I'm reading Dirichlet Forms and Symmetric Markov Processes by M. Fukushima, Y. Oshima, and M. Takeda. In Appendix A.2, where they discuss the construction of a random variable, there is the statement:
"..., we need a non-negative random variable $Z(\omega)$ on $(\Omega,\mathcal{M},P_{x})$ which is of exponential distribution with mean $1$; independent of $(X_{t})_{t \ge 0}$ under $P_{x}$ for every $x \in S$ satisfying $Z(\theta_{s}(\omega))=(Z(\omega)-s)\vee 0$."
In the statement, $(\Omega, \mathcal{M}, (X_{t})_{t \ge 0}, (P_{x})_{x \in S})$ is a Hunt process with state space $S$ and $(\theta_{t})_{t \ge 0}$ is a shift operator of this Hunt process.
This book says,
"This requirement is fulfilled by replacing $\Omega$ by its direct with $[0,\infty)$ if necessary. To see this, let $Z(\xi)$ be a non-negative random variable on $([0,\infty), \mathcal{B}([0,\infty)), \lambda)$ of exponent distribution with mean $1$; $\lambda(\left\{ \xi \ge 0; Z(\xi) >a \right\})=e^{-a}$, $\forall a \ge 0$. We can consider the functions $X_{t},\theta_{t}$ and $Z$ as function on $\Omega \otimes [0,\infty)$ by putting $X_{t}(\omega,\xi)=X_{t}(\omega), \theta_{t}(\omega ,\xi)=(\theta_{t}(\omega),(\xi-t)^{+})$ and $Z(\omega,\xi)=Z(\xi)$, $\forall (\omega, \xi) \in \Omega \otimes [0,\infty)$. It is clear, under this identification, that $(\Omega \otimes [0,\infty), \mathcal{M} \otimes \mathcal{B}([0,\infty)), X_{t}, P_{x}\otimes \lambda)$ is a Hunt process with admissible filtration $\mathcal{M}_{t}\otimes \mathcal{B}([0,\infty))$ and that the random variable $Z$ has the desired properties."
Question
I don't understand $Z$ satisfies $Z(\theta_{s}(\omega ,\xi))=(Z(\omega,\xi)-s)\vee 0$. By definition, \begin{align} Z(\theta_{s}(\omega,\xi))&=Z(\theta_{s}(\omega), (\xi-s)^{+}) \\ &=Z((\xi-s)^{+}) \\ &=Z(\omega, (\xi-s)^{+} ). \end{align}
How do I get the equation $Z(\theta_{s}(\omega ,\xi))=(Z(\omega,\xi)-s)\vee 0$ ? Doing somethig wrong?
Thank you in advance.
