I want to prove the following result. Let $S_n$ be a symmetric irreducible random walk on the integers (d=dimension).
Claim: If $x\in A$ and $P_x(T_A=\infty)>0$ then $\forall \epsilon>0\exists y:P_y(T_A=\infty)>1-\epsilon$ ($T_A:=\min\{j\geq0:S_j\in A^c\}$)
I restricted the result for $\epsilon \in (0,1)$ since otherwise the result follows taking an $y\in A^c$. I have troubles to get this result. I tryed stating first that for any $\epsilon\in (0,1)$ we can find a path connecting $x,y$, i.e. For $\epsilon \in (0,1)\exists z_0,...,z_k:P_y(S_k=x)>\delta_{y}(z_0)p(z_0,z_1)...p(z_{k},x)>\epsilon$, but I am not sure that this can help so much to get the lowerbound. Does someone has an hint?
Thanks
Define $f(y)=\mathbb{P}_y(T_A<\infty)$ and let $\alpha=\inf_y f(y)$. Then for all $n\geq 0$: $$\mathbb{P}_{y}(T_{A}\circ\theta_n<\infty)=(P^n f)(y)\geq \alpha,$$ where the left hand side is the chance of hitting $A^c$ after time $n$.
If $f(x)<1$ for some $x$, then the set $A^c$ is transient in the terminology of random walks, so that [1] the walk cannot hit $A^c$ infinitely often, from any starting point. Hence letting $n\to\infty$ gives
$$ 0=\mathbb{P}_{y}(S_n\in A^c\mbox{ infinitely often }) = \lim_n \mathbb{P}_{y}(T_{A}\circ\theta_n<\infty)\geq \alpha.$$ Therefore $\inf_y f(y)=0$.
[1] Principles of Random Walk (2nd edition) by Frank Spitzer (P8, page 298)