The proof for this theorem is quite long and Im reading it. But why is this not immediate? If the function is 0 up to the stopping time and we integrate it up to the stopping time, the integral should be 0.
Does that argument not work because the integral is integrating with respect to every possible brownian and not a single trajectory. Therefore, it's aggregating over all the possible stopping times when calculating f(w,s). So the stopping time up to a single stopping time might be something non-zero?