nondecreasing rearrangement is equimeasurable

Question

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.

Answer 1

Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. This is equivalent to say that $$c_d|y|^d<|\{|f|>t\}|,\qquad \qquad c_d=|B(0,1)|. $$ Assume that $$\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds=0.$$ Then $y\not \in \{|f| > s\}^{*}$ for all $s\geq t$. That is, for all $s\geq t$ we have $$c_d|y|^d\geq |\{|f|>s\}| $$ The Fatou lemma implies $$c_d|y|^d\geq |\{|f|\geq t\}|\geq |\{|f|>t\}|.$$ This is in contradiction with the fact that $c_d|y|^d<|\{|f|>t\}|$. Therefore we deduce $$\int_{t}^{\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds>0.$$ On the other hand, 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,$$ \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) =1 ~~~s\in (0,t)$$

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, $$\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{ x \in \mathbb{R}^n:f^{*}(x)> t \right\}.$$ Conversely, suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. If $t\leq s$ then analogously to relation \eqref{eq-inclu t-s} we get $$y\notin \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}.$$ That is for all $s\geq t$ we get $$\mathbf{1}_{\{ | f| > s\}^*}(y)=0$$ Hence necessarily for all $s>0$ such that $ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}$ one has $0<s\leq t$. 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) > t \right\}. \end{equation} Which ends the prove by taking the complementary.