Let $t_1<t_2 \in [0,T]$, $f(t)\ge0$ and $Z_t$ an increasing process with $Z_0= 0$.

We have clearly that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \mathcal{X}_{t_1<T}(Z_{t_2} - Z_{t_1})\min_{t_1\le s \le t_2}f(s) \end{equation}holds.

In a paper I'm reading the authors state based on this and using the fact that $Z_{t_1}\ge0$ it follows that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \mathcal{X}_{t_1<T}Z_{t_2}(\min_{t_1 \le s \le t_2}(f(s) - f(t_2)) + f(t_2)) \end{equation} holds.

I'm confused how that can be, it seems like an error! Surely since $Z_{t_1}\ge0$ we must have $(Z_{t_2} - Z_{t_1}) \le Z_{t_2}?$ Would appreciate any insight into anything I may have glossed over here!

For reference I am talking about the beginning of page 264 of the following paper: Boetius, Frederik, and Michael Kohlmann. "Connections between optimal stopping and singular stochastic control." Stochastic Processes and their Applications 77.2 (1998): 253-281.

Let $t_1<t_2 \in [0,T]$, $f(t)\ge0$ and $Z_t$ an increasing process with $Z_0= 0$.

We have clearly that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \chi_{t_1<T}(Z_{t_2} - Z_{t_1})\min_{t_1\le s \le t_2}f(s) \end{equation}holds.

In a paper I'm reading the authors state based on this and using the fact that $Z_{t_1}\ge0$ it follows that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \chi_{t_1<T}Z_{t_2}(\min_{t_1 \le s \le t_2}(f(s) - f(t_2)) + f(t_2)) \end{equation} holds.

I'm confused how that can be, it seems like an error! Surely since $Z_{t_1}\ge0$ we must have $(Z_{t_2} - Z_{t_1}) \le Z_{t_2}?$ Would appreciate any insight into anything I may have glossed over here!

For reference I am talking about the beginning of page 264 of the following paper: Boetius, Frederik, and Michael Kohlmann. "Connections between optimal stopping and singular stochastic control." Stochastic Processes and their Applications 77.2 (1998): 253-281.