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I was reading a proof about the Lipschitz approximation of functions $u\in W^{1, p}(\mathbb{R}^n)$. There the author defines a set $$E_{\lambda}=\{x\in\mathbb{R}^n:M|\nabla u|(x)\leq \lambda\}, \quad \lambda>0, $$ where $M$ is the centered maximal function $$ Mf(x)=\sup_{r>0}\frac{1}{|B(x, r)|}\int_{B(x, r)}f(y)\text{d}y. $$

Then he shows that $u$ is $c\lambda$-Lipschitz a.e. in $E_{\lambda}$. Then using McShane extension theorem one can find a $c\lambda$-Lipschitz function $v:\mathbb{R}^n\rightarrow\mathbb{R}$ s.t. $v=u$ a.e. in $E_{\lambda}$. Now we use truncation for the function $v$ by defining a new function $v_{\lambda}=\min\{\lambda, \max\{v, -\lambda\}\}$, or $$ v_{\lambda}(x)=\begin{cases} v(x) & |v(x)|\leq\lambda \\ \lambda & v(x)>\lambda \\ -\lambda & v(x)<-\lambda. \end{cases} $$ Now comes the part I don't really understand. He claims that $v_{\lambda}$ is $2c\lambda$-Lipschitz, but in my opinion $v_{\lambda}$ is $c\lambda$-Lipschitz. Also he states that $v_{\lambda}=u$ a.e. in the set $E_{\lambda}$, but I'm not able to see that myself. Also, does it follow from this that the weak gradients also agree, i.e. $\nabla v_{\lambda}=\nabla u$ a.e. in $E_{\lambda}$?

Any help is appreciated!

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