Let $s$ be a positive decreasing function, why does

$v s^{\leftarrow}(v) + \int_{s^{\leftarrow}(v)}^{\infty} s(u) \: \mathrm{d}u = \int_{0}^{v} s^{\leftarrow}(u) \: \mathrm{d}u \tag{*} $

hold? $s^{\leftarrow}(x)$ is here the generalized inverse function, so

\begin{align} s^{\leftarrow}(x)=\text{inf}\{y:s(y) \ge x \}. \end{align}

I know for $s^{-1}$ as the inverse function of $s$ and $S$ the antiderivative of $s$, we have

\begin{align} \int s^{-1} (u) \: \mathrm{d}u = us^{-1}(u) - S(s^{-1}(u)) + C. \tag{**} \end{align} If we have $s^{\leftarrow}$ as the inverse function and not just the generalized one, I see that $(**)$ implicates $(*)$, but with $s^{\leftarrow}$ as generalized inverse function, I don't see, why $s^{\leftarrow}(0)=\text{inf}\{y:s(y) \ge 0\}$ should be $\infty$.

I hope someone can help me out.


Link 1 (p. 79ff)

Link 2 (p. 162ff)


1 Answer 1


For every $n$ consider the function $s_n(v)=s(v)+\frac1ng(v)$, where $g$ is positive and strictly decreasing and integrable. Then $s_n$ is strictly increasing and so since you proved that (*) holds for strictly increasing functions you have that $$v s^{-1}_n(v)+\int^\infty_{s^{-1}_n(v)}s_n(u)\,du=\int_0^v s^{-1}_n(u)\,du.$$ Now you have to use the fact that as $n\to\infty$, $s^{-1}_n(v)\to s^{\rightarrow}(v)$ (see this link convergence generalized) at every point of continuity of $s^{\rightarrow}$. Since $0\le s_n(v)\le s(v)+g(v)$ for all $n$, we can apply LDTC to conclude that $$v s^{-1}(v)+\int^\infty_{s^{-1}(v)}s(u)\,du=\int_0^v s^{-1}(u)\,du$$ at every point of continuity $v$ of $s^{\rightarrow}$. At the other countably many points you can use the fact that $s^{\rightarrow}$ is right continuous.


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