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Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.