The RMK functional outer content is well defined

Every pointwise-admissible $f$ is nonnegative, so $\mathrm{\Lambda }(f)\ge 0$. The [given, L1] zero function is admissible for every open set, including the empty set; for $U=\mathrm{\varnothing }$ it is the only admissible function. Thus $\rho (U)\in [0,\mathrm{\infty }]$ and $\rho (\mathrm{\varnothing })=0$.

The cutoff supremum is at most the pointwise supremum. Conversely, let [L1, L2] $0\le f\le {\mathbf{1}}_U$ have compact support $K$, and choose $h\in C_c(X)$ with $0\le h\le 1$ and $h=1$ on $K$ by applying [L2] to $K\subseteq X$. For $t>0$ put $f_t=(f-th{)}_{+}$. Then $f_t\prec U$: its support lies in the compact set $\{f\ge t\}\subseteq U$. Moreover $0\le f-f_t\le th$, so positivity gives $\mathrm{\Lambda }(f)\le \mathrm{\Lambda }(f_t)+t\mathrm{\Lambda }(h)$. Letting $t\downarrow 0$ proves that the two displayed suprema defining $\rho (U)$ are equal.

If $U\subseteq V$, every test function admissible for $U$ is admissible for $V$, hence $\rho (U)\le \rho (V)$. The family of open supersets of any $E$ is nonempty because it contains $X$, so ${\mu }^{\ast }(E)$ is well defined in $[0,\mathrm{\infty }]$.

For open $U$, using $U$ itself in the infimum gives [step 2.1] ${\mu }^{\ast }(U)\le \rho (U)$. Conversely, if $U\subseteq V$ with $V$ open, monotonicity gives $\rho (U)\le \rho (V)$; taking the infimum over such $V$ gives the reverse inequality. This also covers $U=\mathrm{\varnothing }$ and $\rho (U)=\mathrm{\infty }$.