Proving technique used to show an equivalence to the definition of a Markov process Let $X=(X_t)_{t\in I}$ be Markov process with values in a Polish space $E$. I want to show, that there exists a stochastic kernel $\kappa:E\times\mathcal{B}(E)^{\otimes I}\to [0,1]$ such that $$\operatorname{E}_x\left[f\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]=\int_{E^I}\kappa\left(X_s,dy\right)f(y)$$ for all $\mathcal{B}(E)^{\otimes I}$-$\mathcal{B}(\mathbb{R})$-measurable, bounded $f:E^I\to\mathbb{R}$, $s\ge 0$ and $x\in E$.

I'm afraid that many people might not be familiar with the notation used above. However, my question is less about the proof of the statement, but more about the proving technique.
I've read that "by the usual approximation arguments, it is enough to consider functions $f$ that depend only on finitely many coordinates $0\le t_1\le\cdots\le t_n$." Why can we do this and how do we prove that?
You can find the statement and proof (as well as the used definition  of a Markov process two sides before) on Google Books.
 A: First note that $\int\kappa\left(X_s,dy\right)f(y)$ is  a $X_s$- measurable function. If it holds 
$$\operatorname{E}_x\left[f\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]=\int_{E^I}\kappa\left(X_s,dy\right)f(y)$$
for every $\hat{f}(y) = f_1(y(t_1))\ldots f_k(y(t_k))$ then consider $f_n(y) \uparrow 1_{A_1}(y(t_1)) \ldots 1_{A_k}(y(t_k))$ (where each $A_i$ is open) then by the monotone convergence theorem you get that for every $B = \{(y(s_1),\ldots, y(s_l)) \in B_l \}$ for $B_l \in \mathcal{B}(E^l)$
$$\int_B \operatorname{E}_x\left[1_{A_1 \times \ldots \times A_k}\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]\, d\Bbb{P}=\int_B \lim_n\operatorname{E}_x\left[f_n\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]\, d\Bbb{P}\\=\int_B \operatorname{E}_x\left[\lim_n f_n\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]\, d\Bbb{P}=\int_B\kappa\left(X_s,{A_1 \times \ldots \times A_k}\right) d\Bbb{P}   $$
Therefore the knowledge of $\operatorname{E}_x\left[f\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]=\int_{E^I}\kappa\left(X_s,dy\right)f(y)$ for finite dimensional functions of the type $\hat{f}$ above is enought to characterize its value on any elementary set of the form  $A_1 \times \ldots \times A_k$ to extend to the general case define $\mathcal{C}=\bigg\{E \in \mathcal{B}(E)^{\otimes I},\operatorname{E}_x\left[1_E\left(\left(X_{s+t}\right)_{t\in I}\right)\mid\mathcal{F}_s\right]=\int_{E^I}\kappa\left(X_s,dy\right)1_E(y)\bigg\}$ Is a $\sigma$-algebra that contains the elementary sets of the form $A_1 \times \ldots \times A_k$ for $A_i$ open. Therefore $\mathcal{C} = \mathcal{B}(E)^{\otimes I}$.
This  ''monotone class'' strategy is useful since it reduces your task to a finite number of coordinates.

The fact that such a kernel exists is more subtle and troublesome. One needs to check that a polish space is a canonical space . this is known in the literature as the kuratowsky fiber theorem and may be found in the first chapter of K. R. PARTHASARATHY Probability measures on metric spaces pgs 25-32.
The short version of this result is that every polish space $(E,\mathcal{B}(E))$ is canonical. Therefore, for every probability $\Bbb{P}$ on $(E,\mathcal{B}(E))$ and $\mathcal{G} $ sub$\sigma$-algebra of $\mathcal{B}(E)$ there is a kernel $\kappa: E\times \mathcal{B}(E) \to [0,1]$ $(\omega, A)\mapsto \kappa_{\mathcal{G}}(\omega,A)$ such that
1) $ \kappa_{\mathcal{G}}(\omega, \cdot): \mathcal{B}(E) \to [0,1]  $ is a probability measure for every $\omega \in E$
2) $\kappa_{\mathcal{G}}(\cdot, A): E \to [0,1]$ is $\mathcal{B}(E)$- measurable for every $A \in \mathcal{B}(E)$
3) $\kappa_{\mathcal{G}}(\cdot, A) = \mathbb{E}[1_A \vert \mathcal{G}](\cdot) $ $\mathbb{P}$-almost surely
$\kappa_{\mathcal{G}}$ is often refered to as the regular conditional probability distribution of $\Bbb{P}$ given $\mathcal{G}$
You could `guess' the result if you tried to consider for each $A \in \mathcal{B}(E)$  the conditional expectation $\mathbb{E}[1_A \vert \mathcal{G}](\cdot) $ the problem is that this is not well defined on a $\Bbb{P}$- null set and as you consider more and more sets $A_1,\ldots A_n$ you will soon be dealing with a lot of exceptional sets. As long as the $\sigma$-algebra has only denumerable sets this is not a problem, but many interesting $\sigma$- algebras have a non denumerable quantity of sets in it.
