Let $X = (X_t)_{t\geq 0} $ be a Markov process adapted to a filtration $\mathbb F = (\mathcal F_t)_{t\geq 0}$ on some probability space. I would like to understand the Markov property
$$ \mathbb E[u(X_{t+s})|\mathcal F_s] = \mathbb E[u(X_t)| X_s], \quad \text{for all} \quad s,t \geq 0, u \quad \text{bounded measurable}. $$
In Rene Schilling's book "Brownian Motion", 3rd edition, page 64 he wrote: "Since we can express $\mathbb E[u(X_t)| X_s]$ as an (essentially unique) function of $X_s$, say $g_{u,s,t+s}(X_s)$, we have
$$ \mathbb E[u(X_{t+s})|\mathcal F_s] = \mathbb E^{s,X_s}[u(X_{t+s})], \quad \text{with} \quad \mathbb E^{s,x}[u(X_t)]:= g_{u,s,t+s}(\cdot).$$
Note that we can always select a Borel measurable version of $g_{u,s,t+s}(\cdot)...."$
Here, I understand that the conditional expection $E[u(X_t)| X_s]$ is always required to be $\sigma(X_s)$-measurable, hence (keeping this version of condition expectation) by a well known theorem there is a measurable function $g_{u,s,t+s}(\cdot)$ such that $E[u(X_t)| X_s] = g_{u,s,t+s}(X_s)$, this is okey. However, I dont get what he meant by "essentially unique". Of course, if all the realizations of $X_s$ is $\mathbb R$ then clearly $g$ is unique (for this fixed version). But what if the realizations of $X_s$ is not $\mathbb R$, say it's only a subset $A$ of $\mathbb R$? When this happens, we can always modify $g$ on the $\mathbb R \setminus A$ to have another $g'$ different from $g$.
Moreover, what does he mean by the last sentence that we can always choose a Borel $g$? Isn't it already Borel measurable? Thank you for all your help :).