I have a question that occurred to me and has been bothering me, because maybe graphically it seems obvious but I don't know how to get there. It has to do with the distribution function and monotone rearrangment.


Given a bounded function $f\colon [a,b] \rightarrow \mathbb{R}$, the (right-continuous) distribution of $f$ is the function $D\_{f}$ definded by


\begin{align}
D\_{f}(y):= \mu(\{t\in I\colon f(t)\leq y \}),\; \forall y\in [\inf f, \sup f].
\end{align}


$\mu$ denotes de Lebesgue-measure. The (left-continuous) rearrangment of $f$, denoted by $f^{\ast}$, is the function:
\begin{align}
f^{\ast}(t):= \inf\{y\in [\inf f, \sup f] \colon D\_{f}(y)\geq t \},\; \forall t\in [0,b-a].
\end{align}


It turns out that $f^{\ast}$ is the left inverse function of $D\_{f}$, moreover it is the generalized inverse. They are both monotone increasing functions. Here is my question:


Let $t'\in [0,b-a]$ be a point such that $f^{\ast}$ is discontinuous in $t'$, this means $D\_{f}$ has a constant segment over $[y\_{j},y\_{j+1})$ of value $t'$, so


\begin{align}
t' = D\_{f}(y\_{j}) = D\_{f}(y), \; \forall y\in [y\_{j},y\_{j+1}).
\end{align}



> 
> **How can I proof that:
> \begin{align}
> \lim\_{t\to t'^{+}} f^{\ast}(t) = y\_{j+1}
> \end{align}**
> 
> 
> 


Clearly:
\begin{align}
\lim\_{t\to t'^{-}} f^{\ast}(t) = f^{\ast}(t') = f^{\ast}(D\_{f}(y\_{j}))=y\_{j}.
\end{align}


I would appreciate your help very much! If I am not being clear on something or you need more details please let me know.



Evaluating a limit at a discontinuity of a monotone rearrangment (distribution function)
$\newcommand{\ep}{\varepsilon}$Let
\begin{equation\*}
 c:=\inf f,\quad d:=\sup f,\quad I:=[a,b],\quad F:=D\_f, 
\end{equation\*}
so that
\begin{equation\*}
 F(y)=\mu(\{t\in I\colon f(t)\le y \})\quad \forall y\in[c,d].
\end{equation\*}
Also introduce the set
\begin{equation\*}
 E\_t:=\{y\in[c,d]\colon F(y)\ge t\}\quad \forall t\in[0,b-a].
\end{equation\*}
Since $F$ is nondecreasing and right-continuous, for all $t\in[0,b-a]$
\begin{equation\*}
 f^\*(t)=\inf E\_t=\min E\_t,\quad E\_t=[f^\*(t),d], 
\end{equation\*}
and, for all $y\in[c,d]$,
\begin{equation\*}
 F(y)\ge t\iff y\ge f^\*(t). \tag{1}
\end{equation\*}


Now take any $t'\in[0,b-a]$ such that $f^\*$ is discontinuous at $t'$. Since $f^\*$ is nondecreasing and left-continuous, we have $f^\*(t'+)>f^\*(t')$. So, $t'<b-a$, and for some real $\ep>0$ and all $t\in(t',b-a]$ we have $f^\*(t)>\ep+f^\*(t')$; so, by (1), $F(\ep+f^\*(t'))<t$ for all $t\in(t',b-a]$ and hence $F(\ep+f^\*(t'))\le t'$. On the other hand, again by (1), $F(f^\*(t'))\ge t'$. Since $F$ is nondecreasing, it follows that $F(y)=t'$ for all $y\in[f^\*(t'),\ep+f^\*(t')]$. So, introducing  

\begin{equation\*}
 Y:=\{y\in[c,d]\colon F(y)=t'\},
\end{equation\*}
\begin{equation\*}
 y\_1:=\inf Y,\quad y\_2:=\sup Y,
\end{equation\*}
we do have $y\_1<y\_2$ and $F(y)=t'$ for all $y\in[y\_1,y\_2)$.


Moreover, for any $y\in[y\_1,y\_2)$ and any $t\in(t',b-a]$, we have $F(y)=t'<t$ and hence, once again by (1), $y<f^\*(t)$. Letting here $y\uparrow y\_2$ and $t\downarrow t'$, we get $f^\*(t'+)\ge y\_2$.


Suppose now that $f^\*(t'+)\ne y\_2$. Then for some real $\ep>0$ we have $f^\*(t'+)>\ep+y\_2$. So, for all $t\in(t',b-a]$ we have $f^\*(t)>\ep+y\_2$, whence, once again by(1), $t>F(\ep+y\_2)$. So, $t'\ge F(\ep+y\_2)\ge F(y\_1)=t'$ and hence $F(\ep+y\_2)=t'$, which contradicts the definition $y\_2:=\sup Y$.


Thus, indeed $f^\*(t'+)=y\_2$.


