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
  
  $$f^*(x):=\int_0^{\infty} \chi_{\{|f|&gt;t\}^*}(x)dt,$$
  
  by comparison to the "layercake" representation of $f$, namely
  $$f(x)=\int_0^{\infty} \chi_{\{f&gt;t\}}(x)dt.$$


They say that it is then an obvious property that 

$$\{x: f^*(x)&gt;t\}=\{x: |f(x)|&gt;t\}^* .$$

But I can't see why/how...
Hint 1: show that if $t_1 &gt; t_2$, then $\{ |f| &gt; t_1\}^* \subseteq \{ |f| &gt; t_2\}^*$. 

Hint 2: use this to show that if $y\in \{ |f| &gt; t \}^*$, then $y \in \{|f| &gt; s\}^*$ 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 $y\not\in \{|f| &gt; t\}^*$, 
$$ \sup \left\{ s \geq 0 ~~|~~ y \in \{|f| &gt; s\}^*\right\} \leq t $$
this implies in particular $f^*(y) \leq t$ (why?).
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
$$ r(t) = C_n\sqrt[n]{\mathcal{L}^n\left(\{ |f| &gt; t \}\right)} $$
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!