Brownian motion hitting a line Consider the line $a+bt$ where $a,b>0$. Let $B(t)$ be Brownian motion and let $\tau=\inf\{t>0:B(t)=a+bt\}$ be the first hitting time of that line, with the understanding that $\tau=\infty$ if the line is never hit. I want to compute the probability that Brownian motion hits that line, i.e. $P(\tau<\infty)$.
There are two steps in my work that I am not confident in.
Define the process $X(t)=B(t)-bt$ and let $T_a=\inf\{t>0:X(t)=a\}$, again with $T_a=\infty$ if $X(t)$ does not hit $a$. Then $\tau$ and $T_a$ have the same distribution (unsure about this) so we may instead calculate $P(T_a<\infty)$.
Let $\tilde{a}>0$ and define $T_{a,-\tilde{a}}=\inf\{t>0:X(t)=a \text{ or } X(t)=-\tilde{a}\}$ with $T_{a,-\tilde{a}}=\infty$ if $\{a,-\tilde{a}\}$ is never hit. The process $e^{2bX(t)}$ is a martingale and Doob's Optional Stopping theorem gives
$
\begin{align*}
E(e^{2bX(T_{a,-\tilde{a}})})&=1, \\
e^{2ba}P(X(T_{a,-\tilde{a}})=a)+e^{-2b\tilde{a}}P(X(T_{a,b})=-\tilde{a})&=1.
\end{align*}
$
Letting $\tilde{a}\to\infty$,
$\begin{align*}
e^{2ba}\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)&=1, \\
\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)&=e^{-2ba}, \\
P(T_a<\infty)&=e^{-2ba}.
\end{align*}
$
The last step where I say $\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)=P(T_a<\infty)$ I am also unsure about.
 A: To answer the question why $\tau$ and $T_a$ have the same distribution: Since the process $X(t)$ is defined as $X(t) = B(t) - bt$, the equations $B(t) = a +bt$ and $X(t) = a$ are equivalent. Therefore, $\tau$ and $T_a$ are the infimum of the same set, hence they are equal as random variables and in particular identically distributed.
As for the second question why
$$\lim_{\tilde a \to \infty} P(X(T_{a,-\tilde{a}}) = a) = P(T_a < \infty)$$
holds: The random variable $T_{a,-\tilde a}$ is the first time that $X(t)$ hits one of the boundaries of the region $-\tilde a < x < a$. The event $X(T_{a,-\tilde a}) = a$ occurs iff $X(t)$ hits the upper bound $a$ before ever hitting the lower bound $-\tilde a$. We can assume that $X(t)$ is continuous since the Brownian motion $B(t)$ is continuous almost surely. Then it is clear that the events $\{\omega : X(T_{a,-\tilde a}) = a\}$ are increasing as $\tilde a \to \infty$ (if $\tilde b > \tilde a$ and $X(t)$ never goes below $-\tilde a$ before hitting $a$, it doesn't go below $-\tilde b$ either) and that their union is $\{\omega: T_a < \infty\}$ (if $X(t)$ hits $a$ at some point, then there exists a lower bound $-\tilde a$ such that $X(t)$ doesn't go below $-\tilde a$ before hitting $a$). This implies 
$$\lim_{\tilde a \to \infty} P(X(T_{a,-\tilde{a}}) = a) = \lim_{n \to \infty} P(X(T_{a,-n}) = a)$$
and the claim follows from the general fact that for a countable increasing sequence of events $A_1 \subseteq A_2 \subseteq \ldots$ one has $P(\bigcup_n A_n) = \sup_n P(A_n)$.
