1
$\begingroup$

Let $A\subset \mathbb{R}$ be measurable such that there are $a,b\in \mathbb{R}$, $a<b$ fulfilling $[b,\infty)\subset A\subset [a,\infty)$. The right rearrangement of $A^{*}$ of $A$ is defined as $A^{*}=[c,\infty)$ where $c:=b-|A\cap [a,b]|$. Now we can define the increasing rearrangement as follows: For a function $u:\mathbb{R}\to[0,1]$ fulfilling $\lim_{x\to -\infty} u(x)=0$ and $\lim_{x\to \infty} u(x)=1$, let the increasing rearrangement $u^{*}:\mathbb{R}\to [0,1]$ be the function which fulfils for all $t\in (0,1)$ \begin{align*} \{x\in\mathbb{R}: u^{*}(x)\geq t\}=\{x\in \mathbb{R}: u(x)\geq t\}^{*}. \end{align*} To show: Assume that $(u_n)_{n\in \mathbb{N}}:\mathbb{R}\to [0,1]$ is a sequence which converges pointwise a.e. to $u$. Then the sequence of increasing rearrangements $(u_n^{*})_{n\in \mathbb{N}}$ is converging a.e. to $u^{*}$.

I asked this question on Math SE: https://math.stackexchange.com/questions/4083976/increasing-rearrangement-convergence

I didn't get any answer. I changed some parts, but I recognized, that it was wrong. Here is my proof: Let $N$ be a nullset such that so that $u_n(x_0)\to u(x_0)$ for all $x_0\in \mathbb{R}\setminus N$. Let $t_0=u(x_0)$. We have $x_0\in \{x\in \mathbb{R}:u^{*}(x)\geq t_0\}=\{x\in \mathbb{R}: u(x)\geq t_0\}^{*}$. By the pointwise convergence, for $\varepsilon >0$ we can choose $N\in \mathbb{N}$ so that for all $n\geq N$ \begin{align*} u_n(x_0)-\varepsilon\leq u(x_0)\leq u_n(x_0)+\varepsilon. \end{align*} And here comes already a mistake: I concluded by the pointwise convergence and the definition of the right rearrangement \begin{align*} x_0\in \{x\in \mathbb{R}: u_n(x)\geq t_0-\varepsilon\}^{*}=\{x\in \mathbb{R}: u_n^{*}(x)\geq t_0-\varepsilon\}. \end{align*} Then I used $x_0\notin \{x\in \mathbb{R}: u^{*}(x)\geq t_0+\delta\}$ for all $\delta >0$. Therefore \begin{align*} x_0\notin \{x\in \mathbb{R}: u^{*}(x)\geq t_0+2\varepsilon\}=\{x\in \mathbb{R}: u(x)\geq t_0+2\varepsilon\}^{*}. \end{align*} I used again pointwise convergence to get \begin{align*} x_0\notin \{x\in \mathbb{R}: u_n^{*}(x)\geq t_0+\varepsilon\}=\{x\in \mathbb{R}: u_n(x)\geq t_0+\varepsilon\}^{*}. \end{align*} I concluded the pointwise convergence of $u_n^{*}$ to $u^{*}$... Is there any hope left to "repair" my proof ?

$\endgroup$
7
  • 1
    $\begingroup$ Sorry, what is $c = b - |A \cap [a,b]|$? $A \cap [a,b]$ maybe an uncountable set, or what is the exact meaning of $c$? $\endgroup$ Apr 17 '21 at 15:12
  • 2
    $\begingroup$ @DieterKadelka: I believe the OP is using $|S|$ to denote the Lebesgue measure of the set $S$. $\endgroup$ Apr 17 '21 at 15:46
  • 1
    $\begingroup$ I don't know what OP is, if you mean $|\cdot|$, then yes, this is the Lebesgue measure. $\endgroup$
    – user99432
    Apr 17 '21 at 15:56
  • 1
    $\begingroup$ I have one problem with your question: What is $A^*$ for $A := \{x \in \mathbb{R} : u(x) \geq t\}$? You have defined $A^*$ if $A$ is bounded from below, but $A$ as defined above need not be bounded from below, even for measurable $u$, F.i. what if $u(x)$ is strictly decreasing from $1$ to $0$? $\endgroup$ Apr 17 '21 at 18:02
  • 1
    $\begingroup$ @user99432 "OP" is internet-speak for "original poster", i.e., the person who made the original post (in this case, you). $\endgroup$
    – Nik Weaver
    Apr 17 '21 at 22:32
2
$\begingroup$

How about $u_n = 1_{[-n-1, -n] \cup [1,\infty)}$ (the characteristic function of $[-n-1, -n] \cup [1,\infty)$)? Then $u_n^* = 1_{[0,\infty)}$ for all $n$ but $u_n$ converges pointwise to $u = 1_{[1,\infty)}$, which has $u^* = 1_{[1,\infty)}$.

$\endgroup$
9
  • $\begingroup$ I think your example is very nice. Honestly I'm confused now. Your example doesn't fit into the definition of right rearrangement since $[-n-1,-n]\cup [1,\infty)$ is not an interval. This right rearrangement was introduced by Alberti and Belletini in link.springer.com/article/10.1007/s002080050159 (Definition 5.5). So it seems that this definition is even not well-defined for $u$ only satisfying this limit conditions... $\endgroup$
    – user99432
    Apr 17 '21 at 23:28
  • 2
    $\begingroup$ I'm baffled by this comment ... the right rearrangement is $1_{[0,\infty)}$ and $[0,\infty)$ is an interval. Is $1_{[-n-1, -n] \cup [1,\infty)}$ somehow not allowed as an initial function? Why? $\endgroup$
    – Nik Weaver
    Apr 18 '21 at 2:04
  • 1
    $\begingroup$ @user99432: Note that $a$ and $b$ in your original definition are not explicitely given. They may be quite arbitrary but lead to the same $A^*$. Is this the problem? $\endgroup$ Apr 18 '21 at 9:47
  • 1
    $\begingroup$ At least $\lambda$-a.e. $\endgroup$ Apr 18 '21 at 11:08
  • 1
    $\begingroup$ But they can be replaced by other values, f.i. $a$ by $a-1$ and $b$ by $b+1$. $\endgroup$ Apr 18 '21 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.