Consider a stochastic process $\{X_s\}_{s\in\mathcal S\subseteq\mathbb R}$ with value in $(\mathbb R,\mathcal B(\mathbb R))$ adapted to a filtration $\{\mathcal F_s\}$ (we can suppose that $\{\mathcal F_s\}$ is the natural filtration). The process has the Markov property if for every $A\in\mathcal B(\mathbb R)$ and every $s\le t$: $$E(X_t\in A|\mathcal F_s)=E(X_t\in A|X_s)\qquad(\ast)$$
Now consider the following proposition:
If $\mathcal S=\{s_1,s_2,\ldots\}$ is a discrete set, the Markov property is equivalent to the following property: For every choice $s_1<s_2<\ldots<s_n<t$ such that $P(X_{s_1}=i_1,\ldots, X_{s_n}=i_n)>0$ we have: $$P(X_t=i|X_{s_1}=i_1,\ldots, X_{s_n}=i_n)=P(X_t=i|X_{s_n}=i_n)\qquad(\ast\ast)$$
I have problems to show that $(\ast)\Rightarrow (\ast\ast)$ and I explain my trouble following Klenke's proof (remark $17.2$ of his book):
Suppose that $A:=\{X_{s_1}=i_1,\ldots, X_{s_n}=i_n\}$, then clearly
$$P(X_t=i,X_{s_1}=i_1,\ldots, X_{s_n}=i_n)=E(E(\chi_{\{X_t=i\}}|\mathcal F_{s_n})\chi_A)$$ by the properties of the conditional expectation (I understand this passage). Moreover by $(\ast)$ we have that $E(E(\chi_{\{X_t=i\}}|\mathcal F_{s_n})\chi_A)=E(E(\chi_{\{X_t=i\}}| X_{s_n})\chi_A)$, but I have problems with the succesive passage, namely:
$$E(E(\chi_{\{X_i=t\}}| X_{s_n})\chi_A)=E(P(X_t=i| X_{s_n}=i_n)\chi_A)$$
Why does the last equality hold?
notation: If $A$ is a set, $\chi_A$ is the characteristic function of $A$.
Thanks in advance.
By definition,
$$\mathbb{E}(\chi_{\{X_t = i\}} \mid X_{s_n}) = \mathbb{P}(X_t = i \mid X_{s_n}). \tag{1}$$
Now recall that for any random variable $Y$ and $B \in \mathcal{B}(\mathbb{R})$ there exists by the factorization lemma a function $h$ such that
$$\mathbb{P}(B \mid Y) = h(Y);$$
this function $h$ is denoted by
$$h(y) = \mathbb{P}(B \mid Y=y).$$
Applying this to $(1)$, we get
$$\mathbb{E}(\chi_{\{X_t = i\}} \mid X_{s_n}) = \mathbb{P}(X_t = i \mid X_{s_n} = x) \bigg|_{x=X_{s_n}}.$$
Since $X_{s_n}(\omega) = i_n$ for any $\omega \in A$, this implies
$$\mathbb{E}(\chi_{\{X_t = i\}} \mid X_{s_n})(\omega) = \mathbb{P}(X_t = i \mid X_{s_n} = i_n)$$
for all $\omega \in A$. Hence,
$$\mathbb{E}(\chi_{\{X_t = i\}} \mid X_{s_n}) \cdot \chi_A = \mathbb{P}(X_t = i \mid X_{s_n} = i_n) \cdot \chi_A.$$