I'm working through Varadhan Stochastic Processes, and I can't follow along with his proof. He defines a Poisson process on page 13 with the following criteria:
- With probability $1$, $N(0)=0$, $N(t)$ is integer valued, right continuous with left limits and non-decreasing in $t$
- For each $h>0$, the distribution of $N(t+h)-N(t)$ is independent of $t$
- The random variables $N(t'_j) - N(t_j)$ are mutually independent if the intervals $[t_j,t_j']$ are non-overlapping.
- $P(N(t+h) - N(t) \ge 2) = P(N(h) \ge 2) = o(h)$
We want to prove that $N(t+s) - N(t)$ or just $N(t)$ is Poisson with parameter $\lambda t$. He seems to break up this interval into $n$ non-overlapping subintervals, and take advantage of property 3 to use moment generating functions. However I am unclear why he is saying:
$$ \lim_{n \to \infty}\left[E \left[\exp\left(-\sigma N\left(\frac{1}{n}\right)\right)\right] \right]^{nt} = \lim_{n \to \infty}\exp\left[(nt)\log\left( E \left[\exp\left(-\sigma N\left(\frac{1}{n}\right)\right)\right]\right) \right] = \exp(-tg(\sigma)) $$ where $g(\sigma) = \lim_{n \to \infty}n\left[1 - E\left[\exp\left(-\sigma N(\frac{1}{n})\right) \right] \right]$
Varadhan: (https://www.math.nyu.edu/faculty/varadhan/spring06/spring06.1.pdf) (we're looking at page 2)
Let $f$ be a differentiable (non-negative) function such that $f(0)=1$. Then
$$\begin{align*} \lim_{n \to \infty} n \log (f(1/n)) &= \lim_{n \to \infty} \frac{\log (f(1/n))-\log (f(0))}{\frac{1}{n}} \\ &= \frac{d}{dx} \log(f(x)) \bigg|_{x=0} \\ &= \frac{1}{f(0)} f'(0) = f'(0). \end{align*}$$
Applying this identity for
$$f(x) := \mathbb{E} \exp \left(- \sigma N (x) \right)$$
gives
$$\lim_{n \to \infty} n \log \left( \mathbb{E} \exp \left[- \sigma N(1/n) \right] \right) = f'(0) = -g(\sigma). \tag{1}$$
Here we have used that
$$f'(0) = \lim_{n \to \infty} \frac{f(1/n)-f(0)}{\frac{1}{n}} = - g(\sigma)$$
by definition of $g$. From $(1)$ the second equality in your question follows easily from the continuity of $\exp$.