Brownian motion and stopping time Let $(W_r)_r$ be a Brownian motion and $\alpha>0.$ Let $\theta_{\alpha}=\inf\{ r \geq 0,W_r=r\alpha+\alpha \}$ and $U_{\alpha}=\inf \{r \geq 0,W_r \geq r\alpha+\alpha\}.$
I want to know (prove rigorously) that $W_{\theta_\alpha \wedge r} \leq \alpha+\alpha(\theta_\alpha \wedge r)$ and why on the event $\{\theta_\alpha<\infty\},$ we have $W_{\theta_{\alpha}}=\theta_{\alpha}\alpha+\alpha.$ These two properties, are they true for $U_{\alpha}?$
 A: These two properties are essentially a consequence of the continuity of $t \mapsto W_t$.  Since that property only holds a.s., we can only prove these properties hold a.s. as well.  Below, we are working exclusively on the event $\{t \mapsto W_t \text{ is continuous}\}$.
First, we will show $W_{\theta_\alpha} = \theta_\alpha \alpha + \alpha$ on the event $\{\theta_\alpha < \infty\}$.  By the definition of $\theta_\alpha$, we can find a sequence $(r_n) \downarrow \theta_\alpha$ with $r_n \ge 0$ and $W_{r_n} = {r_n} \alpha + \alpha$.  Passing to the limit and using continuity of $t \mapsto W_t$, we conclude $W_{\theta_\alpha} = \theta_\alpha \alpha + \alpha$.
To show $W_{\theta_\alpha \wedge r} \le \alpha + \alpha(\theta_\alpha \wedge r)$, it's enough to show $W_t \le \alpha + \alpha t$ for any $t \le \theta_\alpha$.  Suppose to the contrary there exists $t \le \theta_\alpha$ with $W_t > \alpha + \alpha t$, or equivalently $W_t - \alpha t > \alpha$.  Since $f(r) := W_r - \alpha r$ is continuous and $f(0) = 0$ and $f(t) > \alpha$, the intermediate value theorem implies that there exists $s \in (0,t)$ with $f(s) = \alpha$.  However, $\theta_\alpha = \inf\{r : f(r) = \alpha\},$ so $\theta_\alpha \le s < t$.  This contradicts our assumption $t \le \theta_\alpha$, so we conclude that we must have $W_t \le \alpha + \alpha t$ for any $t \le \theta_\alpha$.
These properties do also hold for $U_\alpha$.  The proof is essentially the same, except that when we showed $W_{\theta_\alpha} = \theta_\alpha \alpha + \alpha$ you would instead show $W_{U_\alpha} \ge \theta_\alpha \alpha + \alpha$.  Combining this with the fact that $W_{U_\alpha\wedge r} \le \alpha + \alpha (U_\alpha \wedge r)$ allows us to conclude $W_{U_\alpha} = \alpha + \alpha U_\alpha$ on the event $\{U_\alpha < \infty\}$.
