Equivalence of discrete definition of Markov property in the coutinuous case

In the book, Lectures from Markov Processes to Brownian Motion, it is stated that the oldest definition of Markov property is, for every integer $$n\ge1$$ and $$0\le t_1 and $$f$$ is continuous with compact support, $$E[f(X_u)|X_t,X_{t_n},\cdots,X_{t_1}] = E[f(X_u)|X_t].$$

Another definition is, for any compactly supported $$f$$, $$E[f(X_u)|\mathcal{F}_t]=E[f(X_u)|X_t].$$

I have no idea how to pass $$\sigma$$-algebra generated by "discrete" process to $$\mathcal{F}_t,$$ especially when such stuff is in the condition (behind "|").

Any hint is appreciated.

RHS is measurable w.r.t $$\mathcal F_t$$. Hence equality will follows if we show that the integral of RHS over $$A$$ equals $$\int_A E(f(X_u)|\mathcal F_t)dP$$ for any $$A \in \mathcal F_t$$. But sets of the form $$X_t^{-1}(B)\cap X_{t_1}^{-1}(B_1)\cap \cdots \cap X_{t_1}^{-1}(B_1)$$ where $$t$$ and $$t_i$$'s are as stated in the book and $$B,B_1,\cdots,B_n$$ are Borel sets generate the sigma algebra $$\mathcal F_t$$. When $$A$$ is of this type equality follows from what you have called the discrete version of Markov property. Now use Monotone Class Theorem to complete the proof.