Dgdrxrt

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
$$ P(S^lambda_n in (na,nnu]) ge
Pleft(S^lambda_n in ((a_0-epsilon)n,(a_0+epsilon)n] right) cdot P left(X_n^lambdain ((a-a_0+epsilon)n,(a-a_0+2epsilon)n]right) ge
frac12 Pleft(X^lambda_n in ((a-a_0+epsilon)n,(a-a_0+epsilon)(n+1)]right)$$

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 $P(X^lambda_1in ((a-a_0+epsilon)n, (a-a_0+epsilon)(n+1)])$ behave like $e^na$ for some $a<0$. But it does not work.

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 $Pleft(S^lambda_n in ((a_0-epsilon)n,(a_0+epsilon)n] right) to 1$ as $n to infty$, and thus for sufficiently large $n$ it is at least $frac12$. It seems like the choice of $frac12$ 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 $eta < r$.