If $\mathbb E[f\circ X_t\circ\theta_s\mid F_s]= \mathbb E_{X_s}[f\circ X_t]$, can we show $\mathbb E[g\circ\theta_s\mid F_s]=\mathbb E_{X_s}[g]$? Let

*

*$(\Omega,\mathcal A)$ be a measurable space;

*$(E,\mathcal E)$ be a measurable space;

*$\pi$ be a Markov kernel with source $(E,\mathcal E)$ and target $(\Omega,\mathcal A)$;

*$\operatorname P_\mu:=\mu\pi$ for every probability measure $\mu$ on $(E,\mathcal E)$ and $\operatorname P_x:=\operatorname P_{\delta_x}=\pi(x,\;\cdot\;)$ for $x\in E$;

*$(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;

*$(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$ with $$\operatorname P_x\left[X_0=x\right]=1\tag1$$ and $$\operatorname E_x\left[f(X_{s+t})\mid\mathcal F_s\right]=(\kappa_tf)(X_s)\tag2$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $s,t\ge0$;

*$\theta_t:\Omega\to\Omega$ for $t\ge0$ with $$X_t\circ\theta_s=X_{s+t}\tag3$$ for all $s,t\ge0$.

Note that $$\operatorname E\left[f(X_t)\right]=(\kappa_tf)(x)\tag4$$ for all $t\ge0$ and $$\operatorname E_x\left[f\circ X_t\circ\theta_s\mid\mathcal F_s\right]=(\kappa_tf)(X_s)=\operatorname E_{X_s}\left[f\circ X_t\right]\tag5$$ for all $s,t\ge0$ for all $x\in E$ and bounded $\mathcal E$-measurable $f:E\to\mathbb R$.
I would like to show $$\operatorname E_\mu\left[g\circ\theta_s\mid\mathcal F_s\right]=\operatorname E_{X_s}[g]$$ for all bounded $\mathcal F^X_\infty$-measurable $g:\Omega\to\mathbb R$, where $$\mathcal F^X_\infty:=\sigma(X_t,t\ge0).$$
It is clearly sufficient to show that the claim is true for every finite $I\subseteq[0,\infty)$ and $\mathcal F^X_I$-measurable $g:\Omega\to\mathbb R$, where $\mathcal F^X_I:=\sigma(X_t,t\in I)$.
If $|I|=1$, then the claim is easy to verify using $(5)$. So, assume the claim is true for $|I|=n$ and consider $|I|=n+1$. Then $$I=\underbrace{\{t_1,\ldots,t_n\}}_{=:\:\tilde I}\cup\{t_{n+1}\}$$ for some $0\le t_1<\cdots<t_{n+1}$. Consider $g:=1_A$ for some $A\in\mathcal F^X_I$. Then, $$A=\underbrace{X_{\tilde I}^{-1}(B_1\times\cdots\times B_n)}_{=:\:\tilde A\:\in\:\mathcal F^X_{\tilde I}\:\subseteq\:\mathcal F_{t_n}}\cap X^{-1}_{t_{n+1}}(B_{n+1})\tag6$$ and hence \begin{equation}\begin{split}\operatorname E_\mu\left[1_A\circ\theta_s\mid\mathcal F_s\right]&=\operatorname E_\mu\left[\left(1_{\tilde A}\circ\theta_s\right)\left(1_{B_{n+1}}\circ X_{t_{n+1}}\circ\theta_s\right)\mid\mathcal F_s\right]\\&=\operatorname E_\mu\left[\underbrace{\left(1_{\tilde A}\circ\theta_s\right)\underbrace{\operatorname E_\mu\left[1_{B_{n+1}}\circ X_{t_{n+1}}\circ\theta_s\mid\mathcal F_{s+t_n}\right]}_{=\:\operatorname E_{X_{s+t_n}}\left[1_{B_{n+1}}\:\circ\:X_{t_{n+1}-t_n}\right]}}_{=\:g\:\circ\:\theta_s}\mid\mathcal F_s\right]\\&=\operatorname E_{X_s}[g]\\&=\operatorname E_{X_s}\left[1_{\tilde A}\underbrace{\operatorname E_{X_{t_n}}\left[1_{B_{n+1}}\circ X_{t_{n+1}-t_n}\right]}_{=\:\operatorname E_{X_s}\left[1_{B_{n+1}}\:\circ\:X_{t_n}\mid\mathcal F_{t_n}\right]}\right]\\&\color{red}=\operatorname E_{X_s}\left[1_{\tilde A}\operatorname E_{X_s}\left[1_{B_{n+1}}\circ X_{t_n}\right]\right],\end{split}\tag7\end{equation} where $$g:=1_{\tilde A}\operatorname E_{X_{t_n}}\left[1_{B_{n+1}}\circ X_{t_{n+1}-t_n}\right]$$ is clearly $\mathcal F^X_{\tilde I}$-measurable and hence the induction hypotheses can be used. I'm not sure whether the last equality in $(7)$ really holds (I've asked a separate question for this). However, even if it does, how do we show that it is equal to $$\operatorname E_{X_s}\left[1_{\tilde A}1_{B_{n+1}}\left(X_{t_{n+1}}\right)\right]=\operatorname E_{X_s}[1_A]\tag8?$$
 A: I'll give an alternative proof that makes use of the kernel $\kappa$. I'm just copying the proof for the countable time case and making the proper notational adjustments.
We are given a bounded measurable function $g : E^{[0, \infty)} \to \mathbb{R}$ and a probability measure $\mu$ and $s \in [0, \infty)$, and we want to show that
$$E_\mu(g(X_{s + \cdot}) \mid \mathcal{F}_s) = E_{X_s}(g(X)).$$
By approximation arguments, it is sufficient to show this for $g = 1_{A}$ with $A \in \mathcal{E}^{\otimes [0, \infty)}$. By the $\pi$-$\lambda$ theorem, it suffices to consider the case where $g = 1_A$ depends on finitely many coordinates $t_0 = 0 < t_1 < \dots < t_n \in [0, \infty)$. Now suppose $f$ is bounded and $\mathcal{E}^{\otimes [0, s]}$-measurable and only depends on finitely many coordinates $s_0 = 0 < \dots < s_m = s$. Then
\begin{align}
E_{\mu}(f(X)g(X_{s + \cdot})) &= E_{\mu}(f(X_0, X_{s_1}, \dots, X_{s})g(X_{s}, X_{s + t_1}, \dots, X_{s + t_n})) \\
&= \int_{E} \mu(dx_0)\int_{E}\kappa_{s_1}(x_0, dx_{s_1})\dots\int_{E}\kappa_{s - s_{m - 1}}(x_{s_{m - 1}}, dx_s)\int_{E}\kappa_{t_1}(x_s, dx_{s + t_1})\dots\int_{E}\kappa_{t_n - t_{n - 1}}(x_{s + t_{n - 1}}, dx_{s + t_n})f(x_0, x_{s_1}, \dots, x_{s})g(x_{s}, x_{s + t_1}, \dots, x_{s + t_n}) \\
&= \int_{E} \mu(dx_0)\int_{E}\kappa_{s_1}(x_0, dx_{s_1})\dots\int_{E}\kappa_{s - s_{m - 1}}(x_{s_{m - 1}}, dx_s)f(x_0, x_{s_1}, \dots, x_{s})\int_{E}\kappa_{t_1}(x_s, dx_{s + t_1})\dots\int_{E}\kappa_{t_n - t_{n - 1}}(x_{s + t_{n - 1}}, dx_{s + t_n})g(x_{s}, x_{s + t_1}, \dots, x_{s + t_n}) \\
&= \int_{E} \mu(dx_0)\int_{E}\kappa_{s_1}(x_0, dx_{s_1})\dots\int_{E}\kappa_{s - s_{m - 1}}(x_{s_{m - 1}}, dx_s)f(x_0, x_{s_1}, \dots, x_{s})E_{x_s}(g(X_0, \dots, X_{t_n})) \\
&= \int_{E} \mu(dx_0)\int_{E}\kappa_{s_1}(x_0, dx_{s_1})\dots\int_{E}\kappa_{s - s_{m - 1}}(x_{s_{m - 1}}, dx_s)f(x_0, x_{s_1}, \dots, x_{s})E_{x_s}(g(X)) \\
&= E_{\mu}(f(X_0, X_{s_1}, \dots, X_{s})E_{X_s}(g(X))) \\
&= E_{\mu}(f(X)E_{X_s}(g(X))).
\end{align}
By the $\pi$-$\lambda$ theorem, the above holds for $f = 1_{B}$ for every $B \in \mathcal{E}^{\otimes [0, s]}$. This completes the proof.
