"To every Q-matrix $q$ corresponds a unique Markov process." I'm trying to understand Klenke's proof of the "existence" part of this proposition, namely that given a Q-Matrix $q$, there exists a Markov process $\mathfrak{X}$, whose Q-Matrix is $q$. (Theorem 17.25, see below)
Listed below is the beginning of his proof (the rest, which is irrelevant to my current question, can be found here). What i fail to figure out is how to show that $\mathfrak{X}$ (defined in the proof below) is a Markov process. I recalled Klenke's definition of a Markov process in a previous post.
Relevant definitions
Let $E$ be a non-empty, countable set.
A Q-Matrix in $E$ is a function $q:E\times E\rightarrow\mathbb{R}$ such that
i) $q(e,d)\geq0$ for all $e,d\in E$,
ii) $q(e,e)=-\sum_{e\neq d}q(e,d)$,
iii) $0<\lambda:=\sup_{e\in E}|q(e,e)|<\infty$
Given an $E$-valued stochastic process $X=(X_t)_{t\in[0,\infty)}$ and a function $q:E\times E\rightarrow\mathbb{R}$, $q$ is the Q-matrix of $X$ iff for all $e,d\in E$ $\lim_{t\downarrow0}\frac{1}{t}\left(\mathrm{P}_e[X_t=d]-\delta_{e,d}\right)=q(e,d)$ ($\delta$ is Kronecker's delta)
Theorem 17.25 $q$ is a Q-matrix $\implies$ $q$ is the Q-matrix of a unique Markov process.
Proof (This is only the beginning of Klenke's proof; the rest can be found here.)
Let $I$ be the unit matrix on $E$. Define $p(e,d):=\frac{1}{\lambda}q(e,d)+I(e,d)\space\space\mathrm{for\, }e,d\in E.$
Then $p$ is a stochastic matrix and $q=\lambda(p-I)$.
Let $\left(Y=(Y_n)_{n\in\mathbb{N}_0},(\mathrm{P}_e^Y)_{e\in E}\right)$ be a discrete Markov chain over the measurable space $S_Y=(\Omega_Y,\mathcal{A}_Y)$ with transition matrix $p$ and let $\left(T=(T_t)_{t\geq0},(\mathrm{P}_n^T)_{n\in\mathbb{N}_0}\right)$ be a Poisson process over the measurable space $S_T=(\Omega_T,\mathcal{A}_T)$ with rate $\lambda$. We may assume w.l.g. that $Y$ and $T$ are defined over the product space $S_Y\otimes S_T$.
Set $X_t:=Y_{T_t}$ and $\mathrm{P}_e:=\mathrm{P}_e^Y\otimes\mathrm{P}_0^T$. Then $\mathfrak{X}:=\left(X=(X_t)_{t\geq0},(\mathrm{P}_e)_{e\in E}\right)$ is a Markov process and $p_t(e,d):=\mathrm{P}_e[X_t=d]=\sum_{n=0}^\infty\mathrm{P}_0^T[T_t=n]\mathrm{P}_e^Y[Y_n=d]=e^{-\lambda t}\sum_{n=0}^\infty \frac{\lambda^nt^n}{n!}p^n(e,d)$