The Riemann–Stieltjes Integral

Show that when $\alpha(x) = x$ , the Riemann–Stieltjes upper/lower sums coincide exactly with the Riemann upper/lower sums, and prove that if one integral exists then so does the other, with the same value.

\Delta\alpha_i = x_i - x_{i-1} = \Delta x_i \;\Rightarrow\; U(f, P, \alpha) = U(f, P)

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

\alpha(x) = x

,

n = 6

: the top and bottom panels draw identically —

U = \frac{7}{12} \approx 0.5833

,

L = \frac{5}{12} \approx 0.4167

, exact

\frac12

.

Show proof

When $\alpha(x) = x$ , $\Delta\alpha_i = \alpha(x_i) - \alpha(x_{i-1}) = x_i - x_{i-1} = \Delta x_i$ . Hence the RS upper sum $U(f, P, \alpha) = \sum M_i\,\Delta\alpha_i$ matches $\sum M_i\,\Delta x_i = U(f, P)$ term by term, and likewise for the lower sums.
So the RS upper integral $\overline{\int_a^b} f\,d\alpha = \inf_P U(f, P, \alpha)$ equals the Riemann upper integral $\inf_P U(f, P)$ , and likewise below. If $f$ is Riemann integrable, the upper and lower integrals agree, hence the RS integral exists with $\int_a^b f\,d\alpha = \int_a^b f\,dx$ — and conversely.
In other words, $\alpha(x) = x$ is the choice that does not change the "measuring stick" — which is why, in the scene with $\alpha$ set to identity, the lower $\alpha$ -panel overlaps the upper Riemann panel exactly.

Let $\alpha$ have a single jump of size $J$ at $c$ and be constant elsewhere, and let $f$ be continuous at $c$ . Prove completely, by a squeezing argument on the interval straddling the jump, that $\int_a^b f\,d\alpha = f(c)\,J$ .

U(f, P, \alpha) = M_I J, \quad L(f, P, \alpha) = m_I J \;\Rightarrow\; \int_a^b f\,d\alpha = f(c)\,J

Step

\alpha

(

c = \frac12

,

J = 1

),

n = 6

: all zero-width bars vanish, leaving only

[\frac26, \frac36]

straddling the jump —

U = \frac12

,

L = \frac13

, exact

f(\frac12) \cdot 1 = \frac12

.

Show proof

For any partition $P$ , intervals with $\Delta\alpha_i = 0$ contribute nothing to the RS sums. Whether $c$ is a partition point $x_k$ or lies inside an interval, exactly one interval has $\Delta\alpha_i = J$ — the interval $I = [x_{k-1}, x_k]$ straddling $c$ (the value of $\alpha$ changes at only one endpoint). Hence $U(f, P, \alpha) = M_I\,J$ and $L(f, P, \alpha) = m_I\,J$ , where $M_I, m_I$ are the sup/inf of $f$ on $I$ .
Since $f$ is continuous at $c$ , for every $\varepsilon > 0$ there is $\delta > 0$ with $|x - c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon/J$ . If the partition mesh is below $\delta$ , then $I \subset (c - \delta, c + \delta)$ , so $|M_I - f(c)| \le \varepsilon/J$ and $|m_I - f(c)| \le \varepsilon/J$ . Hence $|U(f, P, \alpha) - f(c)J| \le \varepsilon$ and $U - L = (M_I - m_I)J \le 2\varepsilon$ .
Since $\varepsilon$ is arbitrary, both the RS upper and lower integrals converge to $f(c)J$ — the integral exists and $\int_a^b f\,d\alpha = f(c)\,J$ . A jump collapses the integral to "function value at the point × jump size": why the scene's step $\alpha$ gives $f(\frac12)\cdot 1 = \frac12$ .

Let $\alpha$ be $C^1$ and increasing on $[a, b]$ and $f$ continuous. Write $\Delta\alpha_i = \alpha'(\xi_i)\,\Delta x_i$ by the mean value theorem, and complete the standard argument — using uniform continuity to trade the $\xi_i$ for sample points — proving $\int_a^b f\,d\alpha = \int_a^b f(x)\,\alpha'(x)\,dx$ .

U(f, P, \alpha) = \sum M_i\,\alpha'(\xi_i)\,\Delta x_i \;\xrightarrow[\text{mesh} \to 0]{}\; \int_a^b f(x)\,\alpha'(x)\,dx

\alpha(x) = x^2

,

n = 12

:

U

and

L

converge to the exact

\int_0^1 x \cdot 2x\,dx = \frac23 \approx 0.6667

— the Riemann integral weighted by

\alpha' = 2x

.

Show proof

By the mean value theorem, each $[x_{i-1}, x_i]$ holds $\xi_i \in (x_{i-1}, x_i)$ with $\Delta\alpha_i = \alpha'(\xi_i)\,\Delta x_i$ ( $\alpha'$ is continuous since $\alpha$ is $C^1$ ). Hence $U(f, P, \alpha) = \sum M_i\,\alpha'(\xi_i)\,\Delta x_i$ ; $\alpha$ increasing gives $\alpha' \ge 0$ and $A = \max_{[a,b]} \alpha' < \infty$ .
First, integrability: $U - L = \sum (M_i - m_i)\,\alpha'(\xi_i)\,\Delta x_i \le A\sum (M_i - m_i)\,\Delta x_i = A\,(U(f, P) - L(f, P))$ . Since $f$ is continuous it is Riemann integrable, and $U(f, P) - L(f, P) \to 0$ as the mesh $\to 0$ — so $U(f, P, \alpha) - L(f, P, \alpha) \to 0$ , and the RS integral $\int_a^b f\,d\alpha$ exists.
Now compare $U(f, P, \alpha)$ with the Riemann sum $S(P) = \sum f(t_i)\,\alpha'(t_i)\,\Delta x_i$ (sample points $t_i \in [x_{i-1}, x_i]$ , e.g. $t_i = x_{i-1}$ ). From $\lvert M_i\,\alpha'(\xi_i) - f(t_i)\,\alpha'(t_i)\rvert \le \lvert M_i - f(t_i)\rvert\,\alpha'(\xi_i) + \lvert f(t_i)\rvert\,\lvert\alpha'(\xi_i) - \alpha'(t_i)\rvert$ , with $F = \max_{[a,b]} |f|$ , we get $\lvert U(f, P, \alpha) - S(P)\rvert \le A\,\omega_f(\delta)\,(b-a) + F\,\omega_{\alpha'}(\delta)\,(b-a)$ , where $\omega_f, \omega_{\alpha'}$ are the moduli of continuity from uniform continuity and $\delta$ is the mesh.
As $\delta \to 0$ , $\omega_f(\delta), \omega_{\alpha'}(\delta) \to 0$ , so $|U(f, P, \alpha) - S(P)| \to 0$ . Since $f\,\alpha'$ is continuous, $S(P) \to \int_a^b f(x)\,\alpha'(x)\,dx$ , and $U(f, P, \alpha) \to \int_a^b f\,d\alpha$ (RS integrability shown above); the two limits must agree: $\int_a^b f\,d\alpha = \int_a^b f(x)\,\alpha'(x)\,dx$ . Every step is complete — nothing is left at idea level.

The Riemann–Stieltjes Integral

Examples & Exercises