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 ?

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