Positive open-set and metric-ball volume

Choose $p\in O$ and a chart around it restricted inside $O$. For $n\ge 1$, continuity and $r_x(x(p))>0$ give a relative coordinate neighborhood where $r_x\ge c=r_x(x(p))\mathrm{/}2>0$. This neighborhood contains a closed nondegenerate Euclidean box $Q$ in the interior of the half-space: even if $p$ is on the face, move its last coordinate a sufficiently small positive distance and choose a still smaller box. Hence ${\mu }_r(O)\ge {\int }_Q{r}_x\ge c\mathrm{vol}(Q)>0$.

For $n=0$ the singleton $\{p\}$ is open with measure $r(p)>0$, so again ${\mu }_r(O)>0$. The empty manifold has no nonempty open subset, making this clause vacuous.

If $q\in B_d(p,R)$, then $ϵ=R-d(p,q)>0$ and the triangle inequality gives $B_d(q,ϵ)\subset B_d(p,R)$. Thus the ball is open in the metric topology, hence in the manifold topology and Borel. It contains $p$ since $d(p,p)=0<R$, so the preceding positivity applies.

Suppose $K=\overline{B_d(p,R)}$ is compact. Each point of $K$ has a chart neighborhood contained in a bounded coordinate box on whose closure the smooth coefficient $r_x$ has a finite upper bound; the chart formula and finite Euclidean box measure make that neighborhood's measure finite. In dimension zero a singleton chart has finite weight instead. A finite subcover of $K$ therefore gives ${\mu }_r(K)<\mathrm{\infty }$. Monotonicity yields ${\mu }_r(B_d(p,R))\le {\mu }_r(K)<\mathrm{\infty }$.