Sufficient condition for the Markov property for a process in continuous time with values in a countable space

I would like to verify the folowing: Let $$(X_s)_{s \in [0,\infty)}$$ be a stochastic process in continuous time with values in a countable space $$E$$ and let $$\mathcal F_s=\sigma(X_r: 0 \le r \le s)$$

Given that for all $$n \in \mathbb N$$, all $$0 \le s_1< ... and all $$i_1,...,i_n \in E$$ such that $$P[X_{s_1}=i_1,...X_{s_n } =i_n ]>0$$, we have that

$$P[X_t=i|X_{s_1}=i_1,...X_{s_n } =i_n]=P[X_t=i|X_{s_n }=i_n]$$

Then for any $$A \in \mathcal B(E)$$ and any $$s \le t$$

$$P[X_t \in A |\mathcal F_s]=P[X_t \in A |\sigma(X_s)]$$

My approach to this was to try to show that given the above $$P[X_t \in A |\mathcal F_s]$$ is a version of $$P[X_t \in A |\sigma(X_s)]$$ and that $$P[X_t \in A |\sigma(X_s)]$$ is a version of $$P[X_t \in A |\mathcal F_s]$$. Is this correct?

Thus for the first case we have that since $$\sigma(X_s) \subset \mathcal F_s$$ and from the definition of $$P[X_t \in A |\mathcal F_s]$$, for any $$B \in \sigma(X_s)$$

$$E[1_B P[X_t \in A |\mathcal F_s]]=E[1_B 1_{X_t \in A } ]$$ So that one of the two conditions for $$P[X_t \in A |\mathcal F_s]$$ to be a version of $$P[X_t \in A |\sigma(X_s)]$$ is satisfied. But I'm not sure of the other condition which is about measurability?

For the second case one immediatly gets the measurability condition but I'm not sure about the "Expectation condition".