 Solution

For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.