Im trying to show that for an open set $U\subset\mathbb{R}^n$ and a function $u\in W^{1,1}(U)$, also the positive part $u^+$ is in $W^{1,1}(U)$.
My idea is the following:
Let $E\subseteq U$ defined by $$E := \{ x\in U|\,u(x) > 0.$$ It is clear from $L^p$-theory that $u^+\in L^1(U)$. What remains to show is that the weak derivative $\partial_i u^+\in L^1(U)$, and the function I defined indeed is the weak derivative.
I defined $$\partial_i u^+(x) = \begin{cases}\partial_i u(x),&x\in E\\0,&x\not\in E\end{cases}$$ where $\partial_i u$ is the weak derivative of $u$. Then it holds that $$\begin{eqnarray*}\|\partial_iu^+\|_{L^1(U)} &=& \int_U |\partial_iu^+(x)|\,dx\\&=&\int_E |\partial_i u(x)|\,dx + \int_{U\setminus E}0\,dx\\&=&\int_E |\partial_i u(x)|\,dx\\&\leq& \int_U |\partial_i u(x)|\,dx\\&\overset{\partial_iu\in L^1(U)}<&\infty\end{eqnarray*}$$
What remains is just to show that $\partial_iu^+$ is indeed the weak derivative of $u^+$. Therefore we need to show $$\int_U \varphi(x)\partial_iu^+(x)\,dx = -\int_U \partial_i\varphi(x) u^+(x)\,dx,\quad\forall\varphi\in C^\infty_0(U).$$
Now I just calculate $$\begin{eqnarray*}\int_U\varphi(x)\partial_iu^+(x)\,dx &=& \int_E \varphi(x)\partial_iu(x)\,dx\\&\overset{\text{P.I.}}=& \underbrace{\int_{\partial E}\varphi(x)u(x)\,dS}_{=0\text{ as }u_{|\partial E}=0} - \int_E D_i\varphi(x)u(x)\,dx\\&=&-\int_E \partial_i\varphi(x)u(x)\,dx - \int_{U\setminus E}\varphi(x)\cdot 0\,dx\\&=& - \int_U \partial_i \varphi(x)u^+(x)\,dx\end{eqnarray*}$$
Did I miss something or is this everything? I'm somehow confused because this task gives a lot of points on an exercise-sheet of mine, so maybe it has to be more difficult?
Your approach is only valid if the set $E$ is sufficiently nice. Otherwise you cannot justify the integration by parts on $E$.
In fact, the set $E$ is only defined up to measure zero, but the boundary $\partial E$ may change, if you modify $E$ by a set of measure zero, so your integration by parts cannot hold.