= Solution
For $x<y$, <finite additivity> on the <disjoint> pieces of the half-line gives
$$
h(y)-h(x)=m\bigl(E\cap(x,y]\bigr).
$$
By <monotonicity of a measure> and the <Lebesgue measure> of an <interval>,
$$
0\leq h(y)-h(x)\leq m((x,y])=y-x.
$$
Exchanging the two points if necessary proves
$$
\boxed{|h(y)-h(x)|\leq|y-x|\qquad(x,y\in\mathbb R).}
$$
Thus the <cumulative measure of a finite-measure set> is a <Lipschitz function> with <Lipschitz constant> at most one, hence is \b[<uniformly continuous>]: given $\varepsilon>0$, the choice $\delta=\varepsilon$ 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 $h(-N)\to0$. Since $h$ is nondecreasing,
$$
\lim_{x\to-\infty}h(x)=0,\qquad
\lim_{x\to\infty}h(x)=m(E).
$$
For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The <intermediate value theorem> on $[-R,R]$ gives $t$ with $h(t)=c$. Then
$$
\boxed{A=E\cap(-\infty,t]\text{ is Lebesgue measurable and }m(A)=c.}
$$
The 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.
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