Let $N_t$ be a Poisson process with intensity $\lambda$. Then how to show that $$ \lim_{t\rightarrow 1} (1-t) \int_0^t \frac{2}{(1-s)^2}dN_s = 0 $$ almost surely?
What I have done.
Since for some $\varepsilon>0$, the probability $\mathbb{P}\left\{ N_{1-\varepsilon,1}=\infty \right\}$ or $\mathbb{P}\{dN_1 > 0\}$ is zero, so my intuition says that the above equality should hold. First, I tried to apply Borel-Cantelli Lemma: $$ \sum_{n=1}^{\infty} \mathbb{P}(E_n)<\infty \implies \mathbb{P}(\lim_{n\rightarrow \infty} E_n) = 0. $$ So I define $E_n = \left\{(1-t_n) \int_0^{t_n} \frac{2}{(1-s)^2} dN_s > \varepsilon \right\}$ for some $\epsilon>0$ and $t_n \rightarrow 1$ as $n\rightarrow \infty$, and using Markov's inequality: $$ \mathbb{P}(E_n) \le \frac{\mathbb{E}\left[(1-t_n)\int_0^{t_n}\frac{2}{(1-s)^2}dV_s\right]}{\varepsilon} \\ = \frac{1-t_n}{\varepsilon}\int_0^{t_n}\frac{2}{(1-s)^2}\lambda ds \\ = \frac{2}{\varepsilon}-2\frac{1-t_n}{\varepsilon} $$ is not tight enough to bound the sum $\sum_{n=1}^{\infty} \mathbb{P}(E_n)$. This is same for the higher moment bounds. Then is it possible to show the above equality?
Thanks.
As you observe, $N$ has no jump at time $1$, a.s. Let $T\in[0,1)$ be the last jump of $N$ before time $1$. ($T=0$ means there are no such jumps.) Then because, $s\mapsto(1-s)^{-2}$ is increasing, $$ \int_0^t{2\over (1-s)^2}dN_s\le 2N(t)\cdot(1-T)^{-2}\le 2N(1)\cdot (1-T)^{-2}. $$