In short , I don't understand the step 2 of the answer:
$$ \begin{aligned} \mathbb{E}\left[\sum_{k=1}^{N_t}f(S_k)\right] &= \sum_{n\ge 0} \mathbb{E}\left[ \left.\sum_{k=1}^{n}f(S_k)\right|N_t=n \right] P(N_t=n)\\ &= \sum_{n\ge 0} \frac{n}{t} \int_0^t f(x)dx \frac{(\lambda t)^n}{n!} e^{-\lambda t} \end{aligned} $$ where $f$ is a continuous real valued function, $N_t$ is a Poisson process, $S_k$ is the waiting times for $N_t$'s $k^{th}$ jump.
The step 2 of the answer implies that $$ \mathbb{E}\left[\sum_{k=1}^{n}f(S_k)\right]=\frac{n}{t}\int_0^t f(x)dx, $$ which is my confusing place, how does it come?
Furthermore, the aforementioned formula may implies $\mathbb{E}[f(S_k)]=\frac{1}{t}\int_{0}^{t}f(x)dx$, how does it come?
Does it mean that the pdf of $S_k$ is $\frac{1}{t}$? how to get this result?