The RMK functional outer content is an outer measure

The definition gives ${\mu }^{\ast }(\mathrm{\varnothing })=0$ and monotonicity: an open superset of $F$ is also one of $E$ when $E\subseteq F$.

Let $E\subseteq {\bigcup }_n{E}_n$ and choose open $U_n\supseteq E_n$. If $U={\bigcup }_n{U}_n$ and $0\le f\le {\mathbf{1}}_U$ has compact support $K$, finitely many $U_{n_j}$ cover $K$. By [L1] there are ${\phi }_j\in C_c(X)$ subordinate to those sets with sum $1$ near $K$. Then $f={\sum }_{j}f{\phi }_j$, each summand is admissible for $U_{n_j}$, and positivity and linearity give $\mathrm{\Lambda }(f)\le {\sum }_j\rho (U_{n_j})\le {\sum }_n\rho (U_n)\mathrm{.}$ Taking the supremum over $f$ yields $\rho (U)\le {\sum }_n\rho (U_n)$.

If ${\sum }_n{\mu }^{\ast }(E_n)=\mathrm{\infty }$, countable subadditivity is automatic. Otherwise, for $\epsilon >0$ choose $U_n\supseteq E_n$ with $\rho (U_n)\le {\mu }^{\ast }(E_n)+\epsilon 2^{-n-1}$. Step 1.2 and $E\subseteq U$ give ${\mu }^{\ast }(E)\le \rho (U)\le {\sum }_n{\mu }^{\ast }(E_n)+\epsilon \mathrm{.}$ Letting $\epsilon \downarrow 0$ proves countable subadditivity.