Lipschitz curves and dominated interval vector measures

Define the increment measure on the rational interval algebra. On the algebra generated by rational half-open intervals and $\{0\}$, set ${\nu }_0(\{0\})=0$. For a finite disjoint union $A={⨆}_j(a_j,b_j]$ with rational endpoints in $[0,1]$, put ${\nu }_0(A)={\sum }_j(F(b_j)-F(a_j))$. Common endpoint refinement and telescoping make this representation-independent and finitely additive. By [L1], $\mathrm{\parallel }{\nu }_0(A)\mathrm{\parallel }\le L\lambda (A)$ and, more generally, $\mathrm{\parallel }{\nu }_0(A)-{\nu }_0(B)\mathrm{\parallel }\le L\lambda (A\mathrm{△}B)$.

Extend to every Lebesgue set. The class of Borel sets approximable in symmetric-difference measure by the rational interval algebra is a sigma-algebra: complements preserve the distance, and countable unions reduce by continuity from below in [L3] to one large finite union. It contains the rational intervals and hence all Borel sets; the completion assertion in [L3] adds every Lebesgue set. Use [A1] to select an approximating sequence $A_n$ for $E$. Step 1.1 makes ${\nu }_0(A_n)$ Cauchy, so completeness of $X$ defines ${\nu }_F(E)={\lim }_n{\nu }_0(A_n)$ independently of the approximants. The bound $\mathrm{\parallel }{\nu }_F(E)\mathrm{\parallel }\le L\lambda (E)$ follows by passage to the limit. Finite additivity and this bound show norm countable additivity: for disjoint $E_k$, the unaccounted tail has norm at most $L\lambda ({\bigcup }_{k>N}{E}_k)\to 0$. Thus [L2] applies.

Verify variation, all endpoints, and uniqueness. Summing the bound from step 2.1 over any finite partition gives $\mathrm{\mid }{\nu }_F\mathrm{\mid }(E)\le L\lambda (E)$. Rational endpoints satisfy the increment formula by construction; rational approximation to arbitrary $a,b$, the same measure bound, and continuity from [L1] give it for all endpoints. In particular ${\nu }_F(\{0\})=0$. Any other dominated vector measure agreeing on rational intervals agrees on their algebra, and [L3] plus its domination gives equality on every Lebesgue set.

Recover a Lipschitz curve from a dominated vector measure. Conversely let $\mathrm{\mid }\nu \mathrm{\mid }\le L\lambda$ and define $F(t)=\nu ((0,t])$. Then $F(0)=0$ and, for $a<b$, $\mathrm{\parallel }F(b)-F(a)\mathrm{\parallel }=\mathrm{\parallel }\nu ((a,b])\mathrm{\parallel }\le \mathrm{\mid }\nu \mathrm{\mid }((a,b])\le L(b-a)$, so [L1] makes $F$ Lipschitz. Its increment measure agrees with $\nu$ on intervals and hence, by uniqueness in step 3.1, everywhere.

Combine both directions and record the choice boundary. [A1, step 3.1, step 4.1] Steps 1.1--3.1 and step 4.1 are inverse constructions. If $L=0$, both the curve and measure are zero; the empty interval and singleton endpoints have zero increment. A one-interval algebra element is the defining case. The exact choice cost is [A1] in the countable-algebra approximation supplied by [L3]; all other selections are finite or least-indexed.