Poisson-integral convergence and conditionality on the impaired-integrability space

A compensated Poisson martingale proves the convergence assertion and gives the exact criterion for unconditional convergence.

Let $N$ be a rate-one Poisson process with arrival times $S_1<S_2<\cdots$, and assume

$$f\in L^2([0,\infty))\cap L^\infty([0,\infty)), \qquad F(t)=\int_0^t f(s)\,ds\longrightarrow F_\infty.$$

Define

$$M_t=\int_{(0,t]}f(s)\bigl(dN_s-ds\bigr).$$

The Poisson isometry gives

$$\mathbb E|M_t|^2=\int_0^t|f(s)|^2\,ds\le\|f\|_2^2.$$

Hence $M_t\to M_\infty$ almost surely and in $L^2$. Optional sampling and the same isometry give

$$\mathbb E|M_\infty-M_{S_n}|^2 =\int_0^\infty|f(s)|^2\Pr(S_n<s)\,ds \longrightarrow0.$$

Indeed, $S_n\to\infty$ almost surely and dominated convergence applies. Since the convergent primitive $F$ is bounded, $F(S_n)\to F_\infty$ both almost surely and in $L^2$. Therefore

$$\sum_{j=1}^n f(S_j) =M_{S_n}+F(S_n) \longrightarrow M_\infty+F_\infty$$

almost surely and in $L^2$.

For the sharper unconditionality classification, set

$$H=\sum_{j\ge1}|f(S_j)|.$$

The Poisson Laplace functional gives

$$\mathbb E e^{-H} =\exp\left(-\int_0^\infty(1-e^{-|f(s)|})\,ds\right).$$

Boundedness of $f$ makes $1-e^{-|f|}$ comparable to $|f|$. Thus, if $f\notin L^1$, the expectation above vanishes and $H=\infty$ almost surely. Conversely, if $f\in L^1$, Campbell's formula gives

$$\mathbb EH=\int_0^\infty|f(s)|\,ds<\infty,$$

so $H<\infty$ almost surely. Consequently

$$\sum_{j\ge1}f(S_j)\text{ converges unconditionally almost surely} \quad\Longleftrightarrow\quad f\in L^1.$$

In particular, convergence is conditional precisely for the genuinely improper functions in the stated space outside $L^1$; it cannot be non-unconditional for every function in that space, because bounded integrable functions also belong to it.