# Clarification for a proof about Lipschitz approximation in $W^{1, p}(\mathbb{R}^n)$

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!