Interval Increments and Lebesgue-Stieltjes Premeasures

From Endpoint Increments to a Premeasure

A nondecreasing function records how much its value changes between two points. Assigning that change to the interval between the points is a natural way to define interval mass. The endpoint convention matters: for intervals of the form \((a,b]\), jumps at \(b\) are included, while jumps at \(a\) are excluded. Right continuity is what makes these increments countably additive when an interval is partitioned into countably many smaller intervals.

Let \(F:\mathbb{R}\to\mathbb{R}\) be nondecreasing and right-continuous. Here nondecreasing means \(F(s)\le F(t)\) whenever \(s\le t\), and right-continuity means \(\lim_{h\downarrow0}F(t+h)=F(t)\) for every \(t\in\mathbb{R}\). Let \(\mathcal{S}\) consist of the empty set and all bounded half-open intervals \((a,b]\), where \(a<b\). These intervals are Borel subsets of \(\mathbb{R}\). Define the interval set function \(\mu_F\) on \(\mathcal{S}\) by

The notation \(\mu_F\) distinguishes this set function from Lebesgue measure \(m\). Since \(F\) is real-valued and nondecreasing, each increment is finite and nonnegative. At this stage \(\mu_F\) is defined only on \(\mathcal{S}\); it is not yet a measure on a sigma-algebra.

Finite Additivity and a Covering Inequality

Proof. For a subdivision \(a=x_0<\cdots<x_k=b\), the intervals \((x_{j-1},x_j]\) are disjoint and their union is \((a,b]\). By the definition of \(\mu_F\),

For any finite disjoint interval decomposition of \((a,b]\), order the intervals from left to right. Their union has no gaps, because it is all of \((a,b]\); disjointness and the half-open convention therefore make each interval’s right endpoint the next interval’s left endpoint. The decomposition is a subdivision of the form just considered, so the same telescoping sum proves the assertion. \(\square\)

We will also need an inequality for finite covers, where the covering intervals may overlap.

Proof. Insert into the ordered list \(a=x_0<x_1<\cdots<x_k=b\) every endpoint \(u_j\) and \(v_j\) lying strictly between \(a\) and \(b\), removing repetitions. Each cell \((x_{r-1},x_r]\) is contained in at least one covering interval. Indeed, a point strictly between \(x_{r-1}\) and \(x_r\) is covered, and no endpoint of a covering interval lies inside that cell. The covering interval that contains the point must therefore contain the entire cell. Assign each cell to one such interval. For any fixed covering interval, the assigned cells are disjoint subintervals of it, so finite additivity and nonnegativity of increments show that their total increment is at most the increment of that covering interval: the unassigned gaps, if present, have nonnegative increments. Summing over the \(N\) covering intervals bounds the sum of all cell increments by \(\sum_{j=1}^{N}\mu_F((u_j,v_j])\). Finite additivity on the subdivision gives \(\sum_{r=1}^{k}\mu_F((x_{r-1},x_r])=\mu_F((a,b])\), proving the inequality. \(\square\)

The Premeasure Property

Proof. Suppose \((a,b]=\bigcup_{n=1}^{\infty}(a_n,b_n]\), with the intervals pairwise disjoint. Each interval in the union is a subset of \((a,b]\). Finite additivity, applied after inserting the endpoints of any finite collection of these intervals into a subdivision of \((a,b]\), gives \(\sum_{n=1}^{N}\mu_F((a_n,b_n])\le\mu_F((a,b])\). Taking the supremum over \(N\) yields \[ S:=\sum_{n=1}^{\infty}\mu_F((a_n,b_n])\le F(b)-F(a). \] It remains to prove the reverse inequality.

Fix \(\varepsilon>0\). By right-continuity at \(a\), choose \(a'\) with \(a<a'<b\) and \(F(a')-F(a)<\varepsilon\). For each \(n\), right-continuity at \(b_n\) gives \(\eta_n>0\) such that \(F(b_n+\eta_n)-F(b_n)<\varepsilon 2^{-n}\). The open intervals \((a_n,b_n+\eta_n)\) cover the compact interval \([a',b]\): every \(x\in[a',b]\) belongs to the original union, hence to some \((a_n,b_n]\), and then \(a_n<x\le b_n<b_n+\eta_n\). By compactness, finitely many of these open intervals cover \([a',b]\). Their corresponding half-open intervals \((a_n,b_n+\eta_n]\) cover \((a',b]\), so the finite-cover lemma gives

Adding \(F(a')-F(a)<\varepsilon\) shows \(F(b)-F(a)<S+2\varepsilon\). Since this holds for every \(\varepsilon>0\), \(F(b)-F(a)\le S\). Combined with \(S\le F(b)-F(a)\), this proves the required equality. Also, \(\mu_F(\varnothing)=0\) by definition, so the empty set causes no exception. \(\square\)

The theorem proves countable additivity when the union is itself one interval in \(\mathcal{S}\), which is the premeasure property on this interval semiring. It does not yet assert that \(\mu_F\) is defined on all Borel sets. Nonnegativity, finite additivity, and this countable-additivity property are the ingredients needed before extending the set function to a measure.

Why Right Continuity and Endpoint Conventions Matter

A common error is to use the interval increment formula without checking right continuity. Consider \(G(t)=\chi_{(0,\infty)}(t)\). This function is nondecreasing but not right-continuous at \(0\), since \(G(0)=0\) while \(G(t)=1\) for every \(t>0\). The intervals \(J_n=(1/(n+1),1/n]\) are disjoint and have union \((0,1]\). But \(G(1)-G(0)=1-0=1\), whereas for every \(n\), both endpoints of \(J_n\) are positive, so \(\mu_G(J_n)=G(1/n)-G(1/(n+1))=1-1=0\). Thus the sum of the interval increments is \(0\), not \(1\). This is a concrete failure of countable additivity when right continuity is dropped.

Endpoint conventions also determine which side of a jump contributes to an interval. For \((a,b]\), the increment \(F(b)-F(a)\) includes a jump at \(b\) and excludes a jump at \(a\). Replacing \((a,b]\) by \([a,b)\) while keeping the same formula would generally assign the jump to the wrong interval. The finite-additivity check at a shared endpoint is a useful safeguard: adjacent intervals \((a,c]\) and \((c,b]\) are disjoint, and their increments telescope exactly to \(F(b)-F(a)\).