On rearrangement of level set: $\{f>t\}^* = \{f^*>t\}\,\,\text{?}$ Let $A$ be a subset of $\mathbb{R}^n$ then the rearrangement of $A$ denoted by $A^*$ is the ball $B(0,r)$ having the same volume as $A$ i.e if  $|A| =|B(0,r)|$  with respect to Lebesgue measure then
$$A^*= B(0,r)$$
Let $f$ be a function from $\mathbb{R}^n$ to $\mathbb{R}$. Then its symmetric decreasing rearrangement $f^*$ is the function defined for $x \in  \mathbb{R}^n$ by
$$f^*(x) = \int_0^{\infty} 1_{\{f>t\}^*}(x) dt.$$
Where $1_{\{f>t\}^*}$ is the characteristic function of the set $\{f>t\}^*= B(0,r_t )$ on $\mathbb{R}^n$ for suitable $r_t >0$.
The set $\{f>t\} := \{x \in \mathbb{R}^n: f(x)>t\}$ is called the $t$-level set of the function $f$.

Question. How can I show that
  $$\{f>t\}^* = \{f^*>t\}\,\,\text{?}$$ 

This is mentioned to be easy in the book of Elliott Lieb and Loss (Analysis second edition, Graduate Studies in Mathematical,
vol 14, American mathematical Society, providence, RI 2001).
 A: 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.
A: First, by definition the rearrangement $f^*$ has the same distribution function as $f$, that is
$$|\{f>t\}|=|\{f^*>t\}|.$$
Also, since $f^*$ is a radially decreasing function, its level set $\{f^*>t\}$ is an open ball centered at the origin.
The volume of that ball is $|\{f^*>t\}|=|\{f>t\}|$.
On the other hand, $\{f>t\}^*$ is also an open ball centered at the origin with volume $|\{f>t\}|$. This implies $\{f>t\}=\{f^*>t\}$.
