B_{\mathbf{s}} \cap A \cap A_{s_0} )}{
    {a(n)} \sP( \mathbf{X} / a(n) \in A)} \\
  \label{eq:01}
  & \longrightarrow 
  \frac{\mu(B_{\mathbf{s}} \cap A \cap A_{s_0})}{\mu(A)},\quad (n\to\infty),\end{aligned}$$ where the convergence follows from , as long as $\mu\{ \partial (B_{\mathbf{s}} \cap A \cap A_{s_0})\} = 0$.\
Let $$\begin{aligned}
    \label{def_xi3}
    \xi^{(1)}(t) = \max_{i\in{\mathbb N}} U_i^{(1)} V^{(1)}_i(t),
    \quad t\in T,
  \end{aligned}$$ where $\sum_{i\in{\mathbb N}} \delta_{U_i^{(1)}}$ is a Poisson point process with intensity $(1-p)u^{-2}\rd u$ and let $\mu^{(1)}$ be the exponent measure of the associated max-stable random vector $(\xi^{(1)}(t_0), \ldots, \xi^{(1)}(t_k))$. Then the choice $A =
  (1,\infty)\times [0,\infty)^k$ guarantees that $\mu(\cdot \cap A) =
    \mu^{(1)}(\cdot \cap A)$. Comparing the construction of $\xi^{(1)}$ in with the definition of the exponent measure, we see that $\mu^{(1)}$ is the intensity measure of the Poisson point process $\sum_{i\in{\mathbb N}} \delta_{(U_i^{(1)}
      V_i^{(1)}(t_0),\, \ldots,\, U_i^{(1)} V_i^{(1)}(t_k))}$ on $E$. Hence, $$\begin{aligned}
    \mu(A) &= \int_0^\infty (1-p)u^{-2} \sP(u V^{(1)}(t_0) > 1) \rd u \notag\\    
    &= (1-p)\int_0^\infty u^{-2} \int_{[u^{-1}, \infty)} 
      \sP(V^{(1)}(t_0) \in \rd y) \rd u \notag\\
      &= (1-p)\int_0^\infty y \sP(V^{(1)}(t_0) \in \rd y)
      = (1-p)\sE V^{(1)}(t_0) = 1,
  \end{aligned}$$ where the last equality follows from $\sE V^{(1)}(t_0) = \sE
  V(t_0)/(1-p)$. Furthermore, for $s_0\geq 1$ and ${\mathbf s}\in[0,\infty)^k$ with $\sP(\Delta\mathbf{\tilde V}^{(1)} \in \partial [{\mathbf 0},{\mathbf s}])=0$, $$\begin{aligned}
    &\mu(B_{\mathbf{s}} 