Let $U\subset \mathbb R^n$ be open, $p \in [1,\infty>$ and $k \in \mathbb N$. We know the following:
(*) If $\phi_\epsilon$ is a mollifier and $f\in L^p(U)$ than $f_\epsilon :=f*\phi_\epsilon \in C^\infty\cap L^p(U)$ and $|| f-f_\epsilon ||_{L^p(U)} \rightarrow 0 $ as $\epsilon \rightarrow 0$.
Now if $f\in W^{k,p}(U)$ than we have a result about local approximation which gives us that $f_\epsilon \rightarrow f$ in $W^{k,p}_{loc}(U)$ as $\epsilon \rightarrow 0$ (i.e. it converges in the $W^{k,p}(V)$ norm on all V compactly included in U).
For the proof of the claim first we have to note that for all multiindices $\alpha$ with $\vert{\alpha}\vert\le k$ we have $D^\alpha f_\epsilon(x)=(\phi_\epsilon*D^\alpha f)(x)$ on U. Now the textbook proof notes that convergence of $f_\epsilon$ to $f$ in $W^{k,p}_{loc}(U)$ is equivalent to the convergence in $L^p(V)$ of all $D^\alpha f_\epsilon$ to $D^\alpha f$, for all V compactly contained in U, which is true because $D^\alpha f_\epsilon(x)=(\phi_\epsilon*D^\alpha f)(x)$, and the right-hand side does indeed converge to the weak derivative since it is its mollification.
My question is why can't we use (*) to immediately obtain a global approximation in the same way? Why do we need to restrict ourselves to compactly contained sets for the argument to work? It seems to me that I have some fundamental misunderstanding of how these things work, but I can't work it out by myself.