Let $(X_1,X_2,...)$ be i.i.d random variables, with $P(X_t=1)=P(X_t=-1)=1/2$. Then
$S_t= \frac{1}{t}\sum_{i=1}^{t}X_i $ is a zero mean random walk. Let $\tau$ be the stopping time corresponding to the first time that $S_t$ hits $M \in (0,1/2)$, \begin{equation} \tau=\inf\{t \geq 1 \mid S_t \geq M \} \end{equation} I am trying to show that $P(\tau = \infty)>0$.
For starters, clearly $P(\tau < \infty)>0$ since $P(\tau=1)=1/2$. Also, the probability that $S_t$ exceeds $M$ infinitely often is zero, since $S_t$ converges to $0$. But I would like to show that with positive probability, $S_t$ can converge to zero without ever exceeding $M$.
I'll recast the problem in terms of the random walk $W_n=W_0+\sum_{i=1}^n (X_i-M).$
For $j\geq 0$, define the (identically distributed) stopping times $$\tau_j=\inf\left(n> j: \sum_{i=j+1}^n (X_i-M)>0\right)=\inf(n> j: W_n>W_{j}). $$
If $\mathbb{P}(\tau_0<\infty)=1,$ then $\mathbb{P}(\tau_j<\infty)=1$ for all $j\geq 0$, and hence $\mathbb{P}\left(\cap_{j=0}^\infty [\tau_j <\infty]\right)=1.$ It follows that, with probability 1, the sequence $(W_j(\omega))_{j\geq 0}$ does not achieve its maximum.
But this contradicts the fact, due to the strong law of large numbers, that $W_j\to-\infty$ with probability one.
Therefore $\mathbb{P}(\tau_0<\infty)=1$ is false, that is, $\mathbb{P}(\tau_0=\infty)>0.$