Is the generalized inverse of the gen. inverse of $f$ equal to $f$? Let $f:\mathbb{R}^+\mapsto\mathbb{R}^+\cup\{\infty\}$ be non-decreasing, right-continuous.
define $g(x\in \mathbb{R^+})=\inf\{y:f(y)>x\}$.
It can be shown that $g$ is non-decreasing, right-continuous.
Take any $x\in \mathbb{R}^+$. Let $L=\inf\{y:g(y)>x\}$.
I want to show that $L=f(x)$.

Here's what I've tried:  
Take any $z:g(z)>x$. Then $L\leq z$.
Also, for any $\epsilon_1>0$ there is an $a:g(a)>x$ such that $L+\epsilon_1 > a$.  
Take any $v:f(v)>z$. Then $g(z) \leq v$.
Also, for any $\epsilon_2>0$ there is a $b:f(b)>z$ such that $g(z)+\epsilon_2 > b$.  
But this was just writing out the definition of infimum. The fact that there's so many variables floating around already makes me think this is not the right way to go.
 A: We write $L=L(x)$ for the generalised right-continuous inverse of $g=g(y)$. The aim is to show that $L(x) = f(x)$ for each $x\in [0,\infty)$. 
Let $x$ be fixed. Suppose first that $f(x) < \infty$. By right continuity, $f(x+\epsilon) < \infty$ for $\epsilon>0 $ small. Then
$$ g( f(x+\epsilon)) \overset{\Delta}= \inf\{z : f(z) > f(x+\epsilon)\}\overset{\text{non-decr.}}{\ge} x+\epsilon > x.$$
Therefore by the definition of an infimum,
$$ f(x+\epsilon) \ge \inf\{ y : g(y) > x\} =L(x).$$
Sending $\epsilon \downarrow 0$, we obtain with right continuity of $f$
$$ \fbox{$f(x) \ge L(x).$}$$
For the other inequality, for any $y_0<\infty$ such that $$y_0>L(x)=\inf\{ y : g(y) > x\},$$ we have since $g$ is non-decreasing, and $y_0$ is larger than an infimum,
$$ \inf\{ z:f(z) > y_0\} = g(y_0) >x$$
this implies by the definition of infimum again,
$$ f(x) \le y_0.$$
(If it was true that $f(x) > y_0$, then $\inf\{ z:f(z) > y_0\}\le x.)$
As $y_0$ was arbitrary,
$$\fbox{$f(x) \le L(x).$}$$
Now for the case $f(x) = \infty$: for any $y\in[0,\infty]$, we have
$$g(y) = \inf\{ z : f(z) > y\} \le x.$$
Therefore
$$ L(x)=\inf\{y: g(y) > x\}=\inf\emptyset=\infty = f(x).$$
My source seems to be cliffnotes made from the book of Revuz and Yor by Shiu-Tang Li.
This also looks related.
