= Count-weighted Laplace functional of a Poisson random measure
{title2=$\mathbb E[M(A)e^{-M(f)}]=\mathbb E[e^{-M(f)}]\int_Ae^{-f}\,d\mu$}
Let $M$ be a <Poisson random measure> of intensity $\mu$, let $\mu(A)<\infty$, and let $f\geq0$ be <measurable>. Apply the <Laplace functional of a Poisson random measure> to $f+t\mathbf1_A$ 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 $\mathbf1_A$. 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.
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