I'm having a hard time understanding a step of the following :
Start :
Let $B$ be a Brownian motion starting at 0 and let $\lambda \in \mathbb{R} .$ Define $M_{t}^{\lambda}=e^{\lambda B_{t}-\lambda^{2} t / 2} $ $ \text { and } T_{a}=\inf \left\{t \geq 0: B_{t}=a\right\} $
(1) First, let us remark that for all $t \geq 0, M_{t}^{\lambda} \geq 0$ and $$ \mathbb{E}\left[M_{t}^{\lambda}\right]=e^{-\lambda^{2} \frac{t}{2}} \mathbb{E}\left[e^{\lambda B_{t}}\right]=e^{-\lambda^{2} \frac{t}{2}} e^{\lambda^{2} \frac{t}{2}}=1 $$ Let $s \leq t$ $$ \begin{aligned} \mathbb{E}\left[M_{t}^{\lambda} | \mathcal{F}_{s}\right] &=\mathbb{E}\left[e^{\lambda B_{t}-\lambda^{2} \frac{t}{2}} | \mathcal{F}_{s}\right] \\ &=\mathbb{E}\left[e^{\lambda\left(B_{t}-B_{s}\right)-\lambda^{2} \frac{t}{2}} e^{\lambda B_{s}} | \mathcal{F}_{s}\right] \end{aligned} $$ since $\left(B_{t}-B_{s}\right)$ is independent of $\mathcal{F}_{s}$ and $B_{s}$ is $\mathcal{F}_{s}$ measurable $$ =e^{\lambda B_{s}-\lambda^{2} \frac{t}{2}} \mathbb{E}\left[e^{\lambda\left(B_{t}-B_{s}\right)}\right] $$ $$ \begin{array}{l} {=e^{\lambda B_{s}-\lambda^{2} \frac{t}{2}+\lambda^{2} \frac{t-s}{2}}} \\ {=e^{\lambda B_{s}-\lambda^{2} \frac{s}{2}}} \\ {=M_{s}^{\lambda}} \end{array} $$ Hence $M^{\lambda}$ is a martingale.
(2) Let $m \geq 0$ and let us denote $S_{m}=\inf \left\{T_{a}, T_{0}, m\right\} .$ Hence, $S_{m}$ is a stopping time since it is a minimum of stopping times. Furthermore, $S_{m}$ is bounded, and we have thanks to the optional stopping theorem, $$ \mathbb{E}_{x}\left[M_{S_{m}}^{\lambda}\right]=\mathbb{E}_{x}\left[M_{0}^{\lambda}\right]=e^{\lambda x} $$ Furthermore, if $T_{0}=T_{a}=+\infty, 0 \leq B_{t} \leq a$ and for $\lambda>0, M_{S_{m}}^{\lambda}=$ $\exp \left(\lambda B_{m}-\lambda^{2} \frac{m}{2}\right) \leq \exp \left(\lambda a-\lambda^{2} \frac{m}{2}\right) \rightarrow_{m \rightarrow 0} 0$ When $T_{0}<+\infty$ and $T_{a}<+\infty,$ we have $M_{T_{0}}^{\lambda}=e^{-\lambda^{2} \frac{T_{0}}{2}} \leq 1$ and $$ M_{T_{a}}^{\lambda}=e^{\lambda a-\lambda^{2} \frac{T_{a}}{2}} \leq e^{\lambda a}, \text { and } M_{T_{0}}^{\lambda} \rightarrow \lambda_{\rightarrow 0} 1 \text { and } M_{T_{a}}^{\lambda} \rightarrow \lambda \rightarrow 0 \quad 1, \text { hence } $$ $M_{\mathrm{inf}\left(T_{0}, T_{a}\right)}^{\lambda} \rightarrow_{\lambda \rightarrow 0} 1 .$ Hence we have $$\color{red}{ \mathbb{E}_{x}\left[M_{\mathrm{inf}\left(T_{0}, T_{a}\right)}^{\lambda} \mathbb{1}_{\sup \left(T_{0}, T_{a}\right)<+\infty}\right]+\mathbb{E}_{x}\left[M_{m}^{\lambda} \mathbb{1}_{\inf \left(T_{0}, T_{a}\right)=+\infty}\right]=e^{\lambda x} } $$
By the dominated convergence theorem, $$ \mathbb{E}_{x}\left[M_{m}^{\lambda} \mathbb{1}_{\inf \left(T_{0}, T_{a}\right)=+\infty}\right] \rightarrow_{m \rightarrow \infty} 0 $$
End.
why is $\mathbb{E}_{x}[M_{\inf{\{T_{0}, T_{a}\}}}^{\lambda} \mathbb{1}_{\sup{\{T_{0}, T_{a}\}}<+\infty}]+\mathbb{E}_{x}[M_{m}^{\lambda} \mathbb{1}_{\inf{\{T_{0}, T_{a}\}}= +\infty}] =\mathbb{E}_{x}[M^{\lambda}_{\inf{\{m, T_0, T_a\}}}]$ ?
Shouldn't it be $$ \mathbb{E}_{x}\left[M_{\mathrm{inf}\left(T_{0}, T_{a}\right)}^{\lambda} \mathbb{1}_{\color{red}{\inf \left(T_{0}, T_{a}\right)<+\infty}}\right]+\mathbb{E}_{x}\left[M_{m}^{\lambda} \mathbb{1}_{\inf \left(T_{0}, T_{a}\right)=+\infty}\right]=e^{\lambda x} $$