Let $\left(Y^{(n)}_k\right)_{k\in\mathbb N_0}$ be a Markov chain for $n\in\mathbb N$, $(h_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $h_n\xrightarrow{n\to\infty}\infty$ and $$X^{(n)}_t:=Y^{(n)}_{\lfloor\frac t{h_n}\rfloor}\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$.
Let $n\in\mathbb N$. In Remark 2 below Corollary 206 of these lecture notes, the author is noting that $X^{(n)}$ is not a Markov process.
I don't get that. If $\mathcal F^Z$ denotes the filtration generated by a process $Z$, we should easily obtain $$\mathcal F^{X^{(n)}}_t=\mathcal F^{Y^{(n)}}_{\lfloor\frac t{h_n}\rfloor}\;\;\;\text{for all }t\ge0\tag1$$ and $$\operatorname P\left[X^{(n)}_t\in B\mid\mathcal F^{X^{(n)}}_s\right]=\operatorname P\left[X^{(n)}_t\in B\mid X^{(n)}_s\right]\;\;\;\text{almost surely}\tag2.$$ So, $X^{(n)}$ should be a Markov process. What am I missing?