Thank you LSpice, that is a great suggestion. I am using a workaround for the time being by just writing $\int_{[0,t_2[}f(s)d\xi^*_s \ge \xi^*_{t_2}\min_{0 \le s \le t_2}f(s)$ (since $\xi^*_0=0$).

Yes I agree with you. I am wringing my hands making sense of this hieroglyphic paper for a dissertation, no one seemed to understand my plight. Apparently it has 57 citations, I wonder if the 57+ authors cracked this issue. In any case thanks a lot for your time LSpice, I really appreciate it!

Yes I found it confusing too! I believe they are using it like so: $\chi_{A}=1$ if $A$ is true and $\chi_{A}=0$ otherwise. So yes $\chi_{t_1<T}Z_{t_2}$ is a product.

where $\xi^*_t$ is my increasing process with $\xi^*_0 = 0$ and $\sigma<\sigma^\delta.$

Hi LSpice, to give you more context: When they are writing $J_f \ge \chi_{\sigma<T}(Z^x_\sigma - Z^{x-\delta}_\sigma)f(\sigma) + \delta I_5(\delta) + \delta I_6(\delta),$ by plugging in the terms we supposedly have $\chi_{\sigma^\delta<T}(X^*_{\sigma^\delta} - Z^{x-\delta}_\sigma)f(\sigma^\delta) + \int_{[0,\sigma^\delta[}f(s)d\xi^*_s \ge \chi_{\sigma^\delta<T}(X^*_{\sigma^\delta} - Z^{x-\delta}_{\sigma^\delta})f(\sigma^\delta) + \chi_{\sigma<T}\xi^*_{\sigma^\delta}f(\sigma^\delta) + \chi_{\sigma<T}\xi^*_{\sigma^\delta}\min_{\sigma \le s \le \sigma^\delta}(f(s) - f(\sigma^\delta))$

Thank you LSpice, you are 100% right here

Hi LSpice - thanks for answering - the problem is that this is an argument used in a proof that $b \ge c$, so can't really make that assumption here. I'm seriously thinking this entire argument is a mistake