I'm trying to solve the following exercise from Paolo Baldi - Stochastic Calculus - an introduction through theory and exercises, Exercise 5.20
Let $(B_t)_t$ a Brownian motion and for $\mu>0$, let $X_t= B_t + \mu t$.
Let $a,b>0$ and $\tau$ be the exit time of $(X_t)_t$ from the interval $]-a,b[$.
Show that $\tau$ is a.s. finite.
My attempt:
The solutions says just "by the iterative logarithm law", but I've not had that topic in the lectures, so I'll try an approach "by hand", using just the definition of my stopping time $\tau$ and the law of the Brownian motion.
$$P(\tau>t) = \\P(X_t \in ]-a,b[) =\\ P(B_t + \mu t \in ]-a,b[) = \\ P(B_t \in ]-a - \mu t, b - \mu t[ ) = \\ P(B_1 \in ]\frac{-a-\mu t}{\sqrt{t}},\frac{b-\mu t}{\sqrt{t}}[)$$
The last probability equals $$ \frac{1}{\sqrt{2 \pi}} \int_{ \frac{-a-\mu t}{\sqrt{t}}}^{,\frac{b-\mu t}{\sqrt{t}}} e^{-x^2/2} dx$$
Now, since $P(\tau = \infty) = \lim_{t \rightarrow \infty} P(\tau >t) = \lim_t \frac{1}{\sqrt{2 \pi}} \int_{ \frac{-a-\mu t}{\sqrt{t}}}^{,\frac{b-\mu t}{\sqrt{t}}} e^{-x^2/2} dx =0 $
I conclude that $\tau$ is a.s. finite, where in the last equality I used the fact that the integrand is $C^{\infty}$, and since the function does not explodes as the integration interval becomes smaller and smaller, I can conclude.
Is it okay?
As @KaviRamaMurthy already pointed out, your reasoning is not correct since the first "=" in your calculation fails to hold.
Here is one possible approach: It holds that $$\lim_{t \to \infty} \frac{B_t}{t} = 0 \quad \text{a.s.};$$ there are several ways to show this convergence, e.g. using (a variant of) the strong law of large numbers or the law of iterated logarithm. If $\mu>0$, then the process $X_t := B_t+\mu t$ satisfies
$$\lim_{t \to \infty} \frac{X_t}{t}= \mu>0 \quad \text{a.s.},$$
in particular,
$$\lim_{t \to \infty} X_t = \infty \quad \text{a.s.}$$
Since $(X_t)_{t \geq 0}$ has continuous sample paths (with probability $1$), this means that exit time from any interval $(-\infty,b)$ is finite a.s. for each $b>0$. Hence, $\mathbb{P}(\tau<\infty)=1$.
Another approach: The exit time $\sigma:=\inf\{t>0;B_t \geq b\}$ is finite with probability $1$. Since $X_t \geq B_t$, it follows that $\inf\{t>0;X_t \geq b\}$ is finite almost surely, and so is $\tau$.