Clarification of proof that if Du=0 almost everywhere in a Sobolev space, then u is constant.

Question

From a question in PDEs by Evans, if we have that $U$ is connected and $u\in W^{1,p}(U)$ satisfies $Du = 0$ almost everywhere in $U$, then $u$ is constant in $U$ almost everywhere. The proof I've beens shown is as follows:

*Take a compact set $K \in U$, then $u \in L^1(K)$.

Then the weak derivative $D(u \ast \eta_\epsilon)=Du \ast \eta_\epsilon = 0$. Since $U$ is connected and $u \ast \eta_\epsilon$ is smooth, then $u \ast \eta_\epsilon$ is equal to a constant $c_\epsilon$ which converges to $u$ in $L^1(K)$ as $\epsilon \rightarrow 0$. Then {$c_{1/n}$} forms a cauchy sequence

*since K has finite measure

that converges to a constant $c$. Therefore, $u = c$ almost everywhere in $K$.

*and hence in $U$. $\square$

My issues with the proof are everything I starred. Why does $u$ becomes $L^1$ on a compact set? Why does the fact that $K$ has finite measure matter (I see why it has finite measure though, it's compact so that is clear)? Why does $u=c$ in K imply that $u=c$ in $U$; if U was, say, an open ball I don't see how this could be the case.

Answer 1

1. By Holder's inequality, $$\|u\|_{L^1(K)} = \int_K |u|\cdot 1 \le \|u\|_{L^p}|K|^{1/q}$$
2. If $|K|=\infty$ then unless $c_n= c$ eventually, $c_n$ does not converge to $c$ in $L^1(K)$.
3. Remember that $K$ is an arbitrary compact subset of $U$; every point of $U$ is in one such $K$.

Answer 2

Any compact set has finite Lebesgue measure. On any finite measure space functions in $L^{p}$ are also in $L^{1}$ for $1 \leq p \leq \infty$. Finally we can express $U$ as an increasing countable union of compact subsets. Since $u=c$ almost everywhere on any compact subset it follows that $u=c$ almost everywhere on $U$.