Riemann-Stieltjes Integration and Bounded Variation

Why Bounded Variation Matters

Riemann–Stieltjes integration replaces the increments of time in ordinary Riemann sums by increments of a function. For continuous integrators with bounded variation, the resulting sums converge for every continuous integrand. The reason is quantitative: changing the sample values of the integrand on a fine partition changes the sum by at most its oscillation on that partition times the integrator’s total variation.

Fix a compact interval $[a,b]$. A partition is $P=\{a=t_0<t_1<\cdots<t_n=b\}$, with mesh $|P|=\max_i(t_i-t_{i-1})$. A tagged partition also chooses $\xi_i\in[t_{i-1},t_i]$ for each interval. For functions $f,g:[a,b]\to\mathbb{R}$, its Riemann–Stieltjes sum is $\sum_i f(\xi_i)(g(t_i)-g(t_{i-1}))$. The integrals in this tutorial are deterministic Riemann–Stieltjes integrals, not Ito or Stratonovich integrals.

Definition: The total variation of $g$ on $[a,b]$ is $$ \operatorname{Var}_{[a,b]}(g)=\sup_{P}\sum_{i=1}^n|g(t_i)-g(t_{i-1})|, $$ where the supremum is over all partitions of $[a,b]$. The function $g$ has bounded variation if this supremum is finite. A continuous function $f$ is Riemann–Stieltjes integrable with respect to $g$ if its tagged sums converge to one common limit as the mesh tends to zero, for every choice of partitions and tags; that limit is denoted $\int_a^b f\,dg$.

When $a=b$, the integral is zero by convention. Below, take $a<b$. The important sufficient criterion is that $f$ be continuous and $g$ have bounded variation. This criterion is not a claim that bounded variation is necessary for every individual integrand–integrator pair.

The Bounded-Variation Criterion

Theorem: If $f$ is continuous on $[a,b]$ and $g$ has bounded variation there, then $\int_a^b f\,dg$ exists as a Riemann–Stieltjes integral. In particular, the limit is independent of the partitions and tags.

Proof. Let $V=\operatorname{Var}_{[a,b]}(g)<\infty$, and let $\omega_f(\delta)=\sup\{|f(x)-f(y)|:x,y\in[a,b],\ |x-y|\le\delta\}$ be the modulus of continuity of $f$. Continuity on the compact interval implies uniform continuity, so $\omega_f(\delta)\to0$ as $\delta\to0$. Compare a tagged sum on a partition $P$ with a tagged sum on a refinement $R$ of $P$. On each interval of $P$, the increments of $g$ over the refined subintervals add to the original increment. Subtracting the two sums and using this additivity gives $$ |S_R-S_P|\le \omega_f(|P|)\sum_{J\in R}|g(\sup J)-g(\inf J)| \le \omega_f(|P|)V. $$ For any two tagged partitions $P,Q$, take their common refinement $R$. Applying the bound to each partition gives $$ |S_P-S_Q|\le \big(\omega_f(|P|)+\omega_f(|Q|)\big)V. $$ As both meshes tend to zero, the right-hand side tends to zero. Thus the sums satisfy the Cauchy criterion uniformly over tags and partitions; completeness of $\mathbb{R}$ gives a common limit. $\square$

The theorem applies whether or not $g$ is monotone. If $g$ is nondecreasing, its variation is $g(b)-g(a)$; for a general bounded-variation function, the total variation controls the absolute increments in the proof.

Proposition: If $g$ is continuously differentiable and $f$ is continuous on $[a,b]$, then $$ \int_a^b f\,dg=\int_a^b f(t)g'(t)\,dt, $$ where the right-hand side is the ordinary Riemann integral.

Proof. The mean value theorem gives, on every partition interval, $g(t_i)-g(t_{i-1})=g'(c_i)(t_i-t_{i-1})$ for some $c_i\in[t_{i-1},t_i]$. The corresponding tagged sum is therefore $\sum_i f(\xi_i)g'(c_i)(t_i-t_{i-1})$. The function $fg'$ is continuous. The difference between $f(\xi_i)g'(c_i)$ and any value of $fg'$ on the same interval tends uniformly to zero as the mesh tends to zero, by uniform continuity of $f$ and $g'$, and boundedness of both. Consequently these sums converge to the ordinary Riemann integral of $fg'$. The bounded-variation theorem ensures they also converge to $\int_a^b f\,dg$; the limits agree. $\square$

Worked Example: Integrating Against a Cubic

Compute $\int_0^1 t^2\,d(t^3)$. Here $f(t)=t^2$ is continuous and $g(t)=t^3$ is continuously differentiable, with $g'(t)=3t^2$. The proposition gives $$ \int_0^1 t^2\,d(t^3)=\int_0^1 t^2(3t^2)\,dt =3\int_0^1t^4\,dt =3\left[\frac{t^5}{5}\right]_0^1 =\frac35. $$

Check: Integration by parts gives $\int_0^1t^2\,d(t^3)=[t^2t^3]_0^1-\int_0^1t^3\,d(t^2)=1-\int_0^1 2t^4\,dt=1-\frac25=\frac35$, agreeing with the direct calculation.

Worked Example: An Exponential Integrand

Compute $\int_0^1 e^t\,d(t)$. The integrator $g(t)=t$ has derivative $g'(t)=1$, and $f(t)=e^t$ is continuous. Thus $$ \int_0^1 e^t\,d(t)=\int_0^1e^t\,dt=[e^t]_0^1=e-1. $$

Check: Integration by parts yields $[te^t]_0^1-\int_0^1t e^t\,dt$. The boundary term is $e$, while $\int_0^1t e^t\,dt=[(t-1)e^t]_0^1=0-(-1)=1$. The result is $e-1$, as above.

A Continuous Integrator Need Not Be Smooth

Bounded variation also covers continuous integrators with corners. For a piecewise-linear integrator, one can compute the integral separately on its linear pieces, or use integration by parts when both functions have bounded variation and are continuous. In that setting the identity is $\int_a^b f\,dg=f(b)g(b)-f(a)g(a)-\int_a^b g\,df$.

Worked Example: A V-Shaped Integrator

Let $g(t)=|t-\tfrac12|$ and $f(t)=t$ on $[0,1]$. The function $g$ decreases from $\tfrac12$ to $0$, then increases from $0$ to $\tfrac12$, so its total variation is $\tfrac12+\tfrac12=1$. It is continuous, and $f$ is continuous; the criterion guarantees the integral exists. On $[0,\tfrac12]$, the slope of $g$ is $-1$, and on $[\tfrac12,1]$ its slope is $1$. Therefore $$ \int_0^1 t\,dg =\int_0^{1/2}t(-1)\,dt+\int_{1/2}^1t(1)\,dt =-\left[\frac{t^2}{2}\right]_0^{1/2} +\left[\frac{t^2}{2}\right]_{1/2}^1 =-\frac18+\frac38 =\frac14. $$

Check: The boundary term $f(1)g(1)-f(0)g(0)$ is $\tfrac12-0=\tfrac12$. Also $\int_0^1g\,df=\int_0^1|t-\tfrac12|\,dt=\tfrac18+\tfrac18=\tfrac14$. Integration by parts gives $\tfrac12-\tfrac14=\tfrac14$, confirming the result.

Why Brownian Paths Fail the Criterion

Now work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Brownian paths are continuous almost surely, but continuity alone does not imply bounded variation. The preceding material on Brownian quadratic variation gives the following standard result: for each fixed $T>0$, along the dyadic partitions of $[0,T]$, the sums of squared increments converge to $T$ almost surely.

Proposition: If $g$ is continuous and has bounded variation on $[0,T]$, then for every sequence of partitions $P_n$ whose mesh tends to zero, $$ \sum_{[u,v]\in P_n}(g(v)-g(u))^2\longrightarrow0. $$

Proof. Write $V=\operatorname{Var}_{[0,T]}(g)<\infty$ and $m_n=\max_{[u,v]\in P_n}|g(v)-g(u)|$. Continuity on the compact interval implies uniform continuity, so $m_n\to0$ as the mesh tends to zero. For each partition, $$ 0\le\sum_{[u,v]\in P_n}(g(v)-g(u))^2 \le m_n\sum_{[u,v]\in P_n}|g(v)-g(u)| \le m_nV. $$ The final quantity tends to zero, proving the claim. $\square$

For Brownian motion, the quadratic-variation sums along those dyadic partitions converge almost surely to $T>0$, not to zero. On the probability-one event where this convergence holds and the path is continuous, the proposition rules out finite total variation on $[0,T]$. Thus Brownian paths do not satisfy the bounded-variation hypothesis of the Riemann–Stieltjes criterion, almost surely.

This conclusion is precise but limited: failure of the criterion does not prove that every particular Riemann–Stieltjes integral against a Brownian path fails to exist. For example, a constant integrand has a well-defined sum against any continuous path, because its sums telescope. The conclusion is that bounded variation cannot guarantee existence for every continuous integrand when the integrator is a Brownian path.

Common Pitfall: Confusing the Two Differentials

A Riemann–Stieltjes integral $\int f\,dg$ uses increments of $g$; it is not obtained by multiplying $f$ and $g$ and integrating over time. For the cubic example, the incorrect calculation $\int_0^1 t^2t^3\,dt=\int_0^1t^5\,dt=\tfrac16$ treats the integrator as though its differential were $g(t)\,dt$. The correct differential is $dg=g'(t)\,dt=3t^2\,dt$, giving $\int_0^1 3t^4\,dt=\tfrac35$. Check the distinction by setting $g(t)=t$: then $dg=dt$, and the Riemann–Stieltjes integral reduces to the ordinary integral.

The bounded-variation condition is valuable because it supplies a uniform error bound for all partitions and tags. Smooth paths meet it, as do many nonsmooth paths. Brownian motion does not: its nonzero quadratic variation is incompatible with finite total variation. This is the deterministic boundary that motivates a different integration theory for stochastic paths.

Check Your Understanding

Use the definitions and results above to answer the following questions.

For continuous $f$ and bounded-variation $g$, identify the two factors in the estimate comparing a sum with its refinement. Why does their product tend to zero?
Compute $\int_0^1 t\,d(t^2)$ by reducing it to an ordinary integral, and verify the result by integration by parts.
Find the total variation of $g(t)=|t-\tfrac13|$ on $[0,1]$.
Explain why a continuous bounded-variation path has zero squared-increment sums as the mesh tends to zero.
Does the failure of bounded variation for Brownian paths imply that every Riemann–Stieltjes integral against such a path fails to exist? Explain.