Is it necessarily true that $\sum\limits_{N \in \mathbb N}\sum\limits_{q \in \mathbb Q_{+}}\mathbb P(X>\frac{N-aq}{\sqrt{q}})<\infty$

Question

Is it necessarily true that: $$\sum\limits_{N \in \mathbb N}\sum\limits_{q \in \mathbb Q_{+}}\mathbb P(X>\frac{N-aq}{\sqrt{q}})<\infty (*)$$

where $X$ is a standard normal distribution.

The background to this question is that I want to show, for a Brownian motion $(B_{t})$ that $\sup\limits_{t\geq 0}(B_{t}-at)<\infty$ a.s. and my idea here is to show that the set: $\{ \forall N \in \mathbb N: \exists q \in \mathbb Q_{+} \text{s.t.} B_{q}-aq>N\}$ is a null set. I think this can be done via Borel-Cantelli but I cannot prove that $(*)$ holds.

Answer 1

To prove $\sup_{t \ge 0} (B_t - at) < \infty$ a.s., we can prove $\lim_{t \rightarrow \infty}( B_t - at )= -\infty$ almost surely. Then by continuity of $t \mapsto B_t - at$, we conclude $\sup_{t \ge 0} (B_t - at) < \infty$ a.s.

I will assume $a > 0$, otherwise the answer is false. Note that $B_t - at = t\left(\frac{B_t}{t}-a\right)$, and $\lim_{t \rightarrow \infty} \frac{B_t}{t} = 0$ a.s. This implies that for all $t$ sufficiently large, $\frac{B_t}{t}-a < -\frac{a}{2}$, and hence $$\lim_{t \rightarrow \infty} B_t - at = \lim_{t \rightarrow \infty} t\left(\frac{B_t}{t}-a\right) \le \lim_{t \rightarrow \infty} -\frac a2t = -\infty.$$

Therefore $\sup_{t \ge 0} (B_t - at) < \infty$ a.s. as desired.