Almost-s sure summability of Poisson-arrival double sequences

Let f∈Limp⁡2f\in L^2_{\operatorname{imp}}, let (Ak)(A_k) be pairwise disjoint Borel subsets of R+\mathbb R_+, and define fk=f1Akf_k=f\mathbf 1_{A_k}. Let (Sn)(S_n) be the arrival times of a Poisson process. Double-series summability conjecture. The double sequence fk(Sn)f_k(S_n) is summable almost surely. This concerns the interchange and joint convergence of the decomposition by disjoint spatial sets and the Poisson-arrival index; the preceding finite-partition argument gives a related reordered series but does not settle the original infinite double sequence.

References

Primary source

Jerzy Szulga, “Improper Poisson integral of the sinc function”, arXiv:2504.05574 (2025).

Progress summary

Refreshed
Claimed solved

A reader has proposed a complete counterexample using the sinc function, but it has not been independently verified.

The conjecture asks whether the disjoint-set decomposition of Poisson arrivals is almost surely unconditionally summable. Szulga's 2025 preprint studies improper Poisson integrals for the sinc function and leaves the general case to conjectures.

Posted attempt

An attempted complete disproof uses f(x)=sin⁡(x)/xf(x)=\sin(x)/x and the partition Ak=[k−1,k)A_k=[k-1,k). It claims ∑n∣f(Sn)∣=∞\sum_n |f(S_n)|=\infty almost surely, while the arrival-ordered series converges, so the asserted unconditional double summability fails. The attempt has not been independently verified.

Current status (as of August 2026): a complete counterexample has been claimed, but the conjecture is not verified as false and no independently confirmed resolution was found.

Sources

Solutions 1

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The conjecture is false under the source's explicit definition of “summable” as unconditionally summable.

Let NN be a rate-one Poisson process with arrival times S1<S2<⋯S_1<S_2<\cdots. Take

f(x)={sin⁡(x)/x,x>0,1,x=0,Ak=[k−1,k)(k≥1).f(x)= \begin{cases} \sin(x)/x,&x>0,\\ 1,&x=0, \end{cases} \qquad A_k=[k-1,k)\quad(k\ge1).

The sets AkA_k form a disjoint Borel partition of [0,∞)[0,\infty). Moreover,

f∈L∞∩L2,∫0∞f(x) dx=π2f\in L^\infty\cap L^2, \qquad \int_0^\infty f(x)\,dx=\frac{\pi}{2}

as an improper integral, so ff belongs to the required impaired-integrability space.

Nevertheless,

∫0∞∣sin⁡x∣x dx=∞.\int_0^\infty\frac{|\sin x|}{x}\,dx=\infty.

Indeed, on

Jm=[mπ+π/6, mπ+5π/6]J_m=[m\pi+\pi/6,\ m\pi+5\pi/6]

we have ∣sin⁡x∣≥1/2|\sin x|\ge1/2, giving

∫Jm∣sin⁡x∣x dx≥13(m+1),\int_{J_m}\frac{|\sin x|}{x}\,dx \ge\frac1{3(m+1)},

and the corresponding harmonic series diverges.

Since ∣f∣≤1|f|\le1, we also have

1−e−∣f(x)∣≥(1−e−1)∣f(x)∣.1-e^{-|f(x)|}\ge(1-e^{-1})|f(x)|.

The Poisson Laplace functional therefore yields

Eexp⁡(−∑n≥1∣f(Sn)∣)=exp⁡(−∫0∞(1−e−∣f(x)∣) dx)=0.\mathbb E\exp\left(-\sum_{n\ge1}|f(S_n)|\right) = \exp\left(-\int_0^\infty(1-e^{-|f(x)|})\,dx\right) =0.

Consequently

∑n≥1∣f(Sn)∣=∞almost surely.\sum_{n\ge1}|f(S_n)|=\infty \qquad\text{almost surely}.

Because every arrival belongs to exactly one AkA_k, almost surely

∑k≥1∑n≥1∣f(Sn)1Ak(Sn)∣=∑n≥1∣f(Sn)∣=∞.\sum_{k\ge1}\sum_{n\ge1} \left|f(S_n)\mathbf1_{A_k}(S_n)\right| = \sum_{n\ge1}|f(S_n)| =\infty.

Thus the asserted double family is not unconditionally summable almost surely.

The naturally arrival-ordered series still converges almost surely and in L2L^2: it is the sum of an L2L^2-bounded compensated Poisson martingale and the convergent improper deterministic integral. The failure concerns exactly the stronger unconditional double-summability claim.