TITLE: Does every (generalized?) Markov chain admit transition probabilities?
QUESTION [7 upvotes]: To pose the question let us start by recalling the following notions:

Transition Probabilities.  A  transition probability matrix between two measurable spaces $(S,\mathcal{S})$ and $(V,\mathcal{V})$ is a function $P:S\times\mathcal{V}\to [0,1]$ such that for every $s\in S$, $P(s,\cdot)$ is a probability measure in $\mathcal{V}$ and for every $A\in\mathcal{V}$, $P(\cdot,A)$ is $\mathcal{S}-$measurable. If $(S,\mathcal{S})=(V,\mathcal{V})$ we will speak of a transition probability matrix in $(S,\mathcal{S})$.
Random Elements.  A random element of a measurable space $(S,\mathcal{S})$ is a $\mathcal{F}/\mathcal{S}$ measurable function $\xi:\Omega\to S$ defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ (we can identify two random elements if they coincide at $\mathbb{P}-$a.e. $\omega$, but this definition will be enough for our purposes). Note that, in this definition, the word random comes from the introduction of a probability measure in $(\Omega,\mathcal{F})$, but this measure has no role in the property defining $\xi$.
Markov Chains (with fixed state space). A sequence $(\xi_{k})_{k\in\mathbb{Z}}$  of random elements of $(S,\mathcal{S})$ defined on the same probability space $(\Omega,\mathcal{F},\mathbb{P})$ is a Markov chain if for every $k\in\mathbb{Z}$ there exists a transition probability matrix $P_{k}$ in $(S,\mathcal{S})$ such that for every $A\in\mathcal{S}$,
$$\omega\mapsto {P}_{k}(\xi_{k}(\omega),A)$$
defines a version of $\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]:= E[I_{A}\circ \xi_{k+1}| \sigma(\xi_{j})_{j\leq k}]$ (the conditional expectation is taken with respect to $\mathbb{P}$), where $I_{A}$ is the indicator (or "characteristic'') function of $A$. This implies in particular that for every $A\in\mathcal{S}$ and every $k\in\mathbb{Z}$
$$\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]=\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{k})].$$ 
Generalized (?) Markov Chains. If we can verify the last equation (regardless of whether the family of transitions matrices $(P_{k})_{k\in\mathbb{Z}}$ as before exists), we will say that $(\xi_{k})_{k\in\mathbb{Z}}$ is a generalized Markov chain.

Question: Is every generalized Markov chain a Markov chain? I.e. can we always find, for a generalized Markov chain, a family of transition probability matrices $(P_{k})_{k\in\mathbb{Z}}$ such that $\omega\mapsto P_{k}(\xi_{k}(\omega),A)$ is a version of $\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]$?
Again, I don't know if this question is elementary or its answer is well-known. If such is the case, I would appreciate a reference/explanation of the answer before it gets closed. Note also that the question can be generalized easily to the case in which the state spaces vary with $k$, but an answer to this version is sufficient for my purposes.
Thanks for your attention!

REPLY [5 votes]: Under mild conditions on the state space $(S, \mathcal{S})$, it is true.   For instance, it is sufficient that it be standard Borel.  In that case, each $\xi_k$ admits a regular conditional probability, which is (in your language) a transition probability matrix $Q_k$ such that for each event $B \in \mathcal{F}$, we have $\mathcal{P}(B \mid \xi_k) = Q_k(\xi_k, B)$ almost surely.  Then we can simply let $P_k(x, A) = Q_k(x, \xi_{k+1}^{-1}(A))$.
The existence of regular conditional probabilities is discussed in many advanced probability texts.  For instance, you can find it in Durrett's Probability: Theory and Examples, 4e in Section 5.1.3 (for standard Borel spaces).  If you want a more general version, the standard place to look is Dellacherie and Meyer, or Bogachev's Measure Theory section 10.4.  For instance, I believe regular conditional probabilities still exist if $(S, \mathcal{S})$ is merely Radon; i.e. $\mathcal{S}$ is the Borel $\sigma$-algebra of some topology on $S$ for which every Borel probability measure is Radon.
The result is also a special case of the disintegration theorem, which you can also find treated in depth in the aforementioned references.
The case where the state spaces $S_k$ are distinct also follows, simply by letting $S$ be the disjoint union of all the $S_k$.
But I don't think it's true in full generality.  I think the counterexample from Stoyanov's Counterexamples in Probability can be adapted into a generalized Markov chain. If I ever have some spare time I may try to work it out, or anyone else is welcome to do so.