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For a Lebesgue measurable set $E\subseteq \mathbb{R}$ of finite measure, the displayed function satisfies $|H_E(t)-H_E(s)|\le |t-s|$. Indeed, for $s<t$, finite additivity and monotonicity of a measure give $H_E(t)-H_E(s)=m(E\cap (s,t])\le t-s$. Thus $H_E$ is a nondecreasing Lipschitz function and is uniformly continuous even if $E$ is unbounded.

Continuity from above of a measure gives $H_E(t)\to 0$ as $t\to -\infty$, while continuity from below of a measure gives $H_E(t)\to m(E)$ as $t\to \infty$. Every strictly intermediate value is attained by the intermediate value theorem. The corresponding half-line cut of $E$ realizes any prescribed measure between zero and $m(E)$; the endpoints are realized by the empty set and $E$ itself.

For measurable sets $A\subseteq B$, the disjoint union $B=A\bigsqcup (B\setminus A)$ and finite additivity give $\mu (B)=\mu (A)+\mu (B\setminus A)\ge \mu (A)$. This uses nonnegativity and is valid with infinite values; it does not subtract infinities. Monotonicity of a measure implies that increasing measurable sets have nondecreasing measures, but the equality of their limit with the measure of their union uses the stronger axiom of countable additivity.

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.