# If $X^n$ is a Feller process, $τ_n$ is a stopping time and $h_n\to0$, why can we conclude $|X^n_{τ_n}-X^n_{τ_n+h_n}|→0$ from $|X^n_0-X^n_{h_n}|→0$?

Let $$E$$ be a compact complete separable metric space and $$X^n,X$$ be $$E$$-valued càdlàg strong Markov processes on a probability space $$(\Omega,\mathcal A,\operatorname P)$$.

Question 1: Let $$\tau_n$$ be a finite stopping on $$(\Omega,\mathcal A)$$ time wrt the filtration generated by $$X^n$$ and $$(h_n)_{n\in\mathbb N}\subseteq(0,\infty)$$ with $$h_n\xrightarrow{n\to\infty}0$$. Why it sufficient to show $$d\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}0\;\;\;\text{in probability},\tag1$$ in order to conclude $$d\left(X^n_{\tau_n},X^n_{\tau_n+h_n}\right)\xrightarrow{n\to\infty}0\;\;\;\text{in probability}?\tag2$$ I guess we somehow need to utilize the strong Markov property. (If necessary, assume that $$\sup_{n\in\mathbb N}\sup_{\omega\in\Omega}\tau_n(\omega)<\infty$$.)

I've tried to use the strong Markov property in the following way: Let $$(\kappa^n_t)_{t\ge0}$$ denote the transition semigroup of $$X^n$$ and $$\mathcal F_{\tau_n}$$ the $$\sigma$$-algebra of the $$\tau_n$$-past. Then $$\operatorname P\left[d\left(X^n_{\tau_n},X^n_{\tau_n+h_n}\right)>\varepsilon\mid\mathcal F_{\tau_n}\right]=\kappa^n_{h_n}\left(X^n_{\tau_n},\left\{y\in E:d\left(X^n_{\tau_n},y\right)>\varepsilon\right\}\right)\;\;\;\text{almost surely}\tag0,$$ but that doesn't seem to help.

Question 2: By compactness of $$E$$, we know from Prohorov's theorem that there is a probability measure $$\nu$$ on $$(E,\mathcal B(E))$$ with $$X_0^n\xrightarrow{n\to\infty}\nu\;\;\;\text{in distribution}.\tag3$$ Suppose we know that $$\operatorname E\left[f(X^n_0)g\left(X^n_{h_n}\right)\right]\xrightarrow{n\to\infty}\operatorname E\left[(fg)(X_0)\right]\tag4$$ as long as $$X_0$$ is distributed according to $$\nu$$ from which we could conclude $$\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}(X_0,X_0)\;\;\;\text{in distribution}\tag5$$ and hence $$\rho\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}\rho(X_0,X_0)=0\;\;\;\text{in distribution}\tag6$$ which in turn yields $$(1)$$. Why is this enough, i.e. why is it no restriction to assume that $$X_0$$ is distributed according to $$\nu$$?