I'm wondering if $\int_{a}^{b}g(z)d(-z)=-\int_{a}^{b}g(z)dz$. On the one hand, intuitively this seems true as this looks like we are putting a negative sign on the $\Delta z$ in the Rieman sum, so it should act like we could taking the constant -1 outside the integral. On the other hand, using rigorous change of variable, let $t=-z$ and we will get $\int_{a}^{b}g(z)d(-z)=\int_{a}^{b}g(-t)dt$.
My question is: how to interpret $\int_{a}^{b} g(z)d(-z)$ (or what does $d$ of something represent), and intuitively why you cannot simply take the minus sign out? Thanks!
You should change the domain of the integral as well : $$\int_a^bg(z)d(-z)=\int g(z)\chi_{[a,b]}(z)d(-z)=\int g(-t)\chi_{[-b,-a]}(t)dt=\int_{-b}^{-a}g(-t)dt$$$$=-\int_a^bg(t)dt.$$