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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...

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Your statement is a bit suspect: if $A^*$ has the same measure as $A$, $A^*$ most likely will not be the unit ball. It will just be a ball. – Willie Wong Jan 16 '12 at 14:47
I think that some $t$'s in your definitions should be replaced by $x$'s. – Christian Blatter Jan 16 '12 at 14:53
Fixed, thanks and sorry for that. – Klaus Jan 16 '12 at 14:57
up vote 1 down vote accepted

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?).

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hint 3 is unnecessary. You already proved that the level set of $f^{*}$ is subset of the star set. – TKM Feb 10 '14 at 23:48

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