Consider a nonhomogeneous Poisson process $N(t)$ for $t\ge0$ defined by the instantaneous intensity $\lambda(t)\ge0$ and mean value function $\Lambda(t)=\int_0^t\lambda(s)ds$. We know that $$ \mathbb{P}(N(t)=n)=e^{-\Lambda(t)}\frac{\Lambda(t)^n}{n!},\quad n=0,1,\dots. $$ From this we get the CDF of arrival times $T_n$ as $$ F_{T_n}(t)=\mathbb{P}(T_n\le t)=\mathbb{P}(N(t)\ge n)=1-\frac{\Gamma(n,\Lambda(t))}{\Gamma(n)},\quad t\ge0,\,n=1,2,\dots, $$ where the $\Gamma$ functions in the numerator and denominator are the upper incomplete and complete Gamma functions, respectively. From this we get, correctly, $F_{T_n}(0)=0$ as one "normalization" of the CDF irrespective of the choice of $\lambda(t)$.
Question: now suppose either (i) $\lambda(t)=0$ for $t>T$, in which case $\Lambda(\infty)<\infty$, or (ii) $\lambda(t)$ remains positive for all $t\ge0$ but decays strongly for large $t$ such that $\Lambda(\infty)<\infty$. In these cases $$ F_{T_n}(\infty)=1-\frac{\Gamma(n,\Lambda(\infty))}{\Gamma(n)}<1 $$ and the CDF is not normalized. How can one reconcile this? These choices for $\lambda(t)$ seem perfectly valid to me, still, I find it hard to believe an unnormalized CDF/PDF makes sense.