The Laplace transform of the first hitting time of Brownian motion Let $B_t$ be the standard Brownian motion process, $a &gt; 0$, and let $H_a = \inf \{ t : B_t &gt; a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is
$$\mathbb{E}[\exp(-\lambda H_a)] = \exp (-\sqrt{2\lambda} H_a)$$
by considering the martingale $$M_t = \exp \left(\theta B_t -\frac{1}{2}\theta^2 t\right)$$

There's an obvious argument to follow here: assuming the optional stopping theorem applies, we have
$$1 = \mathbb{E}[M_{H_a}] = \mathbb{E} \left[ \exp \left(\theta a - \frac{1}{2}\theta^2 H_a\right) \right] = \exp(\sqrt{2\lambda} a) \mathbb{E} \left[ \exp(-\lambda H_a) \right]$$
where $\theta = \sqrt{2\lambda}$. This is exactly what we wished to show. However, as far as I can tell, the hypotheses of the optional stopping theorem are not satisfied here. Here is the statement I have:


  If $(X_n)$ is a martingale and $T$ is an a.s. bounded stopping time, then $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.


I think not all is lost yet. $M_t &gt; 0$ for all $t$, so the martingale convergence theorem applies, and $M_t \to M_\infty$ a.s. for some integrable random variable $M_\infty$. For each $t$, $H_a \wedge t = \min \{ H_a, t \}$ is a bounded stopping time, so certainly $\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_0]$. But,
$$\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] + \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}]$$
and clearly what one wants to do is to take $t \to \infty$ on both sides. But here's where I get stuck: I'm sure I need a convergence theorem here in order to conclude that the equation remains valid in the limit. 

Now, $0 &lt; M_{H_a} = \exp \left(\theta a - \frac{1}{2} \theta^2 H_a \right) \le \exp(\theta a)$, so the dominated convergence theorem applies, and so
$$\lim_{t \to \infty} \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a &lt; \infty\}}]$$
and I believe Fatou's lemma gives me that
$$\liminf_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}] \ge \mathbb{E}[M_{\infty} \mathbf{1}_{\{H_a = \infty\}}]$$
but I think what I need is the equality
$$\lim_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}] = \mathbb{E}[M_\infty \mathbf{1}_{\{H_a = \infty\}}]$$
and as far as I can tell neither the monotone convergence theorem nor the dominated convergence theorem applies here. Is there anything I can do to rescue this line of thought?
Use the fact that $0\leqslant M_t\mathbf{1}_{\{H_a&gt;t\}}\leqslant\exp\left(\theta a−\frac12\theta^2t\right)$.
I don't think you need to worry about the case $H_a = \infty$ - for the simple reason that this is an event of probability zero.