= Solution
A <Poisson random measure> with intensity the <sigma-finite measure> $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable <random variable>. For every measurable $A$ with $\mu(A)<\infty$,
$$
\mathbb P(M(A)=j)=e^{-\mu(A)}\frac{\mu(A)^j}{j!},\qquad j=0,1,2,\ldots;
$$
and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are <independent random variables>. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu(A)=0$ implies $M(A)=0$ <almost surely>. The intensity identity is $\mathbb EM(A)=\mu(A)$, also in the extended sense.
For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite <almost surely>, simultaneously for all $m$, so $M$ is a <sigma-finite measure> <almost surely>. If $\mu(A)=\infty$, then $\mu(A\cap E_m)\to\infty$, and for any fixed integer $K$,
$$
\mathbb P(M(A)\leq K)\leq\mathbb P(M(A\cap E_m)\leq K)
=e^{-\mu(A\cap E_m)}\sum_{j=0}^K\frac{\mu(A\cap E_m)^j}{j!}\longrightarrow0.
$$
Hence $M(A)=\infty$ <almost surely>. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general <measurable space>, not just on bounded subsets of Euclidean space.
Back to article page
