Alternative formulation of time-homogeneity of a Markov process Let $(X_t)_{t \geq 0}$ be a time-homogeneous Markov process on a Polish state space $E$ and for a state $x$ denote by $P_x$ and $E_x$ the probability measure resp. expectation for $x$ as initial state.
The time-homogeneity property means that for each fixed $t$ and for each fixed $h$ it holds $E_x[f(X_{t+h}) \mid X_t] = E_{X_t}[f(X_h)]$ $P_x$-a.s. for all bounded measurable functions $f$ on $E$.
How can I show that it moreover holds that
$E_x[f((X_{t+h})_{h \geq 0}) \mid X_t] = E_{X_t}[f((X_h)_{h \geq 0})]$ $P_x$-a.s. for all bounded measurable functions $f$ defined on the product space $E^{[0, \infty)}$?
I think one has to show first, that the time homogeneity property extends to $E_x[f(X_{t_1}, \dots, X_{t_n}) \mid X_t] = E_{X_t} [ f(X_{t-t_1}, X_{t_2-t_1}, \dots X_{t_n-t_{n-1}})]$ $P_x$-a.s. for all bounded measurable functions $f$ on the finite product space $E^n$ with $t < t_1 < \dots < t_n$
and then apply somehow the monotone class argument or Dynkin's lemma. However, I am getting confused with the notation.
 A: Meanwhile I was able to establish the proof. It employs a standard monotonic class argument. Hence, the proof is purely measure-theoretic and does not assume any (topological) regularity on $E$ - that is, $E$ can be a general measurable space.
Fix $x \in E$.
Set $M$ as the set of all those measurable functions $f : E^{[0, \infty)} \to \mathbb{R}$ for which the desired property
$$
E_x[f((X_{t+h})_{h \geq 0}) \mid X_t] = E_{X_t}[f((X_h)_{h \geq 0}))] \quad (*)
$$ holds.
Then $M$ is a monotonic class ($M$ is a vector space, $1 \in M$ and whenever $f_n \in M$ with $f_n \geq 0$ and $f_n \nearrow f$ pointwise to some function $f : E^{[0, \infty)} \to \mathbb{R}$ then $f \in M$). The product $\sigma$-algebra on $E^{[0, \infty)}$ is generated by the collection $\mathcal{C}' := \{ \pi_t^{-1}(B) \mid t \geq 0, B \in \mathcal{E} \}$ where $\pi_t : E^{[0,\infty)} \to E$ is the projection to the $t$-th coordinate. The collection $\mathcal{C}$ of finite intersections of such sets $\pi_t^{-1}(B)$ for $t \geq 0$, $B \in \mathcal{E}$ forms a $\pi$-system (a collection that is closed under finite intersection). Since $\mathcal{C}' \subseteq \mathcal{C} \subseteq \sigma(\mathcal{C}')$ it follows that $\mathcal{C}$ also generates the product $\sigma$-algebra on $E^{[0, \infty)}$. By the monotone class argument, if we can show that the indicator $\chi_{A}$ is contained in $M$ for every $A \in \mathcal{C}$ then $M$ contains all bounded measurable functions $f : E^{[0, \infty)} \to \mathbb{R}$. In other words, the desired property $(*)$ is satisfied for all bounded measurable functions on $E^{[0, \infty)}$.
So, let $A = \pi_{t_1}^{-1}(B_1) \cap \dots \cap \pi_{t_n}^{-1}(B_n)$ with $t_i \in [0, \infty)$ and $B_i \in \mathcal{E}$.
We can assume that $t_1 \leq \dots \leq t_n$.
Let us show this by induction over $n$.
For $n = 1$ this is just the time-homogeneity property: it holds $P_x$-almost surely
$$
E_x[\chi_A((X_{t+h})_{h \geq 0}) \mid X_t]
= P_x(X_{t+t_1} \in B_1 \mid X_t)
= P_{X_t}(X_{t_1} \in B_1)
= E_{X_t}[\chi_A((X_h)_{h \geq 0})].
$$
For the induction step $n \longrightarrow n+1$ assume that $(*)$ is satisfied for $A = \pi_{t_1}^{-1}(B_1) \cap \dots \cap \pi_{t_n}^{-1}(B_n)$ and show that it is also satisfied for $A' := A \cap \pi_{t_{n+1}}^{-1}(B_{n+1})$
where $B_{n+1} \in \mathcal{E}$ and $t_{n+1} \geq t_n$.
From the induction hypothesis it follows that $(*)$ holds for all $\sigma(\pi_{t_1}, \dots, \pi_{t_n})$-simple functions and hence (by taking pointwise limits) for all bounded $\sigma(\pi_{t_1}, \dots, \pi_{t_n})$-measurable functions $f : E^{[0, \infty)} \to \mathbb{R}$ (for any $\sigma$-algebra $\mathcal{A}$, any $\mathcal{A}$-measurable function is the pointwise limit of a sequence of $\mathcal{A}$-simple functions; since $f$ is bounded we can then apply the dominated convergence theorem for conditional expectations). Now, for $P_x$-almost all $\omega$ it holds
\begin{align*}
& E_x[\chi_{A'}((X_{t+h})_{h \geq 0}) \mid X_t](\omega)\\
& = E_x[\prod_{i=1}^{n+1} \chi_{\pi_{t_i}^{-1}(B_i)}((X_{t+h})_{h \geq 0}) \mid X_t](\omega)\\
& = E_x[\prod_{i=1}^{n+1} \chi_{ \{ X_{t+t_i} \in B_i \}} \mid X_t](\omega) \\
& = E_x[ \prod_{i=1}^{n} \chi_{ \{ X_{t+t_i} \in B_i \}} E_x [ \chi_{ \{ X_{t+t_{n+1}} \in B_{n+1} \}} \mid X_t, X_{t+t_1}, \dots, X_{t+t_n} ] \mid X_t ](\omega) \\
& = E_x[ \prod_{i=1}^{n} \chi_{ \{ X_{t+t_i} \in B_i \}} E_x [ \chi_{ \{ X_{t+t_{n+1}} \in B_{n+1} \}} \mid X_{t+t_n} ] \mid X_t ] (\omega) \\
& = E_x[ \prod_{i=1}^{n} \chi_{ \{ X_{t+t_i} \in B_i \}} E_{X_{t+t_n}} [ \chi_{ \{ X_{t_{n+1} - t_n} \in B_{n+1} \}} ] \mid X_t ] (\omega) 
\end{align*}
where we applied the Markov property and the time-homogeneity property.
Set
$$ f : E^{[0, \infty)} \to \mathbb{R}, \quad f(\varphi) := \prod_{i=1}^{n} \chi_{ \{ \varphi_{t_i} \in B_i \}} E_{\varphi_{t_n}} [ \chi_{ \{ X_{t_{n+1} - t_n} \in B_{n+1} \}} ].$$
Then $f$ is $\sigma(\pi_{t_1}, \dots, \pi_{t_n})$-measurable and bounded and it holds
$$
E_x[ \prod_{i=1}^{n} \chi_{ \{ X_{t+t_i} \in B_i \}} E_{X_{t+t_n}} [ \chi_{ \{ X_{t_{n+1} - t_n} \in B_{n+1} \}} ] \mid X_t ] (\omega)
= E_x[ f((X_{t+h})_{h \geq 0}) \mid X_t] (\omega).
$$
From the above conclusion of the induction hypothesis it follows that
\begin{align*}
& E_x[ \prod_{i=1}^{n} \chi_{ \{ X_{t+t_i} \in B_i \}} E_{X_{t+t_n}} [ \chi_{ \{ X_{t_{n+1} - t_n} \in B_{n+1} \}} ] \mid X_t ] (\omega) \\
& = E_x[ f((X_{t+h})_{h \geq 0}) \mid X_t] (\omega)
  = E_{X_t(\omega)}[ f((X_h)_{h \geq 0}) ] \\
& = E_{X_t(\omega)}[ \prod_{i=1}^{n} \chi_{ \{ X_{t_i} \in B_i \}} E_{X_{t_n}} [ \chi_{ \{ X_{t_{n+1} - t_n} \in B_{n+1} \}} ] ].
\end{align*}
On the other hand,
\begin{align*}
& E_{X_t}[\chi_{A'}((X_h)_{h \geq 0})](\omega)
= E_{X_t(\omega)}[\prod_{i=1}^{n+1} \chi_{ \{ X_{t_i} \in B_i \} } ] \\
& = E_{X_t(\omega)}[\prod_{i=1}^{n} \chi_{ \{ X_{t_i} \in B_i \} }
E_{X_t(\omega)} [ \chi_{ \{ X_{t_{n+1}} \in B_{n+1} \} } \mid X_{t_1}, \dots, X_{t_n} ] ] \\
& = E_{X_t(\omega)}[\prod_{i=1}^{n} \chi_{ \{ X_{t_i} \in B_i \} }
E_{X_t(\omega)} [ \chi_{ \{ X_{t_{n+1}} \in B_{n+1} \} } \mid X_{t_n} ] ] \\
& = E_{X_t(\omega)}[\prod_{i=1}^{n} \chi_{ \{ X_{t_i} \in B_i \} }
E_{X_{t_n}} [ \chi_{ \{ X_{t_{n+1}} \in B_{n+1} \} } ] ]
\end{align*}
where we again applied the Markov property and the time-homogeneity property.
Hence, both sides of $(*)$ coincide for $\chi_{A'}$ and the induction step is proved. By the monotonic class theorem, $(*)$ holds for all bounded measurable functions $f$ on $E^{[0, \infty)}$.
