Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $(E,\mathcal E)$ be a measurable space;
- $\delta_x$ denote the Dirac measure at $x\in E$;
- $\pi_I$ denote the projection from $E^{[0,\:\infty)}$ onto $I\subseteq[0,\infty)$ and $\pi_t:=\pi_{\{t\}}$ for $t\ge0$;
- $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued Markov process on $(\Omega,\mathcal A,\operatorname P)$ with time-homogeneous transition semigroup $(\kappa_t)_{t\ge0}$.
Using Kolmogorov's extension theorem, we can show that if $E$ is a Polish space and $\mathcal E=\mathcal B(E)$, then there is a Markov kernel $\kappa$ with source $(E,\mathcal E)$ and target $(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)})$ such that $$\kappa(x,\;\cdot\;)\circ\pi_{t_0,\:\ldots\:,\:t_n}^{-1}=\delta_x\otimes\bigotimes_{i=1}^n\kappa_{t_n-t_{n-1}}\tag1$$ for all $x\in E$, $n\in\mathbb N_0$ and $0=t_0<\cdots<t_n$.
Now assume that $E$ is a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ is a Lévy process. We easily see that $$\kappa_t(x,B)=\operatorname P\left[X_t\in B\right]\;\;\;\text{for all }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0\tag2.$$
Question: Rather than invoking the abstract general result above (which also would force us to assume completeness and separability of $E$), aren't we able to conclude the existence of a kernel $\kappa$ with $(1)$ by simply defining $$\kappa(x,B):=\operatorname P\left[X_t+x\in B\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)^{\otimes[0,\:\infty)}\text{ and }t\ge0;$$ or am I missing something?
The only thing we should need to show is that $\kappa$ is actually a transition kernel. While $\kappa(x,\;\cdot\;)$ is clearly a well-defined probability measure on $\mathcal B(E)^{\otimes[0,\:\infty)}$ for all $x\in E$, how do we obtain the $\mathcal B(E)$-measurability of $\kappa(\;\cdot\;,B)$ for all $B\in\mathcal B(E)^{\otimes[0,\:\infty)}$? Might this be the property which fails to hold?