Let $\lambda:[0,\infty)\rightarrow [0,\infty)$ be a continuous function and $N$ be a Poisson process with rate $1$. Define $\Lambda(t)=\int_0^t{\lambda(x)dx}$ then how do we prove that $$ lim_{s\rightarrow{0}}{\frac{1-P(N(\Lambda(t+s))-N(\Lambda(t))=0)}{t}}=\lambda(t)? $$
I found the following result in many textbooks:
$N(\Lambda(t+s))-N(\Lambda(t))$ is a Poisson random variable with parameter $\int_t^{t+s}\lambda(x)dx$ and if we use this result then to prove the above limit is almost obvious. But I am not sure how to prove the second result or is there any way to get the above limit without using the second result?
Yes the result is obvious since by definition $N(u+v)-N(u)$ is Poisson with parameter $v$, for every nonnegative $u$ and positive $v$, hence $$ \lim_{v\to0}\frac{P(N(u+v)-N(u)\geqslant1)}v=\lim_{v\to0}\frac{1-\mathrm e^{-v}}v=1,$$ which, for $u=\Lambda(t)$ and $v=\Lambda(t+s)-\Lambda(t)$, implies $$ \lim_{s\to0}\frac{P(N(\Lambda(t+s))-N(\Lambda(t))\geqslant1)}{\Lambda(t+s)-\Lambda(t)}=1, $$ and the result in the question then follows from the other limit $$\lim_{s\to0}\frac{\Lambda(t+s)-\Lambda(t)}s=\lambda(t). $$