Non-equivalent definitions of Markov process

Question

As far as I know, there are three definitions of Markov processes (or of Markov chains).

DEFINITION 1 (WEAKER). A process $(X_t)_{t\in[0,\infty)}$ on $(\Omega,\mathcal{F},\mathbb{P})$ with values in an arbitrary measurable space $(E,\mathcal{E})$ is a Markov process iff, for all $s,t\geq 0$, $X_{t+s}|\mathcal{F}^X_t\sim X_{t+s}|X_t$, where $\mathcal{F}^X_t:=\sigma\{X_s\,|\, s\in[0,t]\}$.

DEFINITION 2 (STRONGER). A process $(X_t)_{t\in[0,\infty)}$ on $(\Omega,\mathcal{F},\mathbb{P})$ with values in an arbitrary measurable space $(E,\mathcal{E})$ is a Markov process iff it satisfies Definition 1 and there exists a family $\{P_{t,t+s}|s,t\geq 0\}$ of probability transition kernels $P_{t,t+s}:E\times \mathcal{E}\to [0,1]$ such that, for all $s,t\geq 0$, $P_{t,t+s}$ is a regular version of $X_{t+s}|X_t$ (meaning that for all $A\in \mathcal{E}$ it holds that $\mathbb{E}(1_A(X_{t+s})|X_t)=P_{t,t+s}(A,X_t)$ $\mathbb{P}$-a.e.).

DEFINITION 3 (EVEN STRONGER). Same as Definition 2 but adding the requirement that $P_{t,t}=I$ for all $t\geq 0$ (where $I(x,dy)=\delta_x(dy)$) and that the $P_{t,t+s}$ satisfy the Chapman-Kolmogorov equation, that is $P_{t+s,t+s+r}\circ P_{t,t+s}=P_{t,t+s+r}$ for all $t,s,r\geq 0$ (where, if $P$ and $K$ are kernels on $E\times \mathcal{E}$, $P\circ K$ is defined for $(x,A)\in E\times\mathcal{E}$ as $(P\circ K)(x,A)=\int_E P(y,A)K(x,dy)$).

I have 3 questions:

1. Can anyone provide an example of a process that satisfies Definition 1 but not Definition 2? I am aware that such a process should be valued in some non-Borel-standard space $E$, so I imagine it might be difficult to construct. SOLVED: see Cinlar's Probability and Stochastic, page 446, Remark 11.1 for a counterexample of a process which is Markov but doesn't have a transition kernel.
2. Similarly, can someone provide an example of a process that satisfies Definition 2 but does not satisfy Definition 3 for any transition kernels (possibly different from the ones that satisfy Definition 2)? I suspect such a counterexample to exist but I am not sure.
3. (IMO THE MOST INTERESTING ONE) If a process satisfies Definition 1 and the space $(E,\mathcal{E})$ is Borel-standard, is Definition 3 automatically satisfied? (We know that Definition 2 is, by the existence of regular versions of conditional expectations). SOME PROGRESS: For a Markov Chain the answer is positive since one can just start from $P_{0,1}$,$P_{1,2}$,... and define $P_{i,j}:=P_{i,i+1}\circ...\circ P_{j-1,j}$.

Some very closely related (unanswered) questions on math.stackexchange are this one and this other one.

Answer 1

(To reduce clutter, let $u = t + s \geq t$ and $v = t + s + r \geq u$.)

For Question 3: does Definition 2 imply the Chapman-Kolmogorov equation just by the tower rule? Definition 2 says that for all $t, u$, almost surely $$ P_{t, u}(X_t, A) = \mathbb{E}[1_A(X_u) \mid X_t]. $$ But this should imply that for any bounded $f$, almost surely $$ \mathbb{E}[f(X_u) \mid X_t] = \int_E f(x) \, P_{t, u}(X_t, \mathrm{d}x). $$ (This is the step I'm least confident about—is there a subtlety I'm missing?) From this, we compute $$\begin{aligned} P_{t, v}(A, X_t) &= \mathbb{E}[1_A(X_v) \mid X_t] \\ &= \mathbb{E}[\mathbb{E}[1_A(X_v) \mid X_u] \mid X_t] \\ &= \mathbb{E}[P_{u, v}(X_u, A) \mid X_t] \\ &= \int_E P_{u, v}(x, A) \, P_{t, u}(X_t, \mathrm{d}x) \\ &= (P_{u, v} \circ P_{t, u})(X_t, A). \end{aligned}$$

I think this tells you at least that the kernels $P_{t, u}$ form some semigroupoid with identities $P_{t, t}$ (plug in $t = u < v$ or $t < u = v$). But that semigroupoid might be something like "kernels modulo how they act on parts of $E$ that the process can't reach". So you would need something like "the initial distribution $X_0$ covers all of $E$" to have a hope of saying that Chapman-Kolmogorov is satisfied everywhere.