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Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ of the simple symmetric random walk. The event $\{T_1\le n\}={\bigcup }_{k=0}^n\{S_k=1\}$ belongs to ${\mathcal{F}}_n$, so $T_1$ is a stopping time.

For a positive integer $m$, set ${\tau }_m=T_1\wedge T_{-m}$. This exit time from the finite interval is integrable: in each block of $m+1$ steps there is a probability at least $2^{-(m+1)}$ of all steps being positive, which forces exit if it has not already occurred. Consequently the survival probabilities have a geometric upper bound.

We use the bounded-time optional stopping theorem: if $M$ is a martingale and $\sigma$ a bounded stopping time, then $\mathbb{E}{M}_{\sigma }=\mathbb{E}{M}_0$. Apply it at ${\tau }_m\wedge n$ to $S_n$ and to the square-minus-time martingale of a simple symmetric random walk $S_n^2-n$. The stopped positions lie in $[-m,1]$, so dominated convergence and monotone convergence giveWriting $p_m=\mathbb{P}(T_1<T_{-m})$, the first equation becomes $p_m-m(1-p_m)=0$, hence $p_m=m/(m+1)$. The second gives $\mathbb{E}{\tau }_m=p_m+m^2(1-p_m)=m$. Since ${\tau }_m\le T_1$, its expectation satisfies $\mathbb{E}{T}_1\ge m$ for every $m$. ThereforeNevertheless $\mathbb{P}(T_1<\infty )\ge p_m\to 1$. Thus the infinite mean first passage of a simple symmetric random walk occurs despite almost sure finiteness; using unrestricted optional stopping directly at $T_1$ would be unjustified.

For a simple symmetric random walk, the infinite mean first passage of a simple symmetric random walk gives a finite almost sure first passage $T_1$ to $1$. The stopped martingale $M_n=1-S_{n\wedge T_1}$ is nonnegative and eventually zero almost surely. Nevertheless $\mathbb{E}{M}_n=1$ for every $n$, so it has almost sure convergence but no convergence in L1. It is therefore not uniformly integrable. Before absorption the nearest-neighbour walk is at most zero, ensuring nonnegativity.