Why is it legitimate to assume that the Chapman-Kolmogorov equations hold everywhere? Let


*

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$

*$E$ be a Polish space and $\mathcal E:=\mathcal B(E)$

*$(X_t)_{t\ge0}$ be an $E$-valued $\mathcal F$-Markov process on $(\Omega,\mathcal A,\operatorname P)$

*$\kappa_{s,\:t}$ be a regular version of the conditional probability of $X_t$ given $X_s$, i.e. $\kappa_{s,\:t}$ is a Markov kernel on $(E,\mathcal E)$ with $$\operatorname P\left[X_t\in B\mid X_s\right]=\kappa_{s,\:t}(X_s,B)\;\;\;\text{almost surely for all }B\in\mathcal E\tag1$$ for $s,t\ge0$
By the Markov property and $(1)$, $$\operatorname P\left[X_t\in B\mid\mathcal F_s\right]=\kappa_{s,\:t}(X_s,B)\;\;\;\text{almost surely for all }B\in\mathcal E\text{ and }0\le s\le t.\tag2$$

Usually, we want $(\kappa_{s,\:t}:0\le s\le t)$ to satisfy the Chapman-Kolmogorov equation $$\kappa_{r,\:t}=\kappa_{r,\:s}\kappa_{s,\:t},\tag3$$ where the right-hand side denotes the composition of transiton kernels, for all $0\le r\le s\le t$. However, with the definition of $\kappa_{s,\:t}$ as the conditional probability of $X_t$ given $X_s$, we've only got$^1$ $$\kappa_{r,\:t}(x,B)=(\kappa_{r,\:s}\kappa_{s,\:t})(x,B)\;\;\;\text{for all }B\in\mathcal E\text{ and }\operatorname P\circ\:X_r^{-1}\text{-almost all }x\in E\tag4$$ for all $0\le s\le t$. However, in the literature, one is usually assuming that $(2)$ and $(3)$ hold together. Why is that possible?

I could imagine that the reason is the following: Since $E$ is Polish, given $(\kappa_{s,\:t}:0\le s\le t)$ with $(3)$ there is always a Markov process $\tilde X$ on an other probability space with transition semigroup $(\kappa_{s,\:t}:0\le s\le t)$ and initial distribution $\operatorname P\circ\:X_0^{-1}$. Is that the correct argument?

$^1$ Note that there is a crucial selection of a common null set for all $B$ happening in $(4)$. I guess this is legitimate as long as $\mathcal E$ is countably generated. Maybe someone could comment on this.
 A: My (perhaps naive) interpretation of your question is that it deals with "surely" versus "almost surely" issues. For fixed times $r < s < t$, the following equation only holds in an “almost sure” sense: 
$$ k_{r,t} = k_{r,s} \cdot k_{s,t}  \quad (Eq. 1)$$
To see this, suppose it holds surely (for all corresponding points $x$ and sets  $B$ for which we can evaluate $k_{r,t}(x,B)$).  Let $\tilde{k}_{r,t}$ be another “version” of $k_{r,t}$ that is different from $k_{r,t}$ on a set of measure 0.  Then we want to also have 
$$\tilde{k}_{r,t} = k_{r,s} \cdot k_{s,t} \quad (Eq. 2)$$ surely, but this is impossible since we have changed the left-hand-side of equation (1) without changing the right-hand-side.  The only way it makes sense is that both equations (1) and (2) are only guaranteed to hold in an almost-sure sense.  
An example way to change $k_{r,t}$ to a different version $\tilde{k}_{r,t}$ is this: Suppose we can identify an $x^*$ such that $P[X_r=x^*]=0$.  Then define for all corresponding $x, B$: 
$$ \tilde{k}_{r,t}(x,B) = \left\{ \begin{array}{ll}
k_{r,t}(x,B) &\mbox{ if $x \neq x^*$} \\
\mu(B) & \mbox{ if $x =x^*$} 
\end{array}
\right.$$
where $\mu(B)$ is some measure over events $B$ that is different from $k_{r,t}(x^*,B)$. 
