= Solution
Fix $T>0$ and put $K=\lambda T/4$. Since $0\leq X_n\leq1$, the clock in part (a) is at most $nK$ on $[0,T]$. The <Poisson maximal concentration bound> gives, for every $\delta>0$,
$$
\mathbb P\left(\sup_{t\leq T}|\varepsilon_n(t)|>\delta\right)
\leq2\exp\left[-nK\,h(\delta/K)\right],
\qquad h(u)=(1+u)\log(1+u)-u.
$$
Here $h(u)>0$ for $u>0$, so the probabilities are summable in $n$. The <Borel-Cantelli lemma>, first for positive rational $\delta$ and then for all $\delta$, yields $\sup_{t\leq T}|\varepsilon_n(t)|\to0$ <almost surely>. No independence between the processes for different population sizes is required.
The deterministic <SI model> solves the <logistic equation> $x'=-\lambda x(1-x)$ with $x(0)=a$, giving
$$
\boxed{x(t)=\frac{ae^{-\lambda t}}{1-a+ae^{-\lambda t}}}.
$$
On $[0,1]$, $b$ is <Lipschitz continuous> with constant $\lambda$. Subtracting the two integral equations gives
$$
|X_n(t)-x(t)|\leq\sup_{s\leq T}|\varepsilon_n(s)|
+\lambda\int_0^t|X_n(s)-x(s)|\,ds.
$$
The <Gronwall inequality> implies
$$
\sup_{t\leq T}|X_n(t)-x(t)|
\leq e^{\lambda T}\sup_{t\leq T}|\varepsilon_n(t)|
\longrightarrow0
\quad\text{almost surely}.
$$
This is the <uniform almost-sure fluid limit of the SI model>. One can take the intersection of the probability-one events for integer $T$ to obtain the conclusion on every finite interval. Use population sizes with $an$ integral; if the initial susceptible count is rounded, the extra initial error is at most $1/n$ and the same <Gronwall inequality> proof applies.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-36-si-fluid-limit.png]
{title=Susceptible fractions in stochastic SI epidemics approach the deterministic fluid limit}
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