If we can show that $P^x \left[X_{t+s} \in \Gamma \mid \mathcal{F}_s\right]$ has a $\sigma(X_s)$-msb version then why is it true that $P^x \left[X_{t+s} \in \Gamma \mid \mathcal{F}_s\right]=P^x \left[X_{t+s} \in \Gamma \mid X_s\right], \text{ }P^x\text{-a.s}$?
Why does does establishing the measurability imply that $P^x \left[X_{t+s} \in \Gamma \mid \mathcal{F}_s\right]$ has to be of the form $P^x\left[X_{t+s} \in \Gamma \mid X_s\right]$.
I was reading proposition 2.5.13 in Karatzas and Shreve where they state this fact
You have to assume that $X_t$ is $\mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} \in \Gamma |\mathcal F_s]$ which is measurable w.r.t. $\sigma (X_s)$. To show that $Y=P[X_{t+s} \in \Gamma | X_s]$ all that you have to show is $P[X_{t+s} \in \Gamma, A]=\int_A Y \, dP$ for all $A \in \sigma (X_s)$ which is immediate because $A \in \mathcal F_s$.