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 $0\le t_1<t_2<\cdots<t<u,$ 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.


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