5.5 Riemann-Stieljes Integral
The Riemann–Stieltjes integral replaces the length \(\Delta x_k\) of a subinterval by the increment \(\Delta \alpha _k\) of a function \(\alpha \). It allows sums and integrals to be treated together, and is used for example in probability, where \(\alpha \) is a distribution function.
Throughout, \(\alpha \) is an increasing function on \([a,b]\) and \(f\) is bounded on \([a,b]\). Let \(P: a = x_0 < x_1 < \dots < x_n = b\) be a partition, let \(M_k(f)\) and \(m_k(f)\) be the supremum and infimum of \(f\) on \([x_{k-1},x_k]\) as before, and let \[\Delta \alpha _k = \alpha (x_k) - \alpha (x_{k - 1})\geq 0,\qquad k = 1, 2, \dots , n .\] The upper and lower sums of \(f\) with respect to \(\alpha \) are \[U(P, f, \alpha ) = \sum ^n_{k =1} M_k(f)\, \Delta \alpha _k, \qquad L(P, f, \alpha ) = \sum ^n_{k = 1} m_k(f)\, \Delta \alpha _k,\] and for points \(t_k \in [x_{k-1},x_k]\), the Riemann–Stieltjes sum is \(S(P, f, \alpha ) = \sum ^n_{k = 1} f(t_k)\, \Delta \alpha _k\).
Definition 5.5.1. A bounded function \(f\) on \([a,b]\) is Riemann–Stieltjes integrable with respect to \(\alpha \) if \[\inf _P U(P,f,\alpha ) = \sup _P L(P,f,\alpha ) .\] The common value is denoted by \(\int ^b_af(x)\, d\alpha (x)\) or \(\int ^b_af\, d\alpha \). When \(\alpha (x) = x\) this is the Riemann integral \(\int ^b_af(x)\, dx\).
As for the Riemann integral, refining a partition lowers upper sums and raises lower sums, every lower sum is at most every upper sum, and \(f\) is integrable with respect to \(\alpha \) if and only if for every \(\varepsilon >0\) there is a partition \(P\) with \(U(P,f,\alpha ) - L(P,f,\alpha )<\varepsilon \). The proofs are the same, with \(\Delta x_k\) replaced by \(\Delta \alpha _k\).
Theorem 5.5.2. Let \(\alpha \) be increasing on \([a,b]\) with \(\alpha '\in R[a,b]\), and let \(f\) be bounded on \([a,b]\). Then \(f\) is Riemann–Stieltjes integrable with respect to \(\alpha \) if and only if \(f\alpha '\in R[a,b]\), and in that case \[\int _a^bf\,d\alpha = \int _a^bf(x)\alpha '(x)\,dx .\]
Proof. Let \(|f|\leq M\) and \(\varepsilon >0\). By Riemann’s Criterion there is a partition \(P\) with \(U(P,\alpha ') - L(P,\alpha ')<\varepsilon \), and this remains true for every refinement of \(P\). By the Mean Value Theorem, for each \(k\) there is \(s_k\in (x_{k-1},x_k)\) with \(\Delta \alpha _k = \alpha '(s_k)\Delta x_k\). For any \(t_k\in [x_{k-1},x_k]\), both \(\alpha '(s_k)\) and \(\alpha '(t_k)\) lie between \(m_k(\alpha ')\) and \(M_k(\alpha ')\), so \[\left |\sum _kf(t_k)\Delta \alpha _k - \sum _kf(t_k)\alpha '(t_k)\Delta x_k\right | \leq M\sum _k|\alpha '(s_k) - \alpha '(t_k)|\Delta x_k < M\varepsilon .\] Taking the supremum over the points \(t_k\) gives \(|U(P,f,\alpha ) - U(P,f\alpha ')|\leq M\varepsilon \), and the same holds for every refinement of \(P\). Hence the upper integrals satisfy \(\left |\overline {\int _a^b}f\,d\alpha - \overline {\int _a^b}f\alpha '\right |\leq M\varepsilon \), and since \(\varepsilon \) is arbitrary they are equal. The same argument with infima shows that the lower integrals are equal. So \(f\) is integrable with respect to \(\alpha \) exactly when \(f\alpha '\) is Riemann integrable, and the integrals agree. □
Solution. By Theorem 5.5.2, with \(\alpha (x) = x^2\) and \(\alpha '(x) = 2x\), \[\int _0^1x\,d(x^2) = \int _0^1x\cdot 2x\,dx = \frac {2}{3}.\] With \(\alpha (x) = \sin x\), increasing on \(\left [0,\frac {\pi }{2}\right ]\) with \(\alpha ' = \cos x\), \[\int _0^{\pi /2}\cos x\,d(\sin x) = \int _0^{\pi /2}\cos ^2x\,dx = \frac {\pi }{4}.\]
Example 5.5.4. Let \(a<s<b\), let \(\alpha (x) = 0\) for \(x\leq s\) and \(\alpha (x) = 1\) for \(x>s\) (a unit jump at \(s\)), and let \(f\) be continuous at \(s\). Show that \(\int _a^bf\,d\alpha = f(s)\).
Solution. Let \(s<t<b\) and take the partition \(P = \{a, s, t, b\}\). Then \(\Delta \alpha _1 = \alpha (s) - \alpha (a) = 0\), \(\Delta \alpha _2 = \alpha (t) - \alpha (s) = 1\) and \(\Delta \alpha _3 = 0\), so \[U(P,f,\alpha ) = \sup _{[s,t]}f,\qquad L(P,f,\alpha ) = \inf _{[s,t]}f .\] As \(t\longrightarrow s^+\), both tend to \(f(s)\), by continuity of \(f\) at \(s\). So the upper and lower integrals both equal \(f(s)\). The Riemann–Stieltjes integral with respect to a jump simply evaluates \(f\) at the jump.
Example 5.5.5. Evaluate \(\int _0^3x\,d\lfloor x\rfloor \) and \(\int _0^3x^2\,d\bigl (x + \lfloor x\rfloor \bigr )\), where \(\lfloor x\rfloor \) is the greatest integer not exceeding \(x\).
Solution. On \([0,3]\), \(\lfloor x\rfloor \) has unit jumps at \(1\), \(2\) and \(3\). Write \(\lfloor x\rfloor = \alpha _1 + \alpha _2 + \alpha _3\), where \(\alpha _j\) has a unit jump at \(j\) (the jump at the end point \(3\) is handled in the same way as in the previous example, using the last subinterval \([t,3]\)). The integral is additive in \(\alpha \), as the sums show, so \[\int _0^3x\,d\lfloor x\rfloor = 1 + 2 + 3 = 6 .\] Thus the Riemann–Stieltjes integral turns a sum into an integral: \(\sum _{n=1}^3f(n) = \int _0^3f\,d\lfloor x\rfloor \).
For the second, \(\alpha = x + \lfloor x\rfloor \) is the sum of a smooth part and a jump part, so \[\int _0^3x^2\,d\bigl (x + \lfloor x\rfloor \bigr ) = \int _0^3x^2\,dx + (1 + 4 + 9) = 9 + 14 = 23 .\]
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