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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.

Let $\sigma$ denote surface measure on a sphere on $S^2$ and $\nu$ volume measure on the unit ball. The Poisson mapping theorem can be proved directly here: counts of image points in disjoint sets are counts of the original process in their disjoint preimages. They are therefore independent Poisson variables, and their means are the volumes of those preimages.

For the radius map, a measurable set $A\subset [0,1]$ has meanThis follows from spherical coordinates or by subtracting concentric-ball volumes. The disjoint-preimage argument proves that the radius image is a Poisson process with this mean measure of a point process. There is no mass at zero or at one.

The original process has no point at the origin almost surely, since its mean count there is zero. Thus division by $r$ in the direction map is well-defined almost surely. For a measurable surface set $C\subset S^2$, spherical coordinates giveAgain disjoint surface sets have disjoint cone preimages, giving the required independent Poisson counts. Both pushforward measures are diffuse, so there are almost surely no image collisions; they describe the indicated point sets as well as counting measures.

The two image processes are not independent. They have exactly the same total number of points,Indeed,This alone disproves process independence.

More generally, split the original points into those whose radius lies in $A$ and direction lies in $C$, those satisfying only the first condition, and those satisfying only the second. Counts in these three disjoint regions are independent. The count shared by both images is Poisson with mean $\sigma (C){\int }_A{r}^{2}dr$, soThe fact that Poisson polar projections share their total count explains the dependence despite the product polar intensity $r^{2}dr\sigma (d\omega )$. Conditional on $N$, the points are independent uniform samples from the ball. Their radius density is $3r^2$ and their direction law is $\sigma /(4\pi )$, independently for each point. Thus the two image configurations are independent conditional on $N$; unconditionally the shared nonconstant count couples them.

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.