Let $T_n$ be a sequence $0\equiv T_0 \le T_1 \le T_2 \le \cdots$ of random variables such that $T_n - T_{n-1}$ is distributed to a zero mean unit variance random walk so that $E T_n = n$.
Then by the strong law of large numbers we have $$\frac{T_n}{n} \to_{a.s.} ET_1 = 1.$$
In this case why do we have $$\frac{1}{n} \sup_{k\le n} |T_k - k| \to_{a.s.} 0?$$
Notice this fact: Suppose $\{a_n, n\ge 1\}$ is a sequence of nonegative numbers and \begin{equation*} \lim_{n\to\infty}\frac{a_n}{n}=0,\tag{1} \end{equation*} then \begin{equation*} \lim_{n\to\infty}\frac{\sup_{k\le n}a_k}{n}=0. \tag{2} \end{equation*} Now let $a_n=|T_n(\omega)-n|$, we get \begin{gather*} \Big\{\omega: \frac{|T_n(\omega)-n|}{n}\to 0\Big\} \subset \Big\{\omega:\frac{\sup_{k\le n}|T_k(\omega)-k|}{n}\to 0\Big\},\\ \lim_{n\to\infty}\frac{T_n}{n}=1 \quad \text{a.s.}\implies \lim_{n\to\infty}\frac{\sup_{k\le n}|T_k-k|}{n}=1 \quad \text{a.s.}. \end{gather*}
Proof of (2): For fixed $\epsilon >0$, there exists $n_\epsilon$ such that \begin{equation*} a_k<\epsilon k, \qquad k>n_\epsilon, \end{equation*} hence \begin{gather*} \sup_{k\le n} a_k\le \sup_{k\le n_\epsilon} a_k +\epsilon n,\\ \frac{\sup_{k\le n} a_k}{n}\le \frac{\sup_{k\le n_\epsilon} a_k}{n} +\epsilon,\\ \varlimsup_{n\to\infty} \frac{\sup_{k\le n} a_k}{n}\le \epsilon. \end{gather*} Now, letting $\epsilon \downarrow 0$, (2) holds.