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Let $N$ be a Poisson point process with intensity measure $\mu$, and let $g$ be a real-valued measurable function. For the ordinary Poisson sum $S_g=\int gdN$, the condition $\int |g|d\mu <\infty$ givesIf also $\int g^{2}d\mu <\infty$, thenExpand the square: equal-index terms contribute $\int g^{2}d\mu$, while the Poisson factorial moment measure makes distinct-index terms contribute $(\int gd\mu {)}^2$. Subtracting the squared expected value proves the variance formula. The Campbell first-moment formula also holds for general point processes; this variance identity uses the Poisson assumption.^[ref]

Square integrability alone is insufficient for the ordinary sum on an infinite-intensity space. Take intensity $dx$ on $[1,\infty )$ and $g(x)=1/x$. Then $\int g^{2}dx=1$, but the sum is infinite almost surely: the ordered points satisfy $Z_i/i\to 1$ by the strong law for Poisson arrival times, so comparison with the harmonic series proves divergence.

For this same $g$, the compensated Poisson integral exists as a mean-zero limit under convergence in Lp with $p=2$ and has variance $1$. Its definition is the L2 construction of compensated Poisson integrals, rather than subtraction of two infinite ordinary integrals. When the total intensity is finite, Cauchy-Schwarz inequality makes L2 imply L1, explaining why the shorter variance statement works in that special case.^[ref]

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.