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.