# Evaluating a limit at a discontinuity of a monotone rearrangment (distribution function)

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.

• Cross-posted to Math.SE Nov 5, 2021 at 7:58
• Do you have a response to the answer below? Nov 10, 2021 at 2:46

$$\newcommand{\ep}{\varepsilon}$$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, $$t', and for some real $$\ep>0$$ and all $$t\in(t',b-a]$$ we have $$f^*(t)>\ep+f^*(t')$$; so, by (1), $$F(\ep+f^*(t')) 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 $$y_1 and $$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' and hence, once again by (1), $$y. Letting here $$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$$.