Count-weighted Laplace functional of a Poisson random measure ($\mathbb{E}[M(A)e^{-M(f)}]=\mathbb{E}[e^{-M(f)}]{\int }_A{e}^{-f}d\mu$)

Let $M$ be a Poisson random measure of intensity $\mu$, let $\mu (A)<\infty$, and let $f\ge 0$ be measurable. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$ and differentiate from the right at $t=0$. On the random side, differentiation is dominated by the integrable random variable $M(A)$; on the intensity side, by $1_A$. This proves the displayed identity when $\int (1-e^{-f})d\mu$ is finite. When this integral is infinite, $e^{-M(f)}=0$ almost surely, and the same identity holds with both sides zero.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.