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Background: The limit of superposition of infinite number of independent point processes is a Poisson process under certain conditions. The conditions for the limit process to exist and be Poisson is given in Theorem 11.2.V in "An Introduction to the Theory of Point Processes. Vol 2 - D.J. Daley and D. Vere-Jones". The setting is a triangular array of point processes ${N_{ni}: i=1,...,m_n; n = 1,2,3,...}$ with $m_n rightarrow infty$ as $nrightarrow infty$. The point processes in each row are independent and we consider their superposition $N_n = sum_{i=1}^{m_n} N_{ni}$. This triangular array is called uniformly asymptotically null (u.a.n.) if $$lim_{nrightarrow infty} sup_{i} ;mathcal{P}[N_{ni}(A)>0] = 0$$ for a measurable set $A$ in the space on which the point process is defined. For the limit process $N_n$ to be a Poisson process, the theorem imposes two conditions on u.a.n. array. These are $$lim_{nrightarrow infty} sum_{i=1}^{m_n};mathcal{P}[N_{ni}(A) geq 2] = 0$$ and $$lim_{nrightarrow infty} sum_{i=1}^{m_n};mathcal{P}[N_{ni}(A) geq 1] = mu(A),$$ where $mu$ is some measure. If these two conditions hold for a u.a.n. array, then $lim_{nrightarrow infty} N_n$ is a Poisson process with intensity $mu$.

Question: Consider an u.a.n. array of Poisson point processes on $mathbb{R}$ where $N_{ni}$ is Poisson process of rate $lambda / n$ for $i = 1,...,n$ (here $m_n = n$). Superposition of each row ($N_n = sum_i N_{ni}$) is a Poisson process of rate $lambda$. Hence, the limit process is also a Poisson process of rate $lambda$. This is straightforward using the fact that superposition of independent Poisson process is a Poisson process with rate equal to sum of the rate of component Poisson processes.

Now, I want to establish the same result using the above theorem. We have $mathcal{P}[N_{ni}(A) geq 1] = frac{lambda}{n} mathcal{l}(A) + o(mathcal{l}(A))$ where $mathcal{l}$ is the lebesgue measure. The second condition in the theorem becomes $$ lim_{nrightarrow infty} sum_{i=1}^n ;mathcal{P}[N_{ni}(A) geq 1] = lim_{nrightarrow infty} sum_{i=1}^n left( frac{lambda}{n}mathcal{l}(A) + o(mathcal{l}(A))right) = lambda mathcal{l}(A) + lim_{nrightarrow infty} n, o(mathcal{l}(A)).$$ The term $lambda mathcal{l}(A)$ is exactly what is needed. But, how does it follow that $lim_{nrightarrow infty} n, o(mathcal{l}(A)) = 0 $? As far as I can see, there is no dependence on $n$ in $o(mathcal{l}(A))$.

Alternatively, for this specific case of Poisson processes, the term $mathcal{P},[N_{ni}(A) geq 1]$ can be written explicitly since $N_{ni}(A)$ is Poisson random variable with the mean $frac{lambda}{n} mathcal{l}(A)$. I have verified that writing each term this way, the above limit is in fact $lambda mathcal{l}(A)$. But, this is not what I am looking for. I feel there must be a way to handle $lim_{nrightarrow infty} n o(mathcal{l}(A))$. Expanding $mathcal{P},[N_{ni}(A) geq 1]$ explicitly gives hint that the $o$ term is actually $oleft(frac{mathcal{l}(A)}{n}right)$. But, I don't see how that happens in general.

I will be grateful for any hint/solution/explanation.