Theorem 4.14 of Brownian Motion and Stochastic Calculus I have been reading the proof of
Theorem 4.14 of Karatzas' book. I wonder whether there is a typo in the description of the process $\eta^{(n)}_{t}$ as $\xi^{(n)}_{t+}-\min(\lambda,A_{t})$ whenever $t\in [0,a)$ because in the definition of $T_{n}(\varepsilon)$ it seems that $\eta^{(n)}_{t}$ were defined as $\xi^{(n)}_{t}-\min(\lambda,A_{t})$ (without considering the right hand limit). Now if the correct definition of $\eta^{(n)}_{t}$ is $\xi^{(n)}_{t}-\min(\lambda,A_{t})$ whenever $t\in [0,a)$, why this process turns out to be right-continuous on the whole interval $[0,a]$?
The right-continuity would be necessary in order that $T_{n}(\varepsilon)$ to be a optional time.
 A: The key is to prove that\begin{gather*}
\inf\{t \in [0, a] \mid ξ_{t^+} - B_t > ε\} = \inf\{t \in [0, a] \mid ξ_t - B_t > ε\},\tag{1}
\end{gather*}
where (almost surely):

*

*$ξ_t$ is right-continuous except on a finite set $D$,

*$B_t$ is right-continuous and increasing,

*$ξ_t \geqslant B_t$, and $ξ_t = B_t$ on $D$.

Define $E = \{t \in [0, a] \mid ξ_t - B_t > ε\}$, $F = \{t \in [0, a] \mid ξ_{t^+} - B_t > ε\}$. Note that $E \cap D = \varnothing$, thus\begin{gather*}
E = \{t \in [0, a] \setminus D \mid ξ_t - B_t > ε\}\\
= \{t \in [0, a] \setminus D \mid ξ_{t^+} - B_t > ε\} \subseteq F,
\end{gather*}
which implies that $\inf E \geqslant \inf F$. If $F = \varnothing$, then $\inf E = \inf F = a$. Otherwise denote $t_0 = \inf F$. Since $|D| < ∞$, there exists $δ_0 > 0$ so that $ξ_t$ is right-continuous on $(t_0, t_0 + δ_0)$.
Case 1: $t_0 \notin F$. For any $t \in (t_0, t_0 + δ_0)$, if $t \in F$, then$$
ξ_t - B_t = ξ_{t^+} - B_t > ε \Longrightarrow t \in E.
$$
Therefore $t_0 \geqslant \inf E$, so (1) is true.
Case 2: $t_0 \in F$. Define $ε_1 = ξ_{t_0^+} - B_{t_0} > ε$. Since$$
\lim_{t → t_0^+} (ξ_t - B_t) = \lim_{t → t_0^+} ξ_t - \lim_{t → t_0^+} B_t = ξ_{t_0^+} - B_{t_0} = ε_1,
$$
then there exists $0 < δ_1 < δ_0$ such that$$
|(ξ_t - B_t) - ε_1| < \frac{1}{2} (ε_1 - ε),\quad\forall t \in (t_0, t_0 + δ_1).
$$
Therefore $(t_0, t_0 + δ_1) \subseteq E$, and $t_0 \geqslant \inf E$, so (1) is true.
