Hausdorff measure is an outer measure

The empty cover has cost zero at every scale; a cover of a larger set covers each subset. Hence both functions vanish at the empty set and are monotone. This uses no positive-exponent assumption.

Fix $\delta$ and $(A_j)$. If ${\sum }_j{\mathcal{H}}_{\delta }^s(A_j)=\mathrm{\infty }$, subadditivity is automatic. Otherwise, for $\epsilon >0$ select for each $j$ a cover with cost at most ${\mathcal{H}}_{\delta }^s(A_j)+\epsilon 2^{-j-1}$; select an enumerated cover, so the resulting double family is countable. Empty $A_j$ may use the empty family.

Flatten these covers along an enumeration of the pairs of indices. Every finite subfamily cost is bounded by the corresponding iterated sum, and every finite rectangle is eventually included; thus the nonnegative sums agree. The union has scale cost at most ${\sum }_j{\mathcal{H}}_{\delta }^s(A_j)+\epsilon$. Letting $\epsilon$ decrease to zero proves the fixed-scale assertion.

For finite $\delta$, the same inequality is at most ${\sum }_j{\mathcal{H}}^s(A_j)$. This bound is independent of $\delta$; taking the supremum proves the claimed inequality, including infinite right sides. Together with the first step these are all the outer-measure axioms.