Let $V$ be a positive function defined on a semi-infinite interval of form $(z,\infty)$ that is non-decreasing.
I want to show $\int^\infty_t V(u) u^{-2} - V(t) = \int^\infty_t u^{-1}V(du)$, $t>0$, where I assume $V(du)$ refers to the measure arising from the increasing function $V$. ($V$ doesn't necessarily have density)
How would one show this? The text says to use Fubini's theorem.
For context, I'm trying to understand this line in the book Extreme Values, Regular Variation and Point Processes.
Resnick, Sidney I., Extreme values, regular variation, and point processes, Applied Probability, Vol. 4, New York etc.: Springer-Verlag. XII, 320 p.; DM 145.00 (1987). ZBL0633.60001.
I think there is a slight typo in your formulation of the question. If your intent is to understand the last line of your text, then the objective is to show
$$t\int_t^{\infty} V(u)u^{-2}\,du - V(t) = t\int_t^{\infty}u^{-1}V(du).$$
Towards that end, note that since $V(u) = V(t) + \int_t^uV(dy)$, it holds $$t\int_t^{\infty} V(u)u^{-2}\,du - V(t) = t\int_t^{\infty}\int_t^uu^{-2}V(dy)du.$$ By Fubini's theorem $$t\int_t^{\infty}\int_t^uu^{-2}V(dy)du = t\int_t^{\infty}\int_y^{\infty}\,u^{-2}\,duV(dy) = t\int_t^{\infty}y^{-1}V(dy),$$ which is what the text claims.