A property of Poisson process

Question

Let $Y_t$ be a centered Poisson process, why \begin{equation} \lim_{n \to \infty} \sup_{s<t} |n^{-1}Y(ns)| = 0 \qquad a.s. \qquad \forall t\ge 0 \end{equation} This is a fundamental step in the proof of the law of large number for continuous time Markov Chain. I'm following the proof on the book by Ethier and Kurtz but they don't explain, or show, the previous statement.

Answer 1

Define $M_{n}(s)={Y(ns)\over n}={1\over n}(N(ns)-\lambda ns),$ where $N$ is a Poisson process with parameter $\lambda$. Then $M_n$ is a martingale, so by Doob's maximal inequality with $p=4$, we have $$\mathbb{E}\left(\sup_{0\leq s\leq t} M^4_n(s)\right)\leq c\,\mathbb{E}(M^4_n(t))\leq {c\,(\lambda n t)^2\over n^4}\leq {c\over n^2}.\tag1$$ We use the fact that the normalized fourth moment of a Poisson distribution with mean $\mu$ is bounded above by a constant times $\mu^2$.

Since the right hand side of (1) is summable, we get $\sum_{n=1}^\infty \mathbb{E}\left(\sup_{0\leq s\leq t} M^4_n(s)\right)<\infty,$ so

$$\sum_{n=1}^\infty \,\sup_{0\leq s\leq t} M^4_n(s) <\infty\quad \mbox{ almost surely},$$ which implies $$ \sup_{0\leq s\leq t} M_n(s)\to 0 \quad \mbox{ almost surely}.$$