Characterization for space-homogeneous markov processes i need some help with a proof of the theorem, which states that a $S$ valued stochastic process is space-homogeneous $\mathcal{F}$-Markov process iff it has $\mathcal{F}$-independent increments, where $S$ is a measurable abelian group. $\mathcal{S}$ will denote the $\sigma$-algebra on $S$.
I have problems to understand the "<=" part. The author assumes $X_t-X_s$ is independent of $\mathcal{F}_s$ with distribution $\mu_{s,t}$ and defines the kernel $\mu_{s,t}$ by $\mu_{s,t}(x,B)=\mu_{s,t}(B-x)$ for $B\in\mathcal{S}$. 
Now he states that $$P[X_t\in B \mid \mathcal{F}_s]=P[X_t-X_s\in B-X_s \mid\mathcal{F}_s]=\mu_{s,t}(B-X_s)=\mu_{s,t}(X_s,B)$$ holds by disintegration.
So, basically, a corolarry of the disintegration theorem is that
$$E[f(\xi,\nu \mid \nu)=\int \mu(\nu,du)f(u,\nu)$$ holds where $\mu(\nu,.)=E[f(\xi,\nu)]$ for some probability kernel $\mu$.
The second equation follows now in my oppinion because we have $$\begin{align*} P[X_t-X_s\in B-X_s \mid \mathcal{F}_s] &=P[X_t-X_s\in B-X_s \mid X_s] \\ & =E[1_{X_t-X_s\in B-X_s} \mid X_s] \\ & =\int \mu_{s,t}(X_s,du)1_{u\in B-X_s}(u,X_s) \\ & =\mu_{s,t}(X_s,B+X_s)=\mu_{s,t}(B) \end{align*}$$
and it remains to prove that $P[X_t-X_s\in \cdot \mid X_s]=\mu_{s,t}(X_s,\cdot)$:
$$\begin{align*} P[X_t-X_s\in \cdot  \mid X_s]& =P[X_t-X_s\in \cdot  \mid\mathcal{F}_s] \\ &=P[X_t-X_s\in \cdot]=\mu_{s,t}(.) \end{align*}$$ which is in general not equal to $\mu_{s,t}(.-X_s)$ in my oppinion.
 A: As far as I can see, you are confusing some things. Let me write up the disintegration result in a different way.

Let $X$ and $Y$ be two random variables and $\mathcal{F}$ a sub-$\sigma$-algebra. Suppose that $X$ is independent from $\mathcal{F}$ and that $Y$ is $\mathcal{F}$-measurable. Then $$\mathbb{E}(h(X,Y) \mid \mathcal{F})(\omega) = \mathbb{E}(h(X,y)) \bigg|_{y=Y(\omega)} \quad \text{almost surely}$$ for any "nice" function $h$.

Applying this theorem with $X := X_t-X_s$, $Y := X_s$, $\mathcal{F} := \mathcal{F}_s$ and $h(x,y) := 1_B(x+y)$, we find
$$\begin{align*} \mathbb{P}(X_t \in B \mid \mathcal{F}_s) &= \mathbb{P}((X_t-X_s)+X_s \in B \mid \mathcal{F}_s)  \\ &= \mathbb{P}(X_t-X_s +y \in B) \bigg|_{y=X_s} \\ &= \mathbb{P}(X_t-X_s \in B-y) \bigg|_{y=X_s}. \tag{1}\end{align*}$$
We can rewrite this using $\mu_{s,t}$:
$$\mathbb{P}(X_t \in B \mid \mathcal{F}_s) = \mu_{s,t}(B-y) \bigg|_{y=X_s}.$$
Since
$$\mu_{s,t}(B-y) = \mu_{s,t}(y,B)$$
by the very definition of $\mu_{s,t}(y,\cdot)$, we find
$$\mathbb{P}(X_t \in B \mid \mathcal{F}_s) = \mu_{s,t}(y,B) \bigg|_{y=X_s}.$$
This finishes the proof. 
The point where you are starting to confuse things is in $(1)$. Note that the expression in $(1)$ 
$$\mathbb{P}(X_t-X_s \in B-y) \bigg|_{y=X_s(\omega)} = \int_{\Omega} 1_{\{B-y\}}(X_t(\omega')-X_s(\omega')) \, d\mathbb{P}(\omega') \bigg|_{y=X_s(\omega)}$$
does not equal
$$\mathbb{P}(X_t-X_s \in B-X_s) = \int_{\Omega} 1_{\{B-X_s(\omega')\}}(X_t(\omega')-X_s(\omega')) \, d\mathbb{P}(\omega').$$
