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Background

Let $\left(X_t\right)_{t \in I}$ ($I\subseteq\mathbb R$) be an $E$-valued stochastic process ($E$ being a Polish space with the Borel $\sigma$-algebra $\mathcal{B}\left(E\right)$) equipped with the filtration generated by $X$, $\left(\mathcal F_t\right)_{t\in I}=\left(\sigma\left(X_s\space:\space s\leq t\right)\right)_{t\in I}$. Suppose $E$ is countable.

Question

Why is it the case (as claimed in Klenke, Remark 17.2) that if for all $n\in\mathbb N$, all $s_1<\cdots and all $i_1,\dots,i_n,i\in E$ with $\mathbb{P}\left[X_{s_1}=i_1,\dots,X_{s_n}=i_n\right]>0$ we have

$\mathbb{P}\left[\left.X_t=i\space\right|\space X_{s_1}=i_1,\dots,X_{s_n}=i_n\right]=\mathbb{P}\left[\left.X_t=i\space\right|\space X_{s_n}=i_n\right]$

then the Markov property applies, namely

$\forall s\leq t\in I\bullet\mathbb{P}\left[\left.X_t\in A\space\right|\space\mathcal{F}_s\right]=\mathbb{P}\left[\left.X_t\in A\space\right|\space X_s\right]$

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Fix $s\leq t$. We can consider that $A=\{i_k\}$, using additivity and the fact that the space is countable. We have to check that $\forall B\in\mathcal F_s,\quad\int_B\chi_{\{X_t=i_k\}}dP=\int_BE[\chi_{\{X_t=i_k\}}\mid X_s]dP.$ To see that, note that the equality is true on the finite intersections of sets of the form $\{X_u\in E'\}$, $E'\subset E$ and $u\leq s$, then show that the sets which satisfy the displayed equality is a $\lambda$-system. You will need Dynkin's theorem.

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    I could use your help [here](https://math.stackexchange.com/questions/2540049/markov-property-when-the-state-space-is-countable)2017-11-27