Proof of Theorem 2.7.10 in Durrett (edited) 
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.
 A: 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

for large $n$ by the weak law of large numbers.

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 $\eta < r$.
