$\int_{0}^{\infty}{f(t)dt}=\int_{0}^{\infty}{f^{-1}(t)dt}$ for generalized inverse

Question

Let $f^{-1}$ be the generalized inverse of a decreasing right-continuous function $f:(0, \infty)\to [0, \infty)$ defined by $$f^{-1}(y)=\sup\{x>0:f(x)>y\}.$$ I can prove that $f^{-1}$ is then decreasing and right-continuous. Assuming that $\lim_{t\to \infty}f(t)=0$ why does it hold that $$\int_{0}^{\infty}{f(t)dt}=\int_{0}^{\infty}{f^{-1}(t)dt}?$$ Here it is claimed (for increasing $f$) that $$v f^{-1}(v) + \int_{f^{-1}(v)}^{\infty} f(u) \: \mathrm{d}u = \int_{0}^{v} f^{-1}(u) \: \mathrm{d}u.$$ Hence, is it clear that we have $$v f^{-1}(v)=0\quad \text{as}~ v\to \infty$$ or is there another way to prove the above?

Thanks in advance!

Answer 1

Proving is a bit subtle in the general case. In case you want an intuitive reason why it holds, suppose your function is a strictly decreasing and continuous bijection of $(0,+\infty)$. Make a drawing to convince yourself that the following two domains are the same for $0<a<b$:

$$ \{ (x,y) : a\leq x \leq b, f(b) \leq y \leq f(x)\} $$ and $$ \{ (x,y) : a \leq x \leq f^{-1}(y), f(b) \leq y \leq f(a) \}. $$

Calculating areas: $$ \int_a^b (f(x)-f(b)) \ dx = \int_{f(b)}^{f(a)} (f^{-1} (y) -a)\ dy $$ Setting $v=f(a)$ and letting $b\to +\infty$ you get the stated identity (both sides could be $+\infty$). Also letting $a\to 0$ both sides are monotone increasing, whence converge in $(0,+\infty]$. You get the identity between (improper) Riemann integrals $$ \int_0^\infty f(x) \ dx = \int_0^\infty f^{-1} (y)\ dy .$$