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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.

The first limit in the PDF has a reciprocal error when $\lambda$ denotes point density. The correct strong law for Poisson arrival times isTo prove it, put $N(t)=\Pi ((0,t])$. The unit-interval increments $N(j)-N(j-1)$ are independent and identically distributed random variables with the Poisson distribution of mean $\lambda$. The strong law of large numbers gives $N(m)/m\to \lambda$ along integers. For $m\le t<m+1$,so $N(t)/t\to \lambda$ for all real $t\to \infty$. The continuous intensity gives distinct points almost surely, and local finiteness and infinitely many points give $Y_n\to \infty$. Since $N(Y_n)=n$, the same pathwise limit along $Y_n$ implies $n/Y_n\to \lambda$, proving the reciprocal limit. For instance at density $2$ the limit is $1/2$, not $2$. The printed limit would instead describe a parameter interpreted as mean interarrival time, but not the stated density.

For the Non-homogeneous Poisson point process on $(0,1)$, every closed subinterval strictly inside $(0,1)$ has finite intensity. The intensity is non-atomic, so the process is simple and has no point at $1/2$ almost surely. Both half-intervals have infinite intensity: the density behaves as $x^{-2}$ at zero and as $(1-x{)}^{-1}$ at one. More explicitly, exhaust either half by finite-intensity intervals of means $m_j\to \infty$. For any fixed integer $K$,Thus there are infinitely many points on each side.

Local finiteness around $1/2$ gives a last point on its left and a first point on its right. To see the former directly, choose any left point; between it and $1/2$ there are only finitely many points, so there is a largest one. Repeat below each chosen point to label the left sequence by $X_{-1},X_{-2},\dots$, and similarly label the right sequence by $X_0,X_1,\dots$. No infinite sequence of points can accumulate in the interior, because a compact neighbourhood there has finite count. ConsequentlyThis constructs the required two-sided order statistics of a boundary-singular Poisson process.

Use its integrated intensity to find the rates. Partial fractions giveThe left cumulative-intensity coordinate isIt maps the left half bijectively onto $(0,\infty )$, reversing its order. The image points form a unit-rate Poisson process directly from the count definition: inverse images of disjoint parameter intervals are disjoint, and their intensity equals their parameter lengths. In particular $L(X_{-k})$ is its $k$th arrival, so the result just proved gives $L(X_{-k})/k\to 1$. Also $xL(x)\to 1$ as $x\downarrow 0$, because $x\log x\to 0$. HenceThis is exactly the requested negative-index limit.

For the right side, the increasing coordinate isIts image is another unit-rate Poisson process; $F(X_n)$ is the $(n+1)$st arrival because the first right point has index zero. Thus $F(X_n)/(n+1)\to 1$. As $x\uparrow 1$,Combining these facts gives the positive-index asymptoticsEquivalently $1-X_n=\exp (-n+o(n))$ almost surely. In particular $n^p(1-X_n)\to 0$ for every fixed $p>0$, unlike the inverse-linear scale at the left endpoint.

There is also an exact distributional description, if desired. For $x\in (1/2,1)$, $X_n>x$ means that at most $n$ right-hand points precede $x$, soThus $F(X_n)$ has the gamma distribution with shape $n+1$ and rate one. This makes the indexing shift explicit and describes the fluctuations left unspecified by the logarithmic limit.

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.