nondecreasing rearrangement is equimeasurable Two functions $f(x)$ and $g(x)$ are called equi-measurable if $m(\{x:f(x)>t\})=m(\{x:g(x)>t\})$.
Nondecreasing rearrangement of a function $f(x)$ is defined as $$f^*(\tau)=\inf\{t>0:m(\{x:f(x)>t\}\leq\tau\}.$$
Prove that $f^*(\tau)$ and $f(x)$ are equimeasurable.
 A: 
It is sufficient to prove that $\{f>t\}^* = \{f^*>t\}$ for all $t>0$

Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}$. One can check that for every, $0<s< t$ one has $$\left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s  \right\}$$ this entails that,
\begin{equation}\label{eq-inclu t-s}\tag{I}
\left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s  \right\}^{*}~~~\textrm{for all $s\in ]0,t[$}.
\end{equation}
this implies that,$$ \mathbf{1}_{\left\{ | f| > s \right\}^*}(y)  =1 ~~~s\in (0,t)$$
Therefore, from definition of $f^{*}$,  if $y\in \{|f|>t\}^*$ then we have
$$\begin{align*}
f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
&= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
& = \int_{0}^{t} ds+\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds \\ &>t.
\end{align*}$$
Whence, $$\left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}\subset \left\{ x \in \mathbb{R}^n:f^{*}(x)> t \right\}.$$
On the other hand, if we suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}$ then for all  $s>0$ such that $ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s  \right\}^{*}$ one has $0<s\leq t$.
Indeed, $t>s $ then from  \eqref{eq-inclu t-s} $$y\in \left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,
$$\sup\left\{s>0 : y\in \left\{x \in \mathbb{R}^n: |f(x)| > s  \right\}^{*}\right\}\leq t. $$
We then deduce that,
$$\begin{align*}
f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
&= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \underbrace{\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds}_{=0}\leq=t
\end{align*}$$
that is  $f^*(y)\leq t$ or that $y\notin \left\{x \in \mathbb{R}^n: f^*(x) > t  \right\}$. We've just prove that,
\begin{equation}\label{eq}\tag{II}
\Bbb R^n\setminus \left\{x \in \mathbb{R}^n: |f(x)| > t  \right\}^{*}\subset \Bbb R^n\setminus\left\{x \in \mathbb{R}^n: f^*(x) > s  \right\}~~~\textrm{for all $s\in ]0,t[$}.
\end{equation}
Which end the prove by taking the complementary.
