NOTE: $U\subseteq \mathbb R^n$ is an open, simply connected set.
I am reading L.C Evans's PDE book. I am on page 263. In order to show that Sobolev spaces are a kind of Banach space, we need to show that Cauchy sequences in $W^{k,p}(U)$ converge in $W^{k,p}(U)$.
Evans writes the following:
It remains to show that $W^{k,p}(U)$ is complete. So assume $\{u_m\}_{m\in\mathbb N}$ is a Cauchy sequence in $W^{k,p}(U)$. Then for each multi-index $\alpha$ satisfying $|\alpha|\leq k$, $\{\mathrm D^\alpha u_m\}_{m\in\mathbb N}$ is a Cauchy sequence in $L^p(U)$. Since $L^p(U)$ is complete, there exist functions $u_\alpha\in L^p(U)$ such that $$\mathrm D^\alpha u_m\overset{m\to\infty}{\longrightarrow} u_{\alpha}~~~\text{in}~~L^p(U)$$ for each $|\alpha|\leq k$. In particular $$ u_m\overset{m\to\infty}{\longrightarrow} u_{(0,\dots,0)}\equiv u~~~\text{in}~~L^p(U)$$ We now claim $$u\in W^{k,p}(U)~,~\mathrm D^\alpha u=u_\alpha \\ \forall \alpha~\text{s.t}~|\alpha|\leq k$$ To verify this assertion, let $\phi\in C^\infty_{\text{comp}}(U)$. Then $$\int_U u~\mathrm D^\alpha \phi~\mathrm d\mu^n = \lim_{m\to\infty}\int_{U} u_m~\mathrm D^\alpha \phi~\mathrm d\mu^n \tag{*}$$ $$ =\lim_{m\to\infty}(-1)^{|\alpha|}\int_U \phi~\mathrm D^\alpha u_m~\mathrm d\mu^n \\ =(-1)^{|\alpha|}\int_U u_\alpha~\phi~\mathrm d\mu^n$$
Which proves the claim. My question is about $\boldsymbol{(*)}$. That is, if $\psi$ is a smooth function of compact support in $U$, and $\{u_m\}_{m\in\mathbb N}$ is a Cauchy sequence in $L^p(U)$, is it true that $$\lim_{m\to\infty}\int_U u_m~\psi~\mathrm d\mu^n=\int_U \lim_{m\to\infty}u_m~\psi~\mathrm d\mu^n \tag{$\star$}$$ ? I tried convincing myself this is true, but couldn't. The monotone convergence theorem doesn't apply because the sequence is not necessarily monotone, and I couldn't apply the dominated convergence theorem because I couldn't construct a dominating function - I don't think you can, in general.
My question:
Can someone give a proof that $\boldsymbol{(\star)}$ holds?
