In Villani's book he presents the following notions of dynamical couplings:
Let $(X,d)$ be a Polish space. A dynamical transference plan $\Pi$ is a probability measure on the space $C([0,1]; X)$. A dynamical coupling of two probabilities measures $\mu_0$, $\mu_1 \in P(X)$ is a random curve $\gamma : [0,1]\to X$ such that $law (\gamma_0) = \mu_0$, $law(\gamma_1) = \mu_1$.
If I'm not mistaken, I understood the "random curve" part as being equivalent to the following sentence:
There exist a measure space $(\Omega, \mathbb{P})$ measurable map $\gamma : [0,1]\times \Omega \to X$ such that $\gamma$ is continuous in $[0,1]$ and $(e_i\circ\gamma)_*(\mathbb{P}) = \mu_i$, $i = 0,1$, where $e_t$ is the evaluation map.
We have the usual notion o coupling of measures as:
Couplings by means of random elements: Let $(E_i, \mu_i)_{i\in I}$ be a family of measures spaces, where $I$ is an index set. A coupling of the measures $\mu_i$ is a family of random elements $X_i : (\Omega, \mathbb{P}) \to E_i$ such that $(X_i)_*(\mathbb{P}) = \mu_i$.
Couplings by means of the product space: Let $(E_i, \mu_i)_{i\in I}$ as before, with $\mathcal{A}_i$ being the $\sigma$-algebra of $E_i$. A coupling of the measures is a measure $\pi$ in the product space $(\prod_{i\in I}E_i, \bigotimes_{i\in I} \mathcal{A}_i)$, such that $(proj_i)_*(\pi) = \mu_i$, where $proj_i$ is the projection in $E_i$, and $\bigotimes_{i\in I} \mathcal{A}_i$ is the analogous to the product topology, but for $\sigma$-algebras.
It can be shown that these two definition are equivalent by taking the map $\Gamma : (\Omega,\mathbb{P}) \to (\prod_{i\in I}E_i, \bigotimes_{i\in I} \mathcal{A}_i)$, $\Gamma(\omega) = (X_i(\omega))_{i \in I}$ and defining $\pi : = \gamma_*(\mathbb{P})$. (The other way is trivial by construction)
I want to prove something similar for the case of dynamical couplings. We can follow the same construction above and take the interval $[0,1]$ as the index set $I$. This way we obtain the map $\Gamma(\omega) : = e_\omega(\gamma)$, where $e_\omega$ is the evaluation on the second coordinate. It's clear that the image of $\Gamma$ lies in $C(I,X)$. The problem is to show that $\Gamma$ is measurable with respect to $\mathcal{B}(C(I,X))$ (with respect to the uniform topology). We just have the measurability of with repect to $\bigotimes_{t\in I} (\mathcal{B}(X))$.
So, how to conclude? Did I do anything wrong here? Alternatives ways are also welcomed.
Thank you very much.