Why is $\frac{1}{n} \sup_{k\le n} |T_k - k| \to_{a.s.} 0$ if $T_n/n \to E T_1 = 1$? 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 $$\frac{1}{n} \sup_{k\le n} |T_k - k| \to_{a.s.} 0?$$
 A: 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.
