Positive smooth densities give Radon volume

For each point $p$ in positive dimension, a chart contains a relative closed ball or half-ball $H$ around its coordinate image. Choose it bounded with closure inside the chart image. Its inverse image $K$ is compact and contains a neighborhood $W$ of $p$. The continuous coefficient on $H$ is bounded by a finite $C$, so ${\mu }_r(W)\le {\mu }_r(K)={\int }_H{r}_x\le C{\lambda }_n(H)<\mathrm{\infty }$. For $n=0$ take $W=K=\{p\}$, whose measure is the finite coefficient $r(p)$.

The neighborhoods $W$ cover any compact $K_0$ finitely, giving ${\mu }_r(K_0)\le {\sum }_{j=1}^m{\mu }_r(W_j)<\mathrm{\infty }$. They also show local finiteness. To get a countable cover, take the members of a countable base that are contained in some such $W$; these cover $M$ and individually have finite measure. Enumerating these basis members proves sigma-finiteness without selecting neighborhoods at every point.

The same compact chart neighborhoods show local compactness also at the boundary; Hausdorffness and second countability are standing assumptions. The compact-finite Borel measure therefore satisfies the regularity theorem. Its conclusion includes the outer and open-set inner regularity required by the stated Radon convention.

Apply the completion theorem to $(M,\mathcal{B}(M),{\mu }_r)$ under the standing countable choice. Explicitly, $E=B\cup N$ with $B,Z$ Borel, $N\subset Z$ and ${\mu }_r(Z)=0$ has ${\bar{\mu }}_r(E)={\mu }_r(B)$. Empty $M$ and empty compact sets have mass zero; a singleton in dimension zero has its finite positive weight. No total-mass bound is asserted.