One-dimensional Hausdorff measure on the line is Lebesgue outer measure

Only the reverse inequality needs proof. If ${\lambda }_1^{\ast }(A)=\mathrm{\infty }$, the lower inequality already gives equality. Otherwise choose an elementary-set cover of total volume less than ${\lambda }_1^{\ast }(A)+\epsilon$. Each finite-volume elementary set is a finite disjoint union of bounded half-open intervals; zero-volume empty pieces can be discarded. This is the elementary algebra and its volume appearing in the defining infimum.

Fix $\delta >0$. Subdivide each such interval into finitely many half-open intervals of lengths at most $\delta$, without changing the sum of lengths. Flatten the resulting countable family. It is an admissible Hausdorff cover of cost at most ${\lambda }_1^{\ast }(A)+\epsilon$. Hence ${\mathcal{H}}_{\delta }^1(A)\le {\lambda }_1^{\ast }(A)$ after letting $\epsilon$ decrease to zero. Taking the supremum over $\delta$ proves equality. For empty $A$ use the empty cover.

For every test set $T$ the two outer values in the Carathéodory splitting identity are identical. Thus a set satisfies that identity for one outer measure if and only if it does for the other, and the restricted values agree. Both restrictions are their complete Carathéodory measures.