A markov process which is not strong Markov (follow up) This is a follow up discussion of
https://mathoverflow.net/q/43833
The examples there are interesting, but it seems that they all rely on an "ambiguous" transition function, i.e., there is some state $x$, $p(X_t\in A|X_0=x)$ may have different values, depending on how we arrived at $x$ in the first place.
My question: can someone provide an example of markov process which has well defined transition function but without strong Markov property?
Essentially, I am asking why continuity assumption in Feller process is necessary in proving the strong markov property.
 A: To prove that a Feller process is strong Markov, it is not the continuity 
of $P_tf(x)$ in $t$ that is important, rather the continuity in $x$. That is, we use that $P_t$ maps bounded continuous functions back into bounded continuous functions.
Here is a sketch of the usual proof: Let $T$ be $({\cal G}_{t})$-optional, and let $T_n$ be the standard dyadic approximations to $T$,
 so that, each $T_n$ is $({\cal G}_t)$-optional and $T_n\downarrow T$ on $\{T<\infty\}$.
 Fix $u\geq 0$ and $f\in C_b(E)$. Then, by the Markov property over $({\cal G}_t)$  we have 
 \begin{eqnarray*} \mathbb{E}(f(X_{u+T_n})1_{\{T<\infty\}})
 &=&\sum_{k=1}^\infty \mathbb{E}(f(X_{u+k/2^n})1_{\{T_n=k/2^n\}})\\
 &=&\sum_{k=1}^\infty \mathbb{E}(P_uf(X_{k/2^n})1_{\{T_n=k/2^n\}})\\[5pt]
 &=& \mathbb{E}(P_uf(X_{T_n})1_{\{T<\infty\}}).
 \end{eqnarray*}
Since $P_uf$ is a bounded continuous function, and $(X_t)$ has 
right continuous paths,  letting $n\to\infty$ gives 
$$\mathbb{E}(f(X_{u+T})1_{\{T<\infty\}})= \mathbb{E}(P_uf(X_{T})1_{\{T<\infty\}}).$$
This shows that $(X_t)$ is a strong Markov process.

By way of  contrast, consider the Markov process $(X_t)$ on $E=[0,\infty)$ that (starting at the origin) spends an 
exponential amount of time at the origin, and then moves to the right deterministically at unit speed. It has transition kernel
$$p_t(x,A)=\cases{e^{-t}\delta_0(A)+\int_0^t e^{-z}\delta_{t-z}(A)\,dz&if $x=0$\cr
\delta_{t+x}(A)&if $x>0$.}$$
For a bounded, measurable $f$ on $E$ we have 
$$p_tf(x)=\cases{e^{-t}f(0)+\int_0^t e^{-z} f(t-z)\,dz&if $x=0$\cr
f(t+x)&if $x>0$.}$$
Notice that even if $f$ is continuous, $p_tf(x)$ will typically have a 
discontinuity at $x=0$ for $t>0$. The right hand limit is $\lim_{x\downarrow 0}p_tf(x)=f(t)$, while its value $p_tf(0)$ is a weighted average of $f$ 
over the interval $[0,t]$. 
This process $(X_t)$ has continuous sample paths, is not Feller, nor is it strong Markov.
