[0, \infty)^{k}$. Define a random vector ${\mathbf W}=(W^{(1)}, \ldots, W^{(k)})$ via its cumulative distribution function $$\begin{aligned}
    \label{def_W}
    \sP( {\mathbf W}\leq \mathbf{s}) = \mu(B_{\mathbf{s}} \cap A), 
    \quad \mathbf{s}\in [0,\infty)^{k},
  \end{aligned}$$ where $A = (1,\infty)\times [0, \infty)^{k}$ and $B_{\mathbf s} =
    \{{\mathbf x}\in [0,\infty)^{k+1}: \, (x^{(1)},\ldots,x^{(k)}) \leq
      x^{(0)} {\mathbf s}\}$. Then, $\Xi$ allows for an incremental representation with ${\mathbf W}_i$, $i\in{\mathbb N}$, being independent copies of ${\mathbf W}$.

First, we note that indeed defines a valid cumulative distribution function. To this end, consider the measurable transformation $$\begin{aligned}
   T: (0, \infty)\times [0, \infty)^{k} 
     \to (0, \infty)\times [0, \infty)^{k}, \ (x_0,\dots, x_k)
   \mapsto \left(x_0, \frac{x_1}{x_0}, \dots, \frac{x_k}{x_0}\right).
 \end{aligned}$$ Then, $ T(B_{\mathbf{s}} \cap A) = (1, \infty) \times [{\mathbf 0}, {\mathbf s}]$ and the measure $\mu^T(\cdot) = \mu(T^{-1}((1,\infty)\times \,\cdot\,))$ is a probability measure on $[0, \infty)^{k}$. Since $$\begin{aligned}
   \mu(B_{\mathbf{s}} \cap A) 
   = \mu(T^{-1}((1,\infty)\times[{\mathbf 0}, {\mathbf s}]))
   = \mu^T([{\mathbf 0}, {\mathbf s}]),
 \end{aligned}$$ the random vector ${\mathbf W}$ is well-defined and has law $\mu^T$.

By definition of the exponent measure, we have $\Xi \eqdist \max_{i
   \in {\mathbb N}} {\mathbf X}_i$, where $\Pi = \sum_{i\in{\mathbb N}} \delta_{{\mathbf X}_i}$ is a PPP on $E$ with intensity measure $\mu$. Then, the transformed point process $T\Pi = \sum_{i\in{\mathbb N}} \delta_{(X_i^{(0)},\,
   