, but I don't understand the last step where this happens. I would be very grateful if someone could clear up my confusion.