Constructing Lebesgue-Stieltjes Measures from Cumulative Functions

From Interval Increments to a Borel Measure

A nondecreasing right-continuous function assigns a nonnegative number to every bounded interval by taking an increment. The central question is whether these local assignments fit together as a genuine countably additive measure. The interval-increment result gives the required premeasure; the Caratheodory extension theorem then constructs a measure on the Borel sets. A separate calculation, using larger and larger bounded intervals, determines whether that measure has finite total mass.

Throughout, let \(F:\mathbb{R}\to\mathbb{R}\) be finite-valued, nondecreasing, and right-continuous. “Nondecreasing” means \(F(a)\le F(b)\) whenever \(a\le b\), and right continuity means \(\lim_{y\downarrow x}F(y)=F(x)\) for every \(x\in\mathbb{R}\). The function \(F\) need not be nonnegative: adding a constant to \(F\) does not change any interval increment. We construct a measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\mathcal{B}_{\mathbb{R}}\) is the Borel sigma-algebra.

The use of \((a,b]\), rather than \([a,b)\), fixes which endpoint receives any mass associated with a jump. Since \(F\) is nondecreasing, every summand is nonnegative. The interval-increments result establishes that this formula is well-defined on \(\mathcal{R}\) and is a premeasure: it is countably additive whenever a disjoint countable union of members of \(\mathcal{R}\) is itself in \(\mathcal{R}\). Right continuity is essential to that premeasure property. In particular, the finite-valued assumption ensures \(\mu_0((a,b])<\infty\) for every bounded interval.

We use the following standard deep result, the Caratheodory extension theorem; its proof is not repeated here. A sigma-finite premeasure on a ring has a measure extension to the sigma-algebra generated by that ring. Here “sigma-finite” means that the underlying space is a countable union of sets of finite measure. The ring \(\mathcal{R}\) generates \(\mathcal{B}_{\mathbb{R}}\): bounded half-open intervals are Borel, and open intervals, hence open sets, can be expressed using countable unions of such intervals.

The Construction Theorem

Proof. Use the interval-increment premeasure \(\mu_0\) on \(\mathcal{R}\) defined above. For each positive integer \(n\), the interval \((-n,n]\) belongs to \(\mathcal{R}\) and has finite premeasure \(F(n)-F(-n)\). Moreover, \(\mathbb{R}=\bigcup_{n\ge1}(-n,n]\). Thus \(\mu_0\) is sigma-finite on \(\mathcal{R}\). By the Caratheodory extension theorem, \(\mu_0\) extends to a measure \(\mu_F\) on \(\sigma(\mathcal{R})=\mathcal{B}_{\mathbb{R}}\). It has the prescribed interval values because it extends \(\mu_0\), and the same cover shows that \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\mu_F)\) is sigma-finite.

The sets \((-n,n]\) increase to \(\mathbb{R}\). Countable additivity implies continuity from below: for increasing measurable sets \(A_n\) with union \(A\), write \(A\) as the disjoint union of \(A_1\) and \(A_{n+1}\setminus A_n\); countable additivity then gives \(\mu_F(A_n)\to\mu_F(A)\). Therefore

Monotonicity gives the extended limits \(F(-\infty)\) and \(F(+\infty)\). The first is either finite or \(-\infty\), since \(F(-n)\le F(0)\) for \(n\ge0\); the second is either finite or \(+\infty\), since \(F(n)\ge F(0)\). If both limits are finite, the displayed difference tends to \(F(+\infty)-F(-\infty)\), which is finite. If \(F(-\infty)=-\infty\), then \(F(n)-F(-n)\ge F(0)-F(-n)\to+\infty\). If \(F(+\infty)=+\infty\), then \(F(n)-F(-n)\ge F(n)-F(0)\to+\infty\). Thus the total mass is finite exactly when both limits are finite, and its value is the stated difference. \(\square\)

Worked Constructions

What the Cumulative Function Records

The total mass formula also clarifies when \(F\) itself is a cumulative distribution function. For each finite \(x\), the half-lines \((-n,x]\), taken for all sufficiently large \(n\), increase to \((-\infty,x]\). Continuity from below and the interval-increment formula give

where the value is \(+\infty\) if \(F(-\infty)=-\infty\). In particular, when the total mass is finite, the cumulative mass is \(F(x)-F(-\infty)\), not necessarily \(F(x)\). The normalization \(F(-\infty)=0\) makes the two agree. This is why adding a constant changes the displayed function but not the measure it constructs.

For the finite example above, \(F(-\infty)=5\), so \(\mu_F((-\infty,0])=F(0)-5=\frac12\), not \(F(0)=\frac{11}{2}\). The normalized function \(F-5\) has the same interval increments and has limit zero at \(-\infty\). In the linear example, \(F(-\infty)=-\infty\), and every half-line \((-\infty,x]\) has infinite mass, even though each bounded interval has finite mass.

Common Pitfalls and the Role of the Hypotheses

Wrong calculation: for \(F(x)=5+\frac12+\frac{1}{\pi}\arctan x\), one might say the total mass is \(F(+\infty)=6\). This confuses the value of the function with its accumulated increment. The correct calculation uses both endpoints at infinity: \(F(+\infty)-F(-\infty)=6-5=1\). As the interval formula shows, replacing \(F\) by \(F+c\) leaves every increment unchanged, so a total-mass formula depending on the additive constant cannot be correct.

Right continuity is not an optional technical detail. For example, define \(G(x)=0\) when \(x\le0\), and \(G(x)=1\) when \(x>0\). This function is nondecreasing but is not right-continuous at \(0\). The interval prescription gives \(G(1/n)-G(0)=1\), hence assigns mass \(1\) to each \((0,1/n]\). These sets decrease to the empty set. A finite measure on \((0,1]\) must have measures tending to zero for such a decreasing sequence: this follows by applying continuity from below to the increasing complements within \((0,1]\). The constant values \(1\) contradict that requirement, so these increments cannot define a measure with the proposed interval values.

Finally, finite mass and sigma-finiteness are different conditions. The construction is always sigma-finite because the real line is covered by countably many bounded intervals of finite measure. It has finite total mass only when both endpoint limits of \(F\) are finite. In particular, the linear example is sigma-finite but not finite.