The mass distribution principle

The function ${\mu }^{\ast }$ vanishes on the empty set, is monotone, and agrees with $\mu$ on Borel sets: inclusion gives the lower inequality, while the set itself is a candidate hull. It is countably subadditive: for finite ${\sum }_j{\mu }^{\ast }(E_j)$ choose Borel supersets $B_j$ of costs at most ${\mu }^{\ast }(E_j)+\epsilon 2^{-j-1}$ and use $\mu ({\bigcup }_j{B}_j)\le {\sum }_j\mu (B_j)$. The latter follows by disjointifying the Borel sets and applying countable additivity. Infinite right sides are automatic. Let $\epsilon$ decrease to zero. Thus it is an outer measure.

Fix $0<\delta <r_0$ and any admissible cover $(U_j)$ of $A$. Subadditivity and the assumed bound give ${\mu }^{\ast }(A)\le {\sum }_j{\mu }^{\ast }(U_j)\le C{\sum }_j(\mathrm{diam}{U}_j{)}^s$. Infimising gives ${\mathcal{H}}_{\delta }^s(A)\ge {\mu }^{\ast }(A)\mathrm{/}C$, even if no cover exists. Taking the supremum and using the critical-exponent criterion proves both conclusions; at $s=0$ the dimension bound is the automatic nonnegativity.

For the ball hypothesis, fix a nonempty $U$ of diameter $d<r_0$ and one $x\in U$. For every $d<r<r_0$, $U\subseteq B(x,r)$, so ${\mu }^{\ast }(U)\le C{r}^s$. Let $r$ decrease to $d$. If $d=0$ and $s>0$ the bound is zero; if $s=0$ it is $C$, as required by the covering-cost convention. This proves the sufficient ball condition without evaluating $\mu$ on a non-Borel set.