Continuous residuals localize on C1 domains

If $h(a_0)>0$ at some $a_0\in \mathrm{\Omega }$, continuity gives a closed ball $\overline{B_r(a_0)}\subset \mathrm{\Omega }$ and a constant $c>0$ with $h\ge c$ on that ball. Let $V=B_r(a_0)$. At each sphere point $p$, choose rigid coordinates centered at $a_0$ with the third axis parallel to $p-a_0$. Near $p$, the sphere is the graph $z=\sqrt{r^2-\mathrm{\mid }y{\mathrm{\mid }}^2}$ over a disk about $y=0$ (reversing the third axis if needed), and $V$ lies on one side; hence $V$ is a bounded $C^1$ subdomain. Choose a nondegenerate closed cube $Q\subset V$. By [F1] the integral over $V$ exists; monotonicity [F2] and the cube-content formula [F3] give ${\int }_{V}1dx\ge {\int }_{Q}1dx=\mathrm{vol}(Q)>0$, hence ${\int }_{V}hdx\ge c{\int }_{V}1dx>0$. This contradicts the zero-integral hypothesis. If instead $h(a_0)<0$, the same ball construction gives $h\le -c$ on $V$ and ${\int }_{V}hdx\le -c{\int }_{V}1dx<0$, contradicting either stated integral hypothesis.

The zero-integral case therefore has neither positive nor negative values, so $h=0$ throughout $\mathrm{\Omega }$. The nonnegative-integral case has no negative values, so $h\ge 0$ throughout $\mathrm{\Omega }$. These are the two asserted conclusions.