(nt)+R}}I\left\{ \mathcal{H}_{n}(\varepsilon
);\tau (n)>nt(1+\varepsilon )\right\} \leq e^{S_{\tau (n)}-S_{\tau
(nt)}}I\left\{ \mathcal{H}_{n}(\varepsilon );\tau (n)>nt(1+\varepsilon
)\right\}  \\
&\leq &e^{S_{\tau (n)}-S_{\tau (nt)}}I\left\{ \tau (n)>nt(1+\varepsilon
)\right\} .\end{aligned}$$Thus, for any $\varepsilon _{1}\in (0,1)$ $$\begin{aligned}
\lefteqn{\mathbf{P}\left( 1-f_{\tau (nt)+R,N}(0)\geq \varepsilon _{1};\mathcal{H}_{n}(\varepsilon );\tau (n)\geq nt\right) }  \notag \\
&\leq &\varepsilon _{1}^{-1}\mathbf{E}\left[ 1-f_{\tau (nt)+R,N}(0);\mathcal{H}_{n}(\varepsilon );\tau (n)\geq nt\right]   \notag \\
&\leq &\varepsilon _{1}^{-1}\mathbf{E}\left[ e^{S_{\tau (n)}-S_{\tau
(nt)}};\tau (n)>nt(1+\varepsilon )\right] .  \label{ProPrel}\end{aligned}$$Hence, using the notation introduced in the proof of Lemma \[Lprospect2\] and the duality principle for random walks it is not difficult to check that$$\begin{aligned}
\lefteqn{\mathbf{E}\left[ e^{S_{\tau (n)}-S_{\tau (nt)}};\tau
(n)>nt(1+\varepsilon )\right] } \\
&=&\sum_{nt(1+\varepsilon )<k\leq n}\mathbf{E}\left[ e^{S_{k}-\min_{0\leq
i\leq nt}S_{i}};\tau (n)=k\right]  \\
&=&\sum_{nt(1+\varepsilon )<k\leq n}\mathbf{E}\left[ e^{\min_{k-nt\leq l\leq
k}S_{l}^{-}};A_{k,n-k}\right]  \\
&\leq &\sum_{nt(1+\varepsilon )<k\leq n}\mathbf{E}\left[ e^{\min_{nt\varepsilon \leq l\leq k}S_{l}^{-}}|A_{k,n-k}\right] \mathbf{P}\left( \tau
(n)=k\right) .\end{aligned}$$Since $S_{l}^{-}\rightarrow -\infty $ $\mathbf{\hat{P}}$-a.s  as $l\rightarrow \infty ,$ we have by Lemma \[Lfran\] $$\lim_{n\rightarrow \infty }\mathbf{E}\left[ e^{\min_{nt\varepsilon \leq
l\leq k}S_{l}^{-}}|A_{k,n-k}\right] =0.$$Now the dominanted convergence theorem gives$$\lim_{n\rightarrow \infty }\mathbf{E}\left[ e^{S_{\tau (n)}-S_{\tau
(nt)}};\tau (n)>nt(1+\varepsilon )\right] =0.$$Combining this fact with (\[ProPrel\]) complete the proof of the lemma.$\square $

\[Lprelim4\]Assume that $A1-$ $A2$ hold. For any $R\in \mathbb{Z}$ and any $t\in (0,1)$$$\left\{ \frac{1-f_{0,\tau (nt)+R}(0)}{e^{S_{\tau (nt)+R}}}|\tau (n)\geq
nt\right\} \overset{d}{\rightarrow }\Theta _{R}$$where $\Theta _{R}$ is the same as in Theorem \[Tclose\].

**Proof**. Let $R\geq 0$ be fixed. We 