Riemann-Stieltjes Integration

⚡ Key Takeaways

One-line statement: The Riemann-Stieltjes integral $\int_a^b f\,d\alpha$ generalizes the Riemann integral by integrating $f$ against an integrator $\alpha$ , weighting increments by $\Delta\alpha$ rather than $\Delta x$ . Key result: When $\alpha$ is of bounded variation and $f$ is continuous, $\int_a^b f\,d\alpha$ exists; it unifies sums, ordinary integrals, and mixed discrete-continuous quantities. Where it applies: Probability (expectation against a CDF), functional analysis (Riesz representation), physics (mass/charge distributions), numerical analysis. Who/when: Defined by Thomas Joannes Stieltjes (1894) in his memoir on continued fractions; later absorbed into measure theory via Lebesgue-Stieltjes integration.

💡 The Core Insight

Integration is fundamentally about accumulating a quantity $f$ against a measure of change in another quantity. The Riemann integral hard-codes that measure of change as ordinary length $dx$ . The Stieltjes generalization frees the "ruler": we may accumulate against $d\alpha$ , where jumps in $\alpha$ contribute point masses and smooth regions contribute densities $\alpha'(x)\,dx$ . One formula thus encodes summation and integration as endpoints of a continuum.

DefinitionRiemann-Stieltjes Sum and Integral

Let $f,\alpha:[a,b]\to\mathbb{R}$ be bounded. A partition $P$ of $[a,b]$ is a finite set

1

\quad P=\{a=x_0<x_1<\cdots<x_n=b\},\qquad \Delta\alpha_i := \alpha(x_i)-\alpha(x_{i-1}).

Derivation hint: a partition simply marks division points; $\Delta\alpha_i$ measures the integrator's change across the $i$ -th subinterval. Choosing tags $t_i\in[x_{i-1},x_i]$ , the Riemann-Stieltjes sum is

2

\quad S(P,f,\alpha)=\sum_{i=1}^{n} f(t_i)\,\Delta\alpha_i.

Derivation hint: replace the Riemann weight $\Delta x_i$ by the integrator weight $\Delta\alpha_i$ in the standard tagged sum. We say $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ , written $f\in\mathscr{R}(\alpha)$ , with integral $A$ , if for every $\varepsilon>0$ there exists a partition $P_\varepsilon$ such that for every refinement $P\supseteq P_\varepsilon$ and every choice of tags,

3

\quad \bigl|\,S(P,f,\alpha)-A\,\bigr|<\varepsilon,\qquad A=:\int_a^b f\,d\alpha.

Derivation hint: this is the Moore-Smith (net) limit over the directed set of partitions ordered by refinement — stronger than a mesh-based limit and the correct definition for non-monotone $\alpha$ .

Symbols: $f$ = integrand; $\alpha$ = integrator (function of bounded variation in the useful theory); $[a,b]$ = domain; $\Delta\alpha_i$ = integrator increment; $t_i$ = sample (tag) point; $A$ = value of the integral.

⚠️ Refinement vs. mesh limits

For the ordinary Riemann integral the "refinement limit" and the "mesh $\to 0$ limit" coincide. For the Stieltjes integral with a general integrator they do not. The standard definition uses refinement (Definition [3]). If one instead demands convergence as $\|P\|=\max\Delta x_i\to 0$ , integrals against step integrators can fail to exist where the refinement definition gives a value. Always state which convention you mean.

DefinitionUpper and Lower Stieltjes Integrals (monotone

\alpha

)

When $\alpha$ is monotone increasing, all $\Delta\alpha_i\geq 0$ , and we may define Darboux-type sums. With $M_i=\sup_{[x_{i-1},x_i]}f$ , $m_i=\inf_{[x_{i-1},x_i]}f$ ,

4

\quad U(P,f,\alpha)=\sum_{i=1}^n M_i\,\Delta\alpha_i,\qquad L(P,f,\alpha)=\sum_{i=1}^n m_i\,\Delta\alpha_i,

Derivation hint: bound $f(t_i)$ above/below on each subinterval; monotonicity of $\alpha$ keeps the weights nonnegative so the inequalities $L\le S\le U$ hold. and the upper/lower integrals $\overline{\int}f\,d\alpha=\inf_P U$ , $\underline{\int}f\,d\alpha=\sup_P L$ . Then $f\in\mathscr{R}(\alpha)$ iff $\overline{\int}=\underline{\int}$ .

TheoremExistence: Continuous Integrand, BV Integrator

If $f$ is continuous on $[a,b]$ and $\alpha$ is of bounded variation on $[a,b]$ , then $f\in\mathscr{R}(\alpha)$ . Moreover

5

\quad \left|\int_a^b f\,d\alpha\right|\le \Bigl(\sup_{[a,b]}|f|\Bigr)\,V_a^b(\alpha),

where $V_a^b(\alpha)$ is the total variation of $\alpha$ . Derivation hint: decompose $\alpha = \alpha_1-\alpha_2$ (Jordan decomposition) into increasing parts; on each, use uniform continuity of $f$ to make $U-L$ small; bound the sum by $\sup|f|$ times $V_a^b(\alpha)$ .

TheoremIntegration by Parts (Symmetry)

If $f\in\mathscr{R}(\alpha)$ on $[a,b]$ , then $\alpha\in\mathscr{R}(f)$ and

6

\quad \int_a^b f\,d\alpha + \int_a^b \alpha\,df = f(b)\alpha(b)-f(a)\alpha(a).

Derivation hint: Abel summation. Write $S(P,f,\alpha)=\sum f(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$ and rearrange by summation by parts to produce $\sum \alpha(\cdot)\Delta f$ plus boundary terms; take the refinement limit.

TheoremReduction to Riemann Integral

If $\alpha$ has a continuous derivative $\alpha'$ (or more generally $\alpha$ is absolutely continuous) on $[a,b]$ and $f\in\mathscr{R}(\alpha)$ , then

7

\quad \int_a^b f\,d\alpha = \int_a^b f(x)\,\alpha'(x)\,dx,

the right side being an ordinary Riemann (resp. Lebesgue) integral. Derivation hint: by the Mean Value Theorem $\Delta\alpha_i=\alpha'(\xi_i)\Delta x_i$ ; substitute into [2], and the resulting sum is a Riemann sum for $f\alpha'$ . Continuity of $\alpha'$ controls the error between $f(t_i)\alpha'(\xi_i)$ and $f(t_i)\alpha'(t_i)$ .

TheoremStep-Function Integrator

Let $\alpha$ be a step function with jumps $j_k=\alpha(c_k^+)-\alpha(c_k^-)$ at interior points $c_k\in(a,b)$ (with appropriate one-sided jumps at endpoints), and let $f$ be continuous at each $c_k$ . Then

8

\quad \int_a^b f\,d\alpha = \sum_k f(c_k)\,j_k.

Derivation hint: only subintervals containing a jump contribute; as the partition refines, the contribution near $c_k$ collapses to $f(c_k)$ times the jump $j_k$ , while flat pieces of $\alpha$ contribute zero.

ProofExistence theorem for continuous

f

, increasing

\alpha

We prove $f\in\mathscr{R}(\alpha)$ when $f$ is continuous and $\alpha$ is increasing (the BV case follows by Jordan decomposition).

Step 1 — Uniform continuity. Since $[a,b]$ is compact and $f$ continuous, $f$ is uniformly continuous. Given $\varepsilon>0$ , set $\eta = \varepsilon/\bigl(1+\alpha(b)-\alpha(a)\bigr)$ and choose $\delta>0$ so that $|s-t|<\delta \Rightarrow |f(s)-f(t)|<\eta$ .

Step 2 — Choose a fine partition. Pick any partition $P$ with mesh $\|P\|<\delta$ . On each subinterval, the oscillation satisfies $M_i-m_i\le\eta$ .

Step 3 — Estimate the Darboux gap. Using [4] and $\Delta\alpha_i\ge 0$ ,

9

\quad U(P,f,\alpha)-L(P,f,\alpha)=\sum_{i=1}^n (M_i-m_i)\,\Delta\alpha_i \le \eta\sum_{i=1}^n \Delta\alpha_i = \eta\,\bigl(\alpha(b)-\alpha(a)\bigr)<\varepsilon.

Derivation hint: factor $\eta$ out of the telescoping sum $\sum\Delta\alpha_i=\alpha(b)-\alpha(a)$ .

Step 4 — Cauchy criterion. Since $U-L$ can be made smaller than any $\varepsilon$ , $\inf_P U=\sup_P L$ , so the upper and lower integrals coincide and $f\in\mathscr{R}(\alpha)$ .

Step 5 — BV case. Write $\alpha=\alpha_1-\alpha_2$ with $\alpha_1,\alpha_2$ increasing (Jordan decomposition; $\alpha_1=$ positive variation, $\alpha_2=$ negative variation). Then $f\in\mathscr{R}(\alpha_1)$ and $f\in\mathscr{R}(\alpha_2)$ , and by linearity in the integrator $f\in\mathscr{R}(\alpha)$ with $\int f\,d\alpha=\int f\,d\alpha_1-\int f\,d\alpha_2$ . The bound [5] follows since $V_a^b(\alpha)=\bigl(\alpha_1(b)-\alpha_1(a)\bigr)+\bigl(\alpha_2(b)-\alpha_2(a)\bigr)$ . $\blacksquare$

∎

mathematician · Thomas Joannes Stieltjes | 1856–1894

A Dutch-French analyst often called "the father of the analytic theory of continued fractions," Stieltjes introduced his integral in the 1894 memoir Recherches sur les fractions continues, where he needed to represent a measure (a distribution of mass on a line) by integration against an increasing function. He died at 38, only months after the work appeared. The integral lay relatively dormant until F. Riesz's 1909 representation theorem revealed it as the natural dual object to continuous functions, after which it became central to functional analysis and probability.

🧠 Intuition

Think of $\alpha$ as describing how mass is distributed along the interval $[a,b]$ : $\alpha(x)$ is the total mass to the left of $x$ . Then $\Delta\alpha_i$ is the mass sitting in the $i$ -th subinterval, and $\int f\,d\alpha$ is the weighted average of $f$ weighted by where the mass actually lives.

If mass is spread uniformly (density 1), then $\alpha(x)=x$ and you recover the ordinary integral.
If mass is spread with density $\rho(x)$ , then $\alpha'=\rho$ and you get $\int f\rho\,dx$ (Theorem [7]).
If a chunk of mass $j_k$ is concentrated at a single point $c_k$ , $\alpha$ jumps, and that point contributes $f(c_k)j_k$ exactly — a discrete term (Theorem [8]).

This is precisely why probabilists love it: take $\alpha=F$ , the cumulative distribution function. Continuous distributions give density integrals; discrete distributions give sums; mixed distributions give both at once — under one symbol $\mathbb{E}[g(X)]=\int g\,dF$ .

Figure

Riemann-Stieltjes sum: integrand sampled at tags, weighted by integrator increments

ExampleMixed integrator: smooth part plus a jump

Given: Integrate $f(x)=x^2$ against

\alpha(x)=\begin{cases} x, & 0\le x<1,\\[2pt] x+2, & 1\le x\le 2.\end{cases}

So $\alpha$ has slope $1$ everywhere except a jump of size $2$ at $x=1$ . Compute $\displaystyle\int_0^2 x^2\,d\alpha$ .

Step 1 — Split into smooth and jump parts. Write $\alpha=\alpha_{\text{cont}}+\alpha_{\text{jump}}$ where $\alpha_{\text{cont}}(x)=x$ (so $\alpha_{\text{cont}}'=1$ ) and $\alpha_{\text{jump}}$ is a step function with a single jump $j=2$ at $c=1$ .

Step 2 — Smooth part (Theorem [7]).

10

\quad \int_0^2 x^2\,d\alpha_{\text{cont}}=\int_0^2 x^2\cdot 1\,dx=\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}.

Derivation hint: since $\alpha_{\text{cont}}'=1$ , Theorem [7] turns the Stieltjes integral into the ordinary integral of $x^2$ .

Step 3 — Jump part (Theorem [8]). Since $f$ is continuous at $c=1$ :

11

\quad \int_0^2 x^2\,d\alpha_{\text{jump}}=f(1)\cdot j = 1^2\cdot 2 = 2.

Derivation hint: a single jump of size $j$ at $c$ contributes $f(c)\,j$ .

Step 4 — Add. $\dfrac{8}{3}+2=\dfrac{14}{3}$ .

\boxed{\displaystyle\int_0^2 x^2\,d\alpha = \tfrac{14}{3}\approx 4.667\ \text{(dimensionless)}}

ExampleExpectation of a mixed random variable

Given: A random variable $X$ has CDF

F(x)=\begin{cases}0,& x<0,\\ \tfrac12 x,& 0\le x<1,\\ 1,& x\ge 1,\end{cases}

i.e. mass $\tfrac12$ spread uniformly on $[0,1)$ plus a point mass of $\tfrac12$ at $x=1$ (since $F$ jumps from $\tfrac12$ to $1$ there). Find $\mathbb{E}[X]=\int_{-\infty}^{\infty} x\,dF(x)$ .

Step 1 — Identify components. Continuous part: density $F'(x)=\tfrac12$ on $(0,1)$ . Discrete part: jump $j=F(1^+)-F(1^-)=1-\tfrac12=\tfrac12$ at $c=1$ .

Step 2 — Continuous contribution (Theorem [7]).

12

\quad \int_0^1 x\cdot \tfrac12\,dx=\tfrac12\cdot\left[\frac{x^2}{2}\right]_0^1=\tfrac12\cdot\tfrac12=\tfrac14.

Derivation hint: integrate against the density $F'=\tfrac12$ over the continuous support.

Step 3 — Discrete contribution (Theorem [8]).

13

\quad x\big|_{x=1}\cdot j = 1\cdot\tfrac12=\tfrac12.

Derivation hint: the point mass at $x=1$ contributes value $1$ times its probability $\tfrac12$ .

Step 4 — Total. $\tfrac14+\tfrac12=\tfrac34$ .

\boxed{\mathbb{E}[X]=\tfrac34=0.75\ \text{(same units as }X\text{)}}

⚠️ Existence is a joint property, not an "and"

A frequent error is to assume " $f$ nice and $\alpha$ nice $\Rightarrow$ integral exists" with each checked independently. Existence depends on how the discontinuities of $f$ and $\alpha$ interact. The clean sufficient condition is: $f$ continuous and $\alpha$ of bounded variation (Theorem [5]). If both have a discontinuity at the same point from the same side, the integral may fail to exist even when each function is "simple."

CounterexampleShared one-sided discontinuity

Let $f=\alpha=H$ on $[-1,1]$ , where $H(x)=0$ for $x<0$ and $H(x)=1$ for $x\ge 0$ (the Heaviside step). Both are of bounded variation and both have a single jump — yet $\int_{-1}^1 f\,d\alpha$ does not exist in the refinement sense. The trouble is at $x=0$ , where both jump. Consider a partition with $0$ as a tag versus $0$ not as a partition point. The only nonzero increment $\Delta\alpha$ occurs across the subinterval containing $0$ ; the value of $f$ at the chosen tag there can be $0$ or $1$ depending on whether the tag falls left or right of $0$ . Refining the partition does not force these to agree, so the Riemann-Stieltjes sums do not converge. The defect is precisely a common discontinuity on the same side: if instead $f$ were continuous at $0$ , or the jumps occurred at distinct points, the integral would exist.

⚠️ Linearity holds in BOTH slots — but bounded variation is needed

$\int f\,d\alpha$ is linear in $f$ (for fixed $\alpha$ ) and linear in $\alpha$ (for fixed $f$ ): $\int f\,d(\alpha+\beta)=\int f\,d\alpha+\int f\,d\beta$ . But the estimate $|\int f\,d\alpha|\le \sup|f|\cdot V_a^b(\alpha)$ uses the total variation $V_a^b(\alpha)$ , not $|\alpha(b)-\alpha(a)|$ . Confusing the two underestimates the bound when $\alpha$ oscillates.

connection

The Riemann-Stieltjes integral sits at a crossroads of several theories.

It generalizes the Riemann Integral (take $\alpha(x)=x$ ) and is itself subsumed by Lebesgue-Stieltjes Integration, where $\alpha$ generates a Borel measure $\mu_\alpha$ and $\int f\,d\alpha=\int f\,d\mu_\alpha$ , removing the awkward shared-discontinuity pathologies.
Via the Riesz Representation Theorem, every bounded linear functional on $C[a,b]$ is $f\mapsto\int_a^b f\,d\alpha$ for some $\alpha$ of bounded variation — making BV functions the dual of continuous functions.
It is the deterministic ancestor of the Itô Integral and stochastic integration; when the integrator has unbounded variation (like Brownian motion), the Stieltjes construction fails and must be replaced by $L^2$ / martingale theory.
Through Functions of Bounded Variation and the Jordan Decomposition, the integrator's structure (jump part, absolutely continuous part, singular part) directly governs the integral's behaviour.
In numerical analysis it underlies Gaussian Quadrature and orthogonal polynomial theory — Stieltjes's original motivation via continued fractions and moment problems.

🔬 Where this appears

⏳ History

- 1854 — Riemann. Bernhard Riemann gives the first rigorous definition of the (ordinary) integral via tagged sums in his Habilitationsschrift, setting the template later generalized.

- 1881 — Jordan. Camille Jordan introduces functions of bounded variation and proves the decomposition of a BV function as a difference of increasing functions — the structural backbone of valid integrators.

- 1894 — Stieltjes. In Recherches sur les fractions continues, Thomas Stieltjes defines

\int f\,d\alpha

to represent distributions of mass, motivated by continued fractions and the moment problem. He dies the same year at 38.

- 1909 — F. Riesz. Frigyes Riesz proves the Riesz representation theorem: bounded linear functionals on

C[a,b]

are exactly Stieltjes integrals against BV functions — elevating the integral to a central object of functional analysis.

- 1913 — Radon. Johann Radon develops measures on

\mathbb{R}^n

, paving the way to recast Stieltjes integration measure-theoretically.

- 1915–1918 — Lebesgue/Daniell. The Lebesgue-Stieltjes integral emerges:

\alpha

(right-continuous, increasing) defines a Borel measure

\mu_\alpha

, absorbing the Riemann-Stieltjes theory and curing its existence pathologies.

- 1925 — Pollard / Young. Refinements clarify existence conditions, and L. C. Young's later work (1936) on integration when both functions have unbounded variation (Young integration) extends the boundary of the theory.

- 1944 — Itô. Kiyosi Itô constructs stochastic integration where the integrator (Brownian motion) has infinite variation, definitively showing the Stieltjes framework's limits and necessitating a new calculus.

open-problem

The bounded-variation barrier. Riemann-Stieltjes integration relies essentially on $\alpha$ having bounded variation; this is what makes the increments controllable and the existence theorems work. The deep frontier is integration against integrators of unbounded variation:

Young integration (1936). If $f\in C^{p}$ -Hölder and $\alpha\in C^{q}$ -Hölder with $p+q>1$ , the Riemann-Stieltjes limit still exists even when both have infinite variation. The case $p+q\le 1$ is the genuinely hard regime.
Rough path theory (Lyons, 1998). Resolves $p+q\le 1$ by enhancing the path with higher-order "iterated integral" data, providing a deterministic theory that contains Itô and Stratonovich calculus as special cases. The precise extent to which a path determines its rough-path lift, and the analytic structure of the resulting integrals, remain active research.
The Stieltjes moment problem's determinacy — whether a moment sequence corresponds to a unique integrator $\alpha$ — is fully understood only under growth conditions (Carleman's condition is sufficient but not necessary); sharp characterizations in the indeterminate case connect to entire function theory and remain subtle.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also