The Probability Generating Functional of a Poisson Point Process In [1], the author wrote:

If $\mathit{\Xi}$ is a Poisson point process (PPP) with intensity function $\lambda(x)$, then
  $$
G_{\mathit{\Xi}}(f) = \exp \left[ \int_{\cal R} (f(x) - 1) \lambda(x) \, dx \right]. \tag{1}
$$

Here $G_{\mathit{\Xi}}(f)$ is the probability generating functional of $\mathit{\Xi}$. $f(x)$ is a function such that $0 < f(x) \leq 1$.
Does anyone know how to prove (1)? Thank you very much.
References
[1] Roy L. Streit, Poisson Point Processes: Imaging, Tracking, and Sensing, Springer, 2010.
 A: The probability generating functional for a point process $\{N(t)\}_t$ is defined as 
$$
G(f) = \mathbb E \Big[ \exp \int \log f(t) \,d N(t) \Big]
$$
For simplicity, write $N (\log f)  = \int \log f(t) \,d N(t)$.
Consider simple functions $f = \sum_{k \le m} a_k 1_{B_k}$ where $B_1,\dots,B_m$ are disjoint. Note that $N(B_1),\dots,N(B_m)$ are independent Poisson variables with
$$ \mathbb E N(B_k) = \int_{B_k} \lambda(x)\,dx =: \mu_k $$
Let $b_k = \log a_k$ so that $\log f = \sum_k b_k 1_{B_k}$. Then,
\begin{align}
G(f) &= \mathbb E \exp N\Big( \sum_k b_k 1_{B_k} \Big)  \\
&= \mathbb E\exp \sum_k b_k N(B_k) \\
&= \prod_k \mathbb E \exp\big(b_k N(B_k)\big)  \quad \text{(By independence)}\\
&= \prod_k \mathbb \exp( \mu_k(e^{b_k} - 1)) \quad \text{(M.G.F. of a Poisson R.V.)}\\
&= \exp \Big( \sum_k \mu_k(a_k - 1)\Big) \\
&= \exp \Big[ \sum_k \int_{B_k}(a_k - 1) \lambda(x) \,dx\Big] \\
&= \exp \Big[ \sum_k \int_{\mathbb R} 1_{B_k }(x)(a_k - 1) \lambda(x) \,dx\Big] \\
&= \exp \Big[  \int_{\mathbb R} \Big( \sum_k a_k 1_{B_k }(x)  - \sum_k 1_{B_k}(x) \Big) \lambda(x) \, dx\Big]  \\
 &= \exp \Big[ \int_{\mathbb R}(f(x) - 1) \lambda(x)\,dx \Big].
\end{align}
This shows that the identity holds for simple functions. Any $f$ such that $0 < f \le 1$ can be written  as $f = e^{-g}$ for some $g \ge 0$. We can extend to any nonnegative measurable $g$ by monotone convergence theorem, and then to any measurable $g$ by dominated convergence theorem. (In terms of $g$ the expression is the Laplace transform of the process). 
