The Riemann–Stieltjes Integral

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.

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 $f(\frac12)\cdot 1 = \frac12$.

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 $\alpha' \ge 0$.

Since $\alpha'$ is continuous on the compact interval $[a,b]$, the extreme value theorem (EVT) says it attains its maximum, so $A = \max_{[a,b]} \alpha' < \infty$ is legitimate.

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 $F = \max_{[a,b]} |f|$ — which exists as an attained maximum by the extreme value theorem (EVT), since $f$ is continuous on the compact $[a,b]$ — 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.