Let $p \in [1,\infty)$, and $W^{1,p} (\mathbb{R}^n)$ be the Sobolev space with functions on $\mathbb{R}^n$, and $W_0^{1,p} (\mathbb{R}^n)$ be the closure of smooth functions in $W^{1,p} (\mathbb{R}^n)$.
I want to show that these two spaces are the same. The idea I have is to show that there is a $R>0$ and a function $v \in W^{1,p} (\mathbb{R}^n)$ such that the support of $v$ is contained in $B_R(0)$, and that $$ \| u - v \|_{W^{1,p}(\mathbb{R}^n)} \le \varepsilon/2$$ and then I can mollify with a smooth function to obtain a function $v_h$ such that $$ \| v - v_h \|_{W^{1,p}(\mathbb{R}^n)} \le \varepsilon/2$$, from which the result should follow, since $v_h \in C^{\infty}_c(\mathbb{R}^n)$.
However, I am stuck at trying to show that such a $v$ exists in the first place. It feels obvious since I can cut the function $u$ off by multiplying by a smooth function, but I am struggling to show that most of the norm is contained in the ball.
One has $$ \|u\|_{W^{1,p}( \mathbb R^n\setminus B_R(0))} \to0 $$ for $R\to \infty$. Then take a cutoff function $\psi_R:\mathbb R^n\to [0,1]$ such that $\psi_R \in C_c^\infty(B_{R+1}(0))$, $\psi_R|_{B_R}=1$, and $|\psi'_R(x)|\le 2$, and define $v:= \psi_R u$. Then $v$ coincides with $u$ on $B_R(0)$, which contains most of the norm of $u$. Now, you need to make sure that the difference $u-v$ has small norm on $B_{R+1}\setminus B_R$: $$ \|u-v\|_{L^p(B_{R+1}\setminus B_R)} =\|(1-\psi_R)u\|_{L^p(B_{R+1}\setminus B_R)}\le \|u\|_{L^p(B_{R+1}\setminus B_R)}, $$ $$ \|\nabla u - \nabla v\|_{L^p(B_{R+1}\setminus B_R)} \le \|(1-\psi_R)\nabla u - u \nabla \psi_R\|_{L^p(B_{R+1}\setminus B_R)} \le \|\nabla u\|_{L^p(B_{R+1}\setminus B_R)} + 2\|u\|_{L^p(B_{R+1}\setminus B_R)} $$ It remains to choose $R$ large enough, that $v$ has the desired approximation property.