Letting $X_1^\lambda,X_2^\lambda,\dots$ be i.i.d. with distribution $F_\lambda$ and $S_n^\lambda=X_1^\lambda+\dots+X_n^\lambda$, we have
$$ P(S^\lambda_n \in (na,n\nu]) \ge
P\left(S^\lambda_n \in ((a_0-\epsilon)n,(a_0+\epsilon)n] \right) \cdot P \left(X_n^\lambda\in ((a-a_0+\epsilon)n,(a-a_0+2\epsilon)n]\right) \ge
\frac12 P\left(X^\lambda_n \in ((a-a_0+\epsilon)n,(a-a_0+\epsilon)(n+1)]\right)$$
This is from the proof of Theorem 2.7.10 in Durrett PTE 5th.
The first term will tends to 1 (since $a_0$ is the mean of $X^{\lambda}_1$ then by weak law of large number). But I can't understand the second inequality, how does $1/2$ come up?
Furthermore,
Here, to prove the limsup is 0 by contradiction. But why limsup<0 will implies $Eexp(\eta X^{\lambda}_1)<\infty$ for some $\eta>0$?
I try to calculate $Eexp(\eta X^{\lambda}_1)$ by applying $P(X^{\lambda}_1\in ((a-a_0+\epsilon)n, (a-a_0+\epsilon)(n+1)])$ behave like $e^{na}$ for some $a<0$. But it does not work.
Best Answer
I have the 4th edition, where this appears to correspond to Theorem 2.6.5. You've omitted the last part of the sentence following the equation, which says
So it is exactly as you say: the weak law of large numbers implies that $P\left(S^\lambda_n \in ((a_0-\epsilon)n,(a_0+\epsilon)n] \right) \to 1$ as $n \to \infty$, and thus for sufficiently large $n$ it is at least $\frac{1}{2}$. It seems like the choice of $\frac{1}{2}$ is arbitrary; we just need some convenient number in between 0 and 1.
For the second part, let $Z = X_1^\lambda / (a-a_0+\epsilon)$. Note that $$e^{\eta Z} \le 1_{\{Z \le 0\}} + \sum_{n=0}^\infty e^{\eta (n+1)} 1_{\{n < Z \le n+1\}}$$ and so $$E[e^{\eta Z}] \le P(Z \le 0) + \sum_{n=0}^\infty e^{\eta (n+1)} P(n < Z \le n+1).$$ Now if the limsup is negative, then there is some $r > 0$ such that $P(n < Z \le n+1) \le e^{-rn}$ for all sufficiently large $n$, which would imply that the sum on the right side converges when $\eta < r$.