= Solution
First prove the <strong law for Brownian motion>, $B_t/t\to0$ with probability one. For each $\varepsilon>0$, the <Gaussian tail bound> gives
$$
\mathbb P(|B_n|>\varepsilon n)\leq2e^{-\varepsilon^2n/2}.
$$
By stationary increments and the <Brownian reflection principle>, followed by the same tail bound,
$$
\mathbb P\left(\sup_{0\leq s\leq1}|B_{n+s}-B_n|>\varepsilon n\right)
\leq4e^{-\varepsilon^2n^2/2}.
$$
Both bounds are summable in $n$. The <First Borel-Cantelli lemma> therefore implies that, eventually, both quantities inside these probability events are at most $\varepsilon n$. For $n\leq t\leq n+1$ this gives $|B_t|/t\leq2\varepsilon$. Intersect over a sequence of positive rational $\varepsilon$ tending to zero to obtain the asserted continuous-time limit.
On this one <almost sure event>,
$$
\frac{B_t+at}{t}\longrightarrow a,
$$
so $B_t+at\to+\infty$ for $a>0$ and to $-\infty$ for $a<0$. For each real $x$, the path is eventually strictly on the corresponding side of $x$. Hence
$$
\boxed{\{t\geq0:B_t+at=x\}\text{ is bounded for every }x\in\mathbb R.}
$$
The same event works simultaneously for all levels $x$, giving the requested transience of <Brownian motion with drift>.
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