Alternative Definition of Markov Property. currently I am trying to solve the following exercise (17.1.1) in Klenke Probability Theory, which states the following:
Let $(X_t)_{t \in I}$ be a stochastic process and denote by $\mathcal{F}_{\leq t} = \sigma(X_s : s \in I, \; s \leq t)$ the sigma-algebra of the past until time $t$, while $\mathcal{F}_{\geq t}$ denotes respectively the sigma-algebra of the future. Then one has the following equivalence:
The process $(X_t)_{t \in I}$ admits the (elementary) Markov property iff both $\mathcal{F}_{\leq t}$ and $\mathcal{F}_{\geq t}$ are independent given $\sigma(X_t)$, where the definition follows:
Conditional Independence
A family of sub-sigma algebras $(\mathcal{A}_i)_{i \in I} \subset \mathcal{F}$ is called independent of the sub-sigma algebra $\mathcal{A} \subset \mathcal{F}$ if for every finite $J \subset I$ one has
$$
P(\cap_{j \in J} A_j \mid \mathcal{A}) = \prod_{j \in J} P(A_j \mid \mathcal{A}) \quad \text{almost surely}.
$$
My try so far:
For the first direction, suppose that $X$ admits the Markov Property then let $(A_j)_{j \in J} \subseteq \sigma(X_s)_{s \leq t}$ be a finite family of events from the sigma algebra of the past, observe that for $A = \cap_{j \in J} A_j$ one has almost surely
$$
P(A \mid \sigma(X_t)) \stackrel{Def.}{=} E[ \mathbf{1}_A \mid \sigma(X_t) ] \stackrel{M.P}{=} E[ \mathbf{1}_A \mid \mathcal{F}_t ] = \mathbf{1}_A = \prod_{j \in J} \mathbf{1}_{A_j} = \prod_{j \in J} P(A_j \mid \mathcal{F}_t) \stackrel{M.P}{=} \prod_{j \in J} P(A_j \mid \sigma(X_t)),
$$
where we used that due to the filtration property $\mathbf{1}_{A_j}$ are $\mathcal{F}_t$-measurable. Now for the sigma algebra of the future let again $(A_j)_{j \in J}$ be a finite family of events of $\mathcal{F}_{t \geq}$ then observe first that by the Markov Property one has
$$
P(A \mid \sigma(X_t) ) \stackrel{M.P}{=} P(A \mid \mathcal{F}_t) \stackrel{Def.}{=} E[\mathbf{1}_{A} \mid \mathcal{F}_t],
$$
then since for all $j \in J$ one has $A_j \in \mathcal{F}_j$ where $j \geq t$ but now I am stuck since I cannot use the tower property in a good way. I am sure there is a connection with independence and conditional expectation but right now I don't see it, any hints?
Thanks in advance.
 A: $(\Rightarrow)$. Suppose $X$ is Markov. Then for any $s\leq t$ and $B \in \mathscr{B}$, $P(X_t \in B|X_s)=P(X_t \in B|\mathscr{F}_s)$ and equivalently for all bounded measurable $f$ we have $E[f(X_t)|\mathscr{F}_s]=E[f(X_t)|X_s]$. We have for $  s \leq  t$ and $A\in \mathscr{B}, F \in \mathscr{F}_s$
$$P(\{X_t\in A\}\cap F|\mathscr{F}_s)=P(X_t \in A|\mathscr{F}_s)\mathbf{1}_{F}=P(X_t \in A|X_s)\mathbf{1}_{F}$$
Since $\sigma(X_s)\subseteq \mathscr{F}_s$, we take the conditional expectations:
$$P(\{X_t\in A\}\cap F|X_s)=P(X_t \in A|X_s)P(F|X_s)$$
So $\sigma(X_t)$ is independent of $\mathscr{F}_s$ conditionally on $\sigma(X_s)$, $\forall t \geq s$. Now, similarly, for $s\leq u\leq t$
$$\begin{aligned}P(\{X_t \in A\}\cap \{X_u \in B\}\cap F|\mathscr{F}_s)&=P(\{X_t \in A\}\cap \{X_u \in B\}|\mathscr{F}_s)\mathbf{1}_{F}=\\
&=E[P(X_t \in A|\mathscr{F}_u)\mathbf{1}_{\{X_u \in B\}}|\mathscr{F}_s]\mathbf{1}_{F}=\\
&=E[P(X_t \in A|X_u)\mathbf{1}_{\{X_u \in B\}}|\mathscr{F}_s]\mathbf{1}_{F}=\\
&=E[P(\{X_t \in A\}\cap\{X_u \in B\}|X_u)|\mathscr{F}_s]\mathbf{1}_{F}=\\
&=E[P(\{X_t \in A\}\cap\{X_u \in B\}|X_u)|X_s]\mathbf{1}_{F}=\\
&=E[P(\{X_t \in A\}\cap\{X_u \in B\}|\mathscr{F}_u)|X_s]\mathbf{1}_{F}=\\
&=P(\{X_t \in A\}\cap\{X_u \in B\}|X_s)\mathbf{1}_{F}\end{aligned}$$
and ultimately
$$P(\{X_t \in A\}\cap \{X_u \in B\}\cap F|X_s)=P(\{X_t \in A\}\cap\{X_u \in B\}|X_s)P(F|X_s)$$
So $(X_{t_1},X_{t_2},...,X_{t_n}),\,s\leq t_1\leq t_2 \leq ... \leq t_n<\infty$ is independent of $\mathscr{F}_s$ conditionally on $\sigma(X_s)$ by a $\pi$-$\lambda$ argument with the rectangle sets $A_1\times A_2\times ... \times A_n\in \mathscr{B}(\mathbb{R}^n)$. Now we can see that
$$\bigcup_{n\in \mathbb{N}}\bigcup_{s\leq t_1\leq ... \leq t_n<\infty}\sigma(X_{t_1},...,X_{t_n})\subseteq\{G \in \mathscr{F}_{\geq s}:P(G\cap F|X_s)=P(G|X_s)P(F|X_s),\,\forall F \in \mathscr{F}_s\}$$
The family on the rhs is straightforwardly seen to be a Dynkin system (i.e. a $\lambda$-system) contained in $\mathscr{F}_{\geq s}$, while the family on the lhs is a $\cap$-stable system (i.e. a $\pi$ system). The lhs generates $\mathscr{F}_{\geq s}=\sigma(t \geq s,X_t)$, and the conclusion follows.
$(\Leftarrow)$. Suppose $P(G \cap F|X_s)=P(G|X_s)P(F|X_s),\,\forall F \in \mathscr{F}_s,\forall G \in \mathscr{F}_{\geq s}$. We have for $F \in \mathscr{F}_s$ (and ofc since $\{X_t \in A\}\in \mathscr{F}_{\geq s}$)
$$\begin{aligned}E[\mathbf{1}_{F}\mathbf{1}_{\{X_t \in A\}}]&=E[E[\mathbf{1}_F|X_s]P(X_t \in A|X_s)]=\\
&=E[\mathbf{1}_FP(X_t \in A|X_s)]\end{aligned}$$
so $P(X_t \in A|\mathscr{F}_s)=P(X_t \in A|X_s)$ a.s. by definition of conditional expectation.
