Is a Markov process uniquely determined? Let


*

*$E$ be a Polish space and $\mathcal E$ be the Borel $\sigma$-algebra on $E$

*$I\subseteq[0,\infty)$ be closed under addition and $0\in I$


Please consider the following result:

Let $(\kappa_t:t\in I)$ be a Markovian semigroup on $(E,\mathcal E)$ $\Rightarrow$ There is a measurable space $(\Omega,\mathcal A)$ and a Markov process $X$ with distributions $(\operatorname P_x)_{x\in E}$ such that $$\operatorname P_x\left[X_t\in B\right]=\kappa_t(x,B)\;\;\;\text{for all }x\in E,B\in\mathcal E\text{ and }t\in I\;.\tag 1$$ Conversely, given a Markov process $X$ with distributions $(\operatorname P_x)_{x\in E}$ on a measurable space $(\Omega,\mathcal A)$, a Markovian semigroup $(\kappa_t:t\in I)$ is defined by $(1)$.

It turns out that $X$ in the first part of the statement can be constructed as the family of coordinate maps on $(\Omega,\mathcal A)=(E^I,\mathcal E^{\otimes I})$.
I've seen that many authors assume that Markov processes are such coordinate maps. Why can they do that?
The statement above doesn't state, that given $(\Omega,\mathcal A)$ there is one unique Markov process, does it? However, the finite-dimensional distributions of $X$, i.e. $$\operatorname P_x\left[X\in\;\cdot\;\right]\circ\pi_J^{-1}\;\;\;\text{for }J\subseteq I\text{ with }|J|<\infty\;,\tag 2$$ where $\pi_J:E^I\to E^J$ are the canonical projections, are uniquely determined by $(1)$.
Maybe $(\operatorname P_x)_{x\in E}$ (not only the finite-dimensional distributions) are uniquely determined by $(1)$, if $I\subseteq \mathbb N_0$ or $I$ is at least almost countable or when $E$ is almost countable.
So, why does the stated result allows us to think about $X$ as being uniquely determined?
 A: The Kolmogorov extension theorem states that there is a probability distribution on $(E^I, \mathcal E^{\otimes I})$ such that $(1)$ holds. 
I claim that this distribution is uniquely determined. 
There are two ways to see this, both appealing to the definition of $\mathcal E^{\otimes I}$. In fact, it is the $\sigma$-algebra generated by cylindrical sets (equivalently, the smallest $\sigma$-algebra under which each evaluation $f\mapsto f(t)$, $t\in I$, is measurable), that is,
$$\mathcal E^{\otimes I} = \sigma(\mathcal P),\text{ where } \mathcal P = \big\{\{f:f(t_1,\dots,t_n)\in B\}, B\in \mathcal E^{\otimes n}, t_1,\dots,t_n\in I, n\ge 1\big\}.$$
Method 1. One of useful consequences of Dynkin's $\pi$-$\lambda$ theorem is the following fact: if two probabilistic measures coincide on a $\pi$-system, then they coincide on the $\sigma$-algebra generated by this system. Now notice that $\mathcal P$ is obviously a $\pi$-system. Therefore, two measures agreeing on $\mathcal P$ (= two distributions of processes having same finite-dimensional distributions) agree on $\sigma(\mathcal P) = \mathcal E^{\otimes I}$.
Method 2. A slightly extended version of my comment. The $\sigma$-algebra $\mathcal E^{\otimes I}$ has a countable description: 
$$
\mathcal E^{\otimes I} = \big\{\{f:f(t_1,t_2,\dots)\in B\}, B\in \mathcal E^{\otimes \mathbb{N}}, t_1,t_2,\dots\in I\big\}.
$$
In other words, each set from $\mathcal E^{\otimes I}$ sits on a countable number of coordinates from $I$. But it is not hard to see (as you correctly noted) that for a countable number of coordinates, the distribution is unique. Therefore, it is unique on $\mathcal E^{\otimes I}$. 
