Under the same assumptions of this early question, consider also a the random time $T_a := \inf\{ t > 0: B_t \geq a\}$ which is a stopping time. Since $M^\lambda$ is a continuous martingale, Doob's optional sampling theorem implies that
$$ \mathbb E \left\{ \exp(\lambda \ B_{T_a\ \wedge\ n} -\frac{\lambda^2}{2}({T_a\ \wedge\ n})) \ |\ \mathcal F_s \right\}=\mathbb E \{M^\lambda _{T_a\ \wedge\ n}\ |\ \mathcal F_s \}= M^\lambda _0=1$$
Then, we can show that conditionally to $\{ T_a < + \infty \}$
$$ \exp(\lambda \ B_{T_a\ \wedge\ n} -\frac{\lambda^2}{2}({T_a\ \wedge\ n})) \underset{n\rightarrow 0}{\overset{\mathbb P -\text{ae}}\longrightarrow} e^{\lambda a}\exp(-\frac{\lambda^2}{2}{T_a}) $$
and conditionally to $\{ T_a = + \infty \}$
$$ \exp(\lambda \ B_{T_a\ \wedge\ n} -\frac{\lambda^2}{2}({T_a\ \wedge\ n})) \underset{n\rightarrow 0}{\overset{\mathbb P -\text{ae}}\longrightarrow} 0 $$
Now, my claim is that this second event has null measure. In other words,
$$\mathbb P \{T_a < + \infty \}=1 .$$ Indeed, since $0 \leq \exp(\lambda \ B_{T_a\ \wedge\ n} -\frac{\lambda^2}{2}({T_a\ \wedge\ n}))\leq e^{\lambda a}$, the dominated convergence theorem implies
$$e^{\lambda a}\mathbb E \left\{\exp(-\frac{\lambda^2}{2}{T_a})\mathbf 1_{\{ T_a < + \infty\}}\ |\ \mathcal F_s \right\}= \mathbb E \left\{\lim_{n \rightarrow +\infty}\exp(\lambda \ B_{T_a\ \wedge\ n} -\frac{\lambda^2}{2}({T_a\ \wedge\ n})) \mathbf 1_{\{ T_a < + \infty\}} \ |\ \mathcal F_s \right\}=1 $$
At this point we can construct a monotonic decreasing sequence $(\lambda_n)$ converging to zero, witch give us by monotone convergence theorem that $$\mathbb P \{T_a < + \infty \}=1 .$$
Therefore, we are able to conclude that $$\mathbb E \left\{\exp(-\frac{\lambda^2}{2}{T_a})\ |\ \mathcal F_s \right\}= e^{-\lambda a}$$
Could any fellow check out my arguments, specially the dominated convergence thm., please?
Thanks in advance.
