# Convergence of increasing rearrangment

Let $$A\subset \mathbb{R}$$ be measurable such that there are $$a,b\in \mathbb{R}$$, $$a fulfilling $$[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 ?

• Sorry, what is $c = b - |A \cap [a,b]|$? $A \cap [a,b]$ maybe an uncountable set, or what is the exact meaning of $c$? Apr 17, 2021 at 15:12
• @DieterKadelka: I believe the OP is using $|S|$ to denote the Lebesgue measure of the set $S$. Apr 17, 2021 at 15:46
• I don't know what OP is, if you mean $|\cdot|$, then yes, this is the Lebesgue measure. Apr 17, 2021 at 15:56
• 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 $0$? Apr 17, 2021 at 18:02
• @user99432 "OP" is internet-speak for "original poster", i.e., the person who made the original post (in this case, you). Apr 17, 2021 at 22:32

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 $$u^* = 1_{[1,\infty)}$$.
• 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... Apr 17, 2021 at 23:28
• 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? Apr 18, 2021 at 2:04
• @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? Apr 18, 2021 at 9:47
• At least $\lambda$-a.e. Apr 18, 2021 at 11:08
• But they can be replaced by other values, f.i. $a$ by $a-1$ and $b$ by $b+1$. Apr 18, 2021 at 14:04