A problem on equivalent definitions of Markov property Suppose that $X, (\Omega,\mathcal{F}),\{P^x\}_{x \in \mathbb{R}^d}$ is a Markov family with shift operators $\{\theta_s\}_{s \ge 0}$ and for every $x \in \mathbb{R}^d,s \ge 0, G \in  \mathcal{F}_s$ and $F \in  \mathcal{F}^X_\infty$ we have
\begin{equation}\label{c''}
P^x[G \cap \theta_s^{-1} F \mid X_s]=P^x[G \mid X_s]P^x[\theta_s^{-1}F \mid X_s]
\end{equation}
Then show that the above implies 
 $P^x[\theta_s^{-1} F \mid \mathcal{F}_s]=P^x[\theta_s^{-1} F \mid X_s] \text{ , } P^x \text{-a.s.}$
This is part of exercise 2.5.17 in Karatzas and Shreve's Brownian motion and stochastic Calculus.
I could manage to show the converse, i.e $P^x[\theta_s^{-1} F \mid \mathcal{F}_s]=P^x[\theta_s^{-1} F \mid X_s] \text{ , } P^x \text{-a.s.}$ implies $$P^x[G \cap \theta_s^{-1} F \mid X_s]=P^x[G \mid X_s]P^x[\theta_s^{-1}F \mid X_s]$$ as follows:
Using properties of conditional expectation we get 
\begin{equation*}
\begin{split}
P^x[G \cap \theta_s^{-1} F \mid X_s]=E^x[ \mathrm{1}_{ \{G \cap \theta_s^{-1} F \}} \mid X_s]=E^x[ \mathrm{1}_{ G} \mathrm{1}_{  \theta_s^{-1} F } \mid X_s]\\=E^x[ E^x[ \mathrm{1}_{ G} \mathrm{1}_{  \theta_s^{-1} F } \mid \mathcal{F}_s] \mid X_s]=E^x[   \mathrm{1}_{ G} E^x[ \mathrm{1}_{  \theta_s^{-1} F } \mid \mathcal{F}_s] \mid X_s]\\= E^x[   \mathrm{1}_{ G} E^x[ \mathrm{1}_{  \theta_s^{-1} F } \mid X_s] \mid X_s]= E^x[ \mathrm{1}_{  \theta_s^{-1} F } \mid X_s] E^x[   \mathrm{1}_{ G}  \mid X_s]\\=P^x[G \mid X_s]P^x[\theta_s^{-1}F \mid X_s]
\end{split}
\end{equation*}
I tried to prove the other direction similarly but it seems that the same technique doesnt workk. Can you give me a hint on how could I go about proving it? I do not want a complete answer as it would defeat the purpose of the exercise.
 A: Recall the following characterization of the conditional expectation:

Let $X \in L^1(\mathbb{P})$ and let $\mathcal{F}$ be a $\sigma$-algebra. An $\mathcal{F}$-measurable random variable $Y \in L^1(\mathbb{P})$ equals almost surely $\mathbb{E}(X \mid \mathcal{F})$ if, and only if, $$\forall G \in \mathcal{F}: \quad \int_G Y \, d\mathbb{P} = \int_G X \, d\mathbb{P}.$$

If we apply this result with $\mathcal{F} := \mathcal{F}_s$, $\mathbb{P}=P^x$, $$X := 1_{\theta_s^{-1}F} \qquad Y := P^x(\theta_s^{-1} F \mid X_s)$$ we find that $P^x(\theta_s^{-1} F \mid \mathcal{F}_s) =  P^x(\theta_s^{-1} F \mid X_s)$ iff $$\forall G\in \mathcal{F}_s: \quad \int_G P^x(\theta_s^{-1} F \mid X_s) \, dP^x = \int_G 1_{\theta_s^{-1} F} \, dP^x.$$
Use the Markov property to verify this condition. Hint: Start  with the right-hand side,
$$\int_G 1_{\theta_s^{-1} F} \, dP^x= E^x \left(1_G 1_{\theta_s^{-1} F}\right) = E^x \left( E^x \big(1_G 1_{\theta_s^{-1} F} \mid X_s \big) \right) = \dots$$ 
Solution:

 \begin{align*}\int_G 1_{\theta_s^{-1} F} \, dP^x= E^x \left(1_G 1_{\theta_s^{-1} F}\right) &\stackrel{\text{tower}}{=} E^x \left( E^x \big(1_G 1_{\theta_s^{-1} F} \mid X_s \big) \right) \\ &\stackrel{\text{Markov}}{=} E^x \left( E^x \big(1_G  \mid X_s \big) E^x(1_{\theta_s^{-1} F} \mid X_s) \right) \\ &\stackrel{\text{pull out}}{=} E^x \left( E^x \big(1_G  E^x(1_{\theta_s^{-1} F} \mid X_s) \mid X_s \big)  \right) \\ &\stackrel{\text{tower}}{=} E^x \left( 1_G   E^x(1_{\theta_s^{-1} F}  \mid X_s) \right) \\ &= \int_G E^x(1_{\theta_s^{-1} F}  \mid X_s) \, dP^x \end{align*} 

