Symmetric-decreasing rearrangement of a function I'm studying section 3.3 of Analysis by Lieb and Loss, about symmetric-decreasing rearrangement of functions.

Let $A\subset \mathbb{R}^n$ a Borel set of finite Lebesgue measure. They define
  $A^*$ to be the ball centered at 0 with the same measure that
  $A$.
The symmetric-decreasing rearrangement of a measurable function $f:\mathbb{R}^n \to \mathbb{R}$ is then defined by
$$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt,$$
by comparison to the "layercake" representation of $f$, namely
  $$f(x)=\int_0^{\infty} \chi_{\{f>t\}}(x)dt.$$

They say that it is then an obvious property that 
$$\{x: f^*(x)>t\}=\{x: |f(x)|>t\}^* .$$
But I can't see why/how...
 A: Hint 1: show that if $t_1 > t_2$, then $\{ |f| > t_1\}^* \subseteq \{ |f| > t_2\}^*$. 
Hint 2: use this to show that if $y\in \{ |f| > t \}^*$, then $y \in \{|f| > s\}^*$ for every $0 \leq s \leq t$. Notice that this implies that $f^*(y) \geq t$ by the definition. 
Hint 3: use hint 1 again to show that if $y\not\in \{|f| > t\}^*$, 
$$ \sup \left\{ s \geq 0 ~~|~~ y \in \{|f| > s\}^*\right\} \leq t $$
this implies in particular $f^*(y) \leq t$ (why?). 
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, if $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  $f^*(y)\leq t$ if $y\notin \left\{x \in \mathbb{R}^n: f^*(x) > t  \right\}$. 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) > t  \right\}.
\end{equation}
Which end the prove by taking the complement.
A: I have been working on this subject for some months now and your question was one of the main obstacles I encountered. I think what makes this particular problem hard to attack is that it seems so obvious and geometrically correct that most would prefer not to bother with a formal proof. Anyway, what worked for me was proving first that
$$ r(t) = C_n\sqrt[n]{\mathcal{L}^n\left(\{ |f| > t \}\right)} $$
where $C_n = \left( \frac{n\Gamma(n/2)}{2\pi^{n/2}} \right)^{\frac{1}{n}}$, is a lower semicontinuous function. Then you can prove that the parameter $t$ in the integral which defines $f^*$ will define a set that contains $x$ just in an OPEN interval. Showing that it is an interval is pretty easy, but the open part is where you need the lower semicontinuity of the function $r$.
Best regards!
A: There is a direct proof that makes the fact "obvious": by the layer cake principle, one has $$f^{*}(x)=\int_{0}^{\infty}\chi_{\{|f|>t\}^{*}}(x)dt=\int_{0}^{\infty}\chi_{\{f^{*}>t\}}(x)dt$$ and whence $\chi_{\{|f|>t\}^{*}}=\chi_{\{f^{*}>t\}}$ by the fact that the two sequences of sets under consideration are respectively monotonic with respect to $t$.
