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.
nondecreasing rearrangement is equimeasurable
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10What are your thoughts on the matter? Do you see a point where to start? – 2011-12-30
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0This proof can be found in "Classical Fourier Analysis", Grafakos, Proposition 1.4.5 (12) – 2016-02-25
1 Answers
You want to prove that $\{f>t\}^* = \{f^*>t\}\,\,\text{?}$
Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. One can check that for every, $0 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\}.$$
On the other hand, if we suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$ then for all $s>0$ such that $ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}$ one has $0
Indeed, $t>s $ then from \eqref{eq-inclu t-s} $$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.