 Solution

A ring of sets is closed under finite unions and differences. A premeasure on it is a nonnegative set function with value zero on the empty set that is countably additive whenever a disjoint union remains in the ring. Caratheodory extension theorem extends a premeasure to a measure on the generated sigma-algebra; this extension is unique there when the premeasure is sigma-finite. It is obtained from the outer measurewhose measurable sets satisfy ${\mu }^{\ast }(E)={\mu }^{\ast }(E\cap B)+{\mu }^{\ast }(E\setminus B)$ for every $E$. The resulting complete measurable sigma-algebra contains the sigma-algebra generated by the ring.

For the required continuity, let $B_n\downarrow \varnothing$ in the bounded-interval ring and fix $ϵ>0$. From each component $(a,b]$ of $B_n$, remove a sufficiently short initial piece and retain the compact interval $[a+\delta ,b]$. The union is a compact set $K_n\subset B_n$, and $B_n\setminus K_n$ is covered by a ring set $C_n$ of total length less than $ϵ2^{-n}$; one can take unions of $(a,a+\delta ]$. We do not incorrectly assume the compact sets themselves belong to the half-open interval ring.

Since ${\bigcap }_n{K}_n=\varnothing$ and $K_1$ is compact, the finite intersection property supplies $N$ with ${\bigcap }_{n=1}^N{K}_n=\varnothing$. For $x\in B_N$, membership in all earlier $B_n$ means some failure of membership in $K_n$ puts $x$ in $C_n$. ThusFinite additivity gives the finite subadditivity used here. Monotonicity now proves $\mu (B_n)\to 0$.

If a disjoint union $A={\bigcup }_j{A}_j$ lies in the ring, its remainders $A\setminus {\bigcup }_{j\le n}{A}_j$ decrease to the empty set. The proved continuity and finite additivity imply countable additivity, so the length function is a premeasure. The ring generates the Borel sigma-algebra and covers the line by bounded intervals of finite measure. Its unique sigma-finite extension is Borel length measure; its completion is Lebesgue measure.