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A sigma-finite measure on a standard Borel space is the intensity measure of a point process with Poisson counts. Partition the space into sets of finite intensity, draw an independent Poisson count on each piece, and conditional on that count draw independent points from its normalized intensity measure. The sum of their Dirac measures is the required Poisson random measure. Non-atomic intensity gives simplicity; local finiteness of the intensity gives local finiteness of the point configuration.

The intensity measure of a point process $\Pi$ is its expected counting measure:It may take the value $+\infty$.^[ref] Countable additivity follows by applying the monotone convergence theorem to the counts of finite unions of pairwise disjoint observation windows. If $\mu$ is locally finite and absolutely continuous with respect to Lebesgue measure, its Radon-Nikodym derivative is the intensity function of a point process $\lambda$:This is an expected number per unit volume, not a probability density function for a randomly selected point. For example, a single deterministic point has $\mu ={\delta }_0$, a Dirac measure, and no Lebesgue measure density. The density representation requires its own assumption.^[ref]

For a bounded Borel set $W$, $\mathbb{E}[N_W]=\mu (W)$, but local finiteness alone does not ensure that this expected value is finite. Conversely, $\mu (W)<\infty$ implies $N_W<\infty$ almost surely, even when $W$ is unbounded. A constant finite intensity function of a point process on all of ${\mathbb{R}}^d$ has infinite total intensity measure, although every bounded window has finite expected value.

Equip $S$ with a sigma-algebra $\mathcal{S}$. A Poisson random measure with intensity measure of a point process $\mu$ is a random counting measure $\Pi$ such that, for every measurable $A$ with $\mu (A)<\infty$,and the counts on every finite collection of pairwise disjoint measurable sets are independent. Thus $\mathbb{E}\Pi (A)=\mu (A)$. For a sigma-finite measure $\mu$, sets of infinite intensity contain infinitely many points almost surely: exhaust such a set by increasing finite-intensity sets and use the Poisson count tails as their means tend to infinity. The count is a measurable random variable for every measurable $A$, which is part of saying that $\Pi$ is a random measure.

Under the usual point representation write $\Pi ={\sum }_j{\delta }_{Z_j}$, counting multiplicities. If $\mu$ is a non-atomic measure, this is a simple Poisson point process; if $\mu$ has atoms, several point occurrences can have the same location, and each occurrence is counted separately. In a topological setting, local finiteness requires that the intensity be finite on compact sets.

Sufficient existence conditions are that $S$ is a standard Borel space and $\mu$ is a sigma-finite measure. This gives existence of a Poisson random measure on a standard Borel space. If a simple locally finite configuration is desired, additionally take a non-atomic, locally finite intensity on a locally compact second-countable Hausdorff space. These conditions are sufficient, not claimed necessary. In particular sigma-finiteness suffices for a countable family of occurrences that can each be independently coloured.

Let $R$ and $G$ be the red and green counting measures. The marking assumption means that, conditional on $\Pi$, the colours of all point occurrences are independent, with the same red probability $r$. For a measurable set $A$ of finite intensity $m=\mu (A)$, condition on its total count. To obtain $a$ red and $b$ green points the total must be $a+b$, so direct multiplication gives the joint red and green Poisson count formulaThus the two counts are independent Poisson variables with respective means $rm$ and $(1-r)m$. This computation includes zero intensity with the usual convention for the probability of zero counts.

For disjoint finite-intensity sets $A_1,\dots ,A_k$, the original counts are independent. Conditional marking in different sets uses disjoint collections of independent colours. Repeating the preceding calculation therefore givesThis proves the disjoint-count definition for each coloured Poisson point process and independence of their two count vectors.

To obtain independence of the whole processes, take any finite list of red test sets and any finite list of green test sets, initially all of finite intensity. Refine their union into the disjoint cells given by their membership patterns. Every test-set count is the sum of the counts of its cells. The entire red cell vector is independent of the entire green cell vector by the product formula, hence so are the corresponding test-set count vectors. For arbitrary measurable test sets, intersect first with a finite-intensity exhaustion of $S$ and pass to the increasing count limits. The count cylinder events generate the two evaluation sigma-algebras, and independence extended from generating pi-systems gives independence of those sigma-algebras. ConsequentlyThis is a direct count proof of the Poisson thinning theorem; no general marking or thinning result has been used to deduce it.

Represent the raindrops by a Poisson random measure on the space of location, impact time and volume. Physical volume is nonnegative, so take the intensity density to vanish for $v<0$, or equivalently work on ${\mathbb{R}}^2\times \mathbb{R}\times [0,\infty )$ and extend it by zero to ${\mathbb{R}}^4$. Assume the radius function is measurable and nonnegative for elapsed times at least zero. Its stated monotonicity means patches grow rather than dry away.

A drop contributes to wetting $\xi$ at time $\tau$ precisely if its impact time is no later than $\tau$ and its current patch contains $\xi$. Thus define the deterministic measurable setThe number of covering drops is $\Pi (A_{\xi ,\tau })$. Its intensity measure of a point process isSince the point is dry if and only if that count is zero, the Poisson count definition gives the void probability for growing Poisson rain patchesHere $e^{-\infty }=0$: when the intensity is infinite, exhaust the covering set by increasing finite-intensity sets; their decreasing zero-count events have probabilities tending to zero. This also covers storms with infinitely many drops that can reach the point. The formula is valid for spatially inhomogeneous intensity. Only if the intensity is spatially constant may the spatial integral be replaced by $\pi r(\tau -t,v{)}^2$ times that intensity. Whether the circular boundary is included makes no difference for a density with respect to Lebesgue measure, since each such boundary has zero spatial area. This construction is a time-dependent Poisson germ-grain model.

For total rainfall, the required general result is the Campbell first-moment formula: if $\Pi$ is a Poisson random measure with intensity $\mu$ and $h$ is any nonnegative measurable function, thenwith infinite values permitted. For a signed $h$, it holds with finite absolute expectation whenever $\int |h|d\mu <\infty$. No square-integrability hypothesis is required for this first-moment identity. The result is stated rather than proved, as allowed here.

The total rainfall volume is $V_{\mathrm{total}}={\sum }_{(X,T,V)\in \Pi }V=\int v\Pi (dxdtdv)$, counting every drop once, independently of overlaps between its patch and other patches. Apply the stated formula with $h(x,t,v)=v$ on the physical mark space. Extending by zero to negative volume givesIf the displayed nonnegative integral is finite, the total volume is integrable and finite almost surely. If it is infinite, its expectation is infinite; this alone does not imply that the random total is infinite almost surely. The radius affects wet-area coverage but does not change this total-volume expectation.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]