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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.

The intensity function of a point process is positive and continuous inside the interval, so its measure is finite on every compact subinterval. Consequently the Poisson point process is locally finite there. Its nonatomic intensity makes it simple and gives no point at $0$ almost surely. At the endpoints,Both half-intervals therefore have infinite intensity. More explicitly, exhaust either half by increasing compact intervals with means $t_j\to \infty$. For any fixed integer $k$,Taking the countable union over $k$ proves infinitely many points on each side almost surely.

Local finiteness excludes accumulation in the interior, including at $0$. Hence there is a largest negative point and a smallest positive point. For example, given any negative point, the compact interval between it and $0$ contains finitely many points and therefore a largest one. Call it $X_0$; call the smallest positive point $X_1$, and enumerate consecutively in both directions. Infinite counts give every integer index, and the lack of interior accumulation implies $X_n\uparrow 1$ as $n\to \infty$ and $X_n\downarrow -1$ as $n\to -\infty$. Thus the specified two-sided labeling exists almost surely.

Use the cumulative-intensity transformationIt has $f(0)=0$, positive derivative, and limits $-\infty ,+\infty$ at the two endpoints, so it is an increasing bijection onto $\mathbb{R}$. One explicit expression isFor any bounded interval $(a,b)$ in the target, its preimage has intensityDisjoint target sets have disjoint preimages, so their counts are independent. This proves directly, or by the Poisson mapping theorem, that the transformed points form a unit-rate Poisson process on the real line.

Put $Y_n=f(X_n)$. On the positive half-line, the successive spacings are independent exponential random variables of mean $1$, so $Y_n=E_1+\cdots +E_n$. The strong law of large numbers gives $Y_n/n\to 1$ almost surely. On the negative half-line, independently, the reflected points satisfy $-Y_{-k}=E_1^{\prime }+\cdots +E_{k+1}^{\prime }$, because $Y_0$ is the first point to the left of zero. Thus $-Y_{-k}/k\to 1$, or $Y_n/n\to 1$ also as $n\to -\infty$. The shift by one has no effect on these limits.

From the explicit formula, or by integrating the endpoint intensity asymptotics,Substitute $x=X_n$ and use the strong law for Poisson arrival times. At the right endpoint,soAt the left endpoint the corresponding result isEquivalently $8|n|(1+X_n)\to 1$. The different powers of $|n|$ reflect the different endpoint singularity orders, an instance of power-law boundary accumulation of a Poisson point process.

For every Borel set $A\subseteq {\mathbb{R}}^d$, letbe its random point count. A Non-homogeneous Poisson point process with intensity function of a point process $\lambda$ is characterized by:

Because $\Lambda$ has a density with respect to Lebesgue measure, it is atomless, so the process is simple: almost surely no location occurs more than once.

A Non-homogeneous Poisson point process with intensity function $\lambda$ is a random counting measure $N$ such that, for every finite family of pairwise disjoint bounded Borel sets $A_1,\dots ,A_k$, the counts $N(A_1),\dots ,N(A_k)$ are independent random variables andThe assumed local integrability makes each Poisson parameter finite. This count law and independence characterize the process.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

If an intensity function of a point process has $\lambda (x)\sim c(b-x{)}^{-\alpha }$ at a finite endpoint, with $c>0$ and $\alpha >1$, the cumulative intensity satisfies $F(x)\sim c(b-x{)}^{1-\alpha }/(\alpha -1)$. The Poisson mapping theorem turns $F(X_n)$ into unit-rate arrivals. The strong law for Poisson arrival times gives $F(X_n)/n\to 1$ almost surely, hence the displayed accumulation rate. Reversing the coordinate gives the same argument at a left endpoint. Different singularity exponents produce different powers of the index.

A point process on ${\mathbb{R}}^d$ is stationary if every translation preserves its probability law. If its mean count in a unit cube is finite, its intensity measure is $\mu (A)=\rho |A|$ for some $\rho \ge 0$, where $|A|$ denotes Lebesgue measure. This follows because stationarity makes the mean measure invariant under translations. A constant intensity function of a point process describes only these mean counts, and by itself asserts neither stationarity of the entire law nor independence.^[ref]