Consider a standard Brownian motion $Y_t$ and the stopping time $$T = \inf \{ t \geq 0: Y_t = -1 \}$$ Show that $P(T < \infty)=1$
I fix $t>0$, and I want to show that $T$ is a.s. bounded.
To this end, I compute, using the fact that $Y_t - N(0,t)$:
$$P( T < t)=P(Y_t > -1)= \int_{-1}^{\infty} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}} dx = \int_{-1}^{0} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}} dx + \int_{0}^{\infty} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}} dx =\int_{0}^{1} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}} dx +\frac{1}{2}$$.
In the last equality I used that the normal density is an even function. Taking the limit as $t \rightarrow + \infty$, I obtain $$P(T< \infty) \geq \frac{1}{2}$$
Now I use a self-similarity argument: I fix a $n \in \mathbb{N}$, and note that for the BM to hit $-n$ starting at $0$ it has to hit $-1, -2 \ldots, -(n-1)$. Moreover, the probability of hitting $-1$ starting at $0$ (call it $\theta$) is larger than $\frac{1}{2}$ and it is equal to the probability of the BM to hit $-2$ starting at $-1$, and so on. So
$$ \frac{1}{2} \leq P(T_{-n} < \infty) =\theta^n $$
Therefore $\theta \geq \frac{1}{2}^{\frac{1}{n}}$ and hence as $n$ goes to infinity I have $\theta \geq 1$, which implies $\theta =1$ since it's a probability measure.
All in all, $$P(T<\infty) = \theta =1$$
Let's set a filtred space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t\geq0}, \mathbb{P})$. We have $B$ a BM issued from $0$, i.e $B_0 = 0$ We define the r.v. \begin{equation} \tilde{T}(\omega) = \inf\{t \geq 0, B(t,\omega) = 1\} \quad \forall\omega \in \Omega \end{equation} We can easily check that $\tilde{T}$ is in fact a stopping time as it is the first hitting time of the continuous process $\mathcal{F}_t$-adpated $B$ in the closed subset $\{1\}$.
Let's work with the geometric Brownian motion defined as : \begin{equation} \forall t \geq 0, \quad M_t^\lambda = \exp\left(\lambda B_t - \frac12\lambda^2t\right) \end{equation} We can easilty show that the process $M$ is a continuous $\mathcal{F}_t$-martingale and that the r.v. $\tilde{T}\wedge n$ with $n\in \mathbb{N}$ is still a stopping time (as the min of two stopping time is still a stopping time) and it is bounded.
Thanks to the Doob's optional stopping theorem, the process $M_{\tilde{T}\wedge n}$ is a martingale and we have:
\begin{equation} \mathbb{E}_0\left[M_{\tilde{T}\wedge n}^\lambda\right] = \mathbb{E}_0\left[M_{0}^\lambda\right] = 1 \quad a.s. \end{equation} Now, we can rewrite the first term as: \begin{equation} \mathbb{E}_0\left[M_{\tilde{T}\wedge n}^\lambda\right] = \mathbb{E}\left[M_{\tilde{T}\wedge n}^\lambda 1_{\tilde{T} < +\infty}\right] + \mathbb{E}\left[M_{\tilde{T}\wedge n}^\lambda 1_{\tilde{T} =+\infty}\right] \tag{1} \end{equation} On the event $\{\tilde{T} < +\infty\}$, we have $\lim_{n\to\infty} M_{\tilde{T}\wedge n}= M_\tilde{T}$. On the event $\{\tilde{T} = +\infty\}$, we have that $B_t \leq 1$ for all $t\geq0$. If $\lambda>0$, then we have that: \begin{equation} \exp\left(\lambda B_{\tilde{T} \wedge n} - \frac12\lambda^2\tilde{T}\wedge n\right) = \exp\left(\lambda B_{n} - \frac12\lambda^2n\right) \to 0 \quad a.s. \end{equation}
Note that we have \begin{equation} \forall n \geq \mathbb{N}^*, \quad 0\leq M_{\tilde{T}\wedge n}^\lambda 1_{\tilde{T} < +\infty} \leq e^{\lambda} \end{equation} Using the Lebgues theorem, we can pass on the limit in $(1)$ (+using the continutiy) and end up with :
\begin{equation} \mathbb{E}\left[M_{\tilde{T}}^\lambda 1_{\tilde{T} < +\infty}\right] = e^{-\lambda}\mathbb{E}\left[\exp\left(-\frac12\lambda^2\tilde{T}\right)1_{\tilde{T} < +\infty}\right] \end{equation} Now using a decreasing sequence $\lambda_n$ to $0$ and using Beppo-levi theorem, we have: \begin{equation} \mathbb{P}(\tilde{T}< +\infty) = 1 \quad a.s. \end{equation}
We know that $B_t = -Y_t$ is still a Brownian motion. Then $\tilde{T} = \inf\{t \geq 0, B_t = 1\} \sim {T} = \inf\{t \geq 0, Y_t = -1\}$. Hence, \begin{equation} \mathbb{P}({T}< +\infty) = 1 \quad a.s. \end{equation}