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 $u^{*}$.
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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 $c$? $\endgroup$– Dieter KadelkaCommented Apr 17, 2021 at 15:12
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2$\begingroup$ @DieterKadelka: I believe the OP is using $|S|$ to denote the Lebesgue measure of the set $S$. $\endgroup$– Willie WongCommented Apr 17, 2021 at 15:46
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1$\begingroup$ I don't know what OP is, if you mean $|\cdot|$, then yes, this is the Lebesgue measure. $\endgroup$– user99432Commented Apr 17, 2021 at 15:56
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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 $0$? $\endgroup$– Dieter KadelkaCommented Apr 17, 2021 at 18:02
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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 WeaverCommented Apr 17, 2021 at 22:32
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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 $u^* = 1_{[1,\infty)}$.
answered Apr 17, 2021 at 22:31
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$\begingroup$ 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... $\endgroup$ Commented Apr 17, 2021 at 23:28
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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? $\endgroup$ Commented Apr 18, 2021 at 2:04
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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? $\endgroup$ Commented Apr 18, 2021 at 9:47
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1$\begingroup$ But they can be replaced by other values, f.i. $a$ by $a-1$ and $b$ by $b+1$. $\endgroup$ Commented Apr 18, 2021 at 14:04