Continuous version of the Law of rare events EDIT: Now I am just looking for a solution to the last part:
Show a realisation of this problem such that a strong limit follows as well for $S_n$.

In the simples version of the Law of rare events we take $n$ i.i.d. Bernoulli$(p(n))$ random variables with mean $p(n)$ small such that $n\cdot p(n)\to\lambda$ as $n\to\infty$, for example $p(n)=\frac\lambda n$. The sum then converges to Poisson$(\lambda)$ in distribution. In this problem we try to mimic this in the continuous world.
Let $X_{n,k},k=1\dots n$ be Uniform$(0,\frac{2\lambda}n)$, so that $\mathbb{E}X_{n,k}=\frac\lambda n$. Assume that these variables are i.i.d. across $k=1\dots n$ for each fixed $n$. Define $S_n=\sum_{k=1}^nX_{n,k}$.

*

*Determine the characteristic function for $X_{n,k}$.

*Identify the weak limit for the distribution of $S_n$.

*Show a realisation of this problem such that a strong limit follows as well for $S_n$.

So far I have some ideas for the first two parts of this question:
$$
\begin{align*}
\varphi_{X_{n,k}}(t)=\mathbb{E}\left(e^{itX_{n,k}}\right)&=\int_0^{\frac{2\lambda}n}\frac n{2\lambda}e^{itx}dx\\
&=\frac n{2\lambda}\left[\frac1{it}e^{itx}\right]_0^{\frac{2\lambda}n}\\
&=\frac{e^{\frac{2it\lambda}n}-1}{\frac{2it\lambda}n}.
\end{align*}
$$
Then we have that
$$
\begin{align*}
\varphi_{S_n}(t)&=\mathbb{E}\left(\exp\left(it\sum_{k=1}^nX_{n,k}\right)\right)\\
&=\prod_{k=1}^n\mathbb{E}\left(e^{itX_{n,k}}\right)=\left(\mathbb{E}\left(e^{itX_{n,k}}\right)\right)^n=\left(\varphi_{X_{n,k}}(t)\right)^n\\
&=\left(\frac{e^{\frac{2it\lambda}n}-1}{\frac{2it\lambda}n}\right)^n\\
&=e^{it\lambda}\left(\frac{\sin\left(\frac{t\lambda}n\right)}{\frac{t\lambda}n}\right)^n\\
&\to e^{it\lambda}\tag{1}\label{eq1}
\end{align*}
$$
I am unsure of exactly how to justify \eqref{eq1}, or if this is even the right answer. I would have expected the characteristic function to converge to the characteristic function of some sort of Poisson variable, but it seems like maybe it doesn't from my workings. Can anybody here check if my answer is correct, and also help out with the next part of the question, which I think has something to do with the SLLN, but I am really unsure especially since the first two parts of the question are making me think that there is something else I am completely overlooking in this question. Any help would be appreciated!

EDIT: Now I am just looking for a solution to the last part:
Show a realisation of this problem such that a strong limit follows as well for $S_n$.
 A: Here is a solution to last part. In the following, the
$\mathsf{E}[S_n-\mathsf{E}S_n]^4$ will be obtained from the characteristic function
$\phi_{S_n-\mathsf{E}S_n}(t)$ of $S_n-\mathsf{E}S_n$ and the strong law of $S_n-\mathsf{E}S_n$
will also be derived from it.
At first,  the MacLaurin expansion for the characteristic function of $S_n-\mathsf{E}S_n$ is
\begin{align*}
&\phi_{S_n-\mathsf{E}S_n}(t)=\mathsf{E}[\exp(it(S_n-\mathsf{E}S_n))]
 =\Big[\frac{n}{\lambda t}\sin\Big(\frac{\lambda t}{n}\Big)\Big]^n\\
&\quad=\Big[1-\frac{1}{3!}\Big(\frac{\lambda t}{n}\Big)^2
 + \frac1{5!}\Big(\frac{\lambda t}{n}\Big)^4 +O(t^6) \Big]^n\\
&\quad = 1-\frac{(\lambda t)^2}{6n}+\Big(\frac{1}{120} 
+\frac{n-1}{72}\Big)\frac{(\lambda t)^4}{n^3} + O(t^6). 
\end{align*}
Hence,
\begin{align*}
 \mathsf{E}(S_n-\mathsf{E}S_n)^2&=\frac{\lambda^2}{3n},\\
 \mathsf{E}(S_n-\mathsf{E}S_n)^4&=4!\Big(\frac{\lambda^4(n-1)}{72 n^3} + \frac{\lambda^4}{120n^3}\Big) \\
 &=\frac{\lambda^4}{3n^2}+O(n^{-3})\le Cn^{-2},
\end{align*}
where $C$ is a constants. Futhermore,
\begin{align*}
 \sum_{n=1}^{\infty}\mathsf{P}(|S_n-\mathsf{E}S_n|>\epsilon)
 & \le \frac{1}{\epsilon^4}\sum_{n=1}^{\infty} \mathsf{E}(S_n-\mathsf{E}S_n)^4\\
 & \le \frac{C}{\epsilon^4} \sum_{n=1}^{\infty} n^{-2}<\infty , \qquad \forall \epsilon>0.
\end{align*}
Meanwhile $\mathsf{E}[S_n]=\lambda $, therefore,
\begin{equation*}
 \lim_{n\to\infty} S_n=\lambda, \qquad \text{a.s.}
\end{equation*}
