. In this section we examine the precise definition of a somewhat more general integral. The familiar intuitive notion of the integral as the limit of the sum of a very large number of very small quantities is central to the discussion.RIEMANN–STIELTJES INTEGRATION
6.1DEFINITION OF THE INTEGRAL
In calculus you did many computational examples and applications involving the Riemann integral . In this section we examine the precise definition of a somewhat more general integral. The familiar intuitive notion of the integral as the limit of the sum of a very large number of very small quantities is central to the discussion.
6.1.1Definitions and Comments
Let ƒ and α be functions from [a, b] to R, and let P: a = x0 < x1 < ... < xn = b be a partition of [a, b]. The size of P is defined as the largest subinterval length; that is,
We assume throughout the discussion that α is increasing and ƒ is bounded. As in the familiar process of making a rectangular approximation for the purpose of computing the area under a curve, we define
(Since ƒ is bounded, Mk and mk are finite.) If tk is a point in [xk−1, xk], k = 1, 2, ..., n (the choice of the tk may be regarded as part of the specification of P ), let
Also define
The upper sum associated with P, ƒ, α is defined as
similarly, the lower sum is
and the Riemann–Stieltjes sum is
If ƒ and α are understood, we write these sums as U(P), L(P), and S(P). When α(x) = x for all x, we have Δαk, = xk − xk−1; in this case, we obtain approximating sums to the Riemann integral.
If ƒ is a bounded function such that U(P) and L(P) approach a common finite limit as |P| → 0, we say that ƒ is Riemann–Stieltjes integrable with respect to α or that the Riemann–Stieltjes integral of ƒ with respect to α exists. The integral is denoted by
If 
is called the Riemann integral of ƒ, denoted by 
.
In numerous practical examples, ƒ is a continuous function. Before showing that the Riemann–Stieltjes integral always exists in this case, we need some preliminaries. By Theorem 4.2.4, ƒ is uniformly continuous on [a, b], so that given ∊ > 0 there is a δ > 0 such that whenever |x − y| < δ we have |f(x) − f(y)| < ∊. Let us prove that if α is increasing and P is a partition of size less than δ,
To establish this, observe that the assumption |P| < δ implies that 0 ≤ Mk − mk ≤ ∊ for all k; hence,
Now consider a refinement of P, in other words, keep the points x0, ..., xn defining P and add new points (possibly in each subinterval) to form a partition P′. If a new point y appears in the subinterval [xk−1, xk] of P so that xk−1 < y < xk, then the term Mk[α(xk) − α(xk−1)], which appears in U(P), may be written as
If 
, the corresponding expression in U(P′) is
(note also that 
; similarly, refinement increases lower sums). The point of all this is to compare U(P) with U(Р′). Since 
, we obtain, as in the proof of (1),
Finally, it follows immediately from the definitions that
It is convenient at this point to make a comment that we need later. Suppose P′ is formed by adding only one pointy to P in the subinterval [xk−1, xk]. The situation we are analyzing involves a fixed y, with the size of P (hence also the size of P′) approaching 0. If ƒ is continuous at y, then Mk, 
, and 
all approach ƒ(y) as |P| → 0. Thus, |U(P) − U(P′)| can be made less than ∊ if |P| is sufficiently small. If α is continuous at y, then α(y) − α(xk−1) and α(xk) − α(y) → 0 as |Ρ| → 0. Since Mk, , and are all bounded by supa≤x≤b ƒ(x), we can again conclude that |U(P) − U(P′)| can be made less than ∊ if |P| is sufficiently small. (A similar analysis can be made for lower sums.) If ƒ and α are discontinuous at the same point, 
does not exist, as you should discover when doing Problem 1.
We are now ready for the main result.
6.1.2THEOREM. If f is continuous and α is increasing on [a, b], then the Riemann–Stieltjes integral 
exists. In particular, if f is continuous on [a, b], then f is Riemann integrable on [a, b].
Proof. Given ∊ > 0, let δ > 0 be such that |x − y| < δ implies |f(x) − f(y)| < ∊. If P1 and P2 are arbitrary partitions with |Ρ1| and |P2| < δ, let P be a common refinement of P1 and P2 (take the union of the points of P1 and the points of P2 to form P). By Eq. (2),
and
Thus,
Since ∊ may be chosen arbitrarily small, it follows that if {Pn} is any sequence of partitions with |Pn| → 0, the sequences {U(Pn)} and {L(Pn)} are Cauchy, and hence converge. By Eq. (1) they converge to the same limit. The observation that U(P1) and U(P2) must be close to each other for small |P1| and |P2| implies that the limit is the same regardless of the particular sequence {Pn}. ■
Note that by Eq. (3), L(Pn) ≤ S(Pn) ≤ U(Pn) for all n; hence, 
as |P| → 0. This may be used as the basis of a generalization of the Riemann–Stieltjes integral in which the monotonicity requirement on α is removed.
6.1.3Definition
Let ƒ : [a, b] → R, α: [a, b] → R, and assume ƒ is bounded. We say that 
exists if S (P, ƒ, α) approaches a finite limit A as |P| → 0. In other words, given ∊ > 0 there is a δ > 0 such that whenever |P| < δ (regardless of the choice of the tk), we have |S(P, f, α) − A| < ∊. Equivalently, if {Pn} is any sequence of partitions with |Pn| → 0, we have S(Pn, f, α) → A.
We know that if ƒ is continuous and α is increasing, then 
exists. The case in which ƒ is continuous and α is decreasing is handled similarly (if we like, we can multiply α by −1 so that −α will be increasing).
Problems for Section 6.1
1.Give an example in which ƒ is Riemann integrable on [a. b] and α is increasing, but 
does not exist.
2.Let ƒ(x) = 1, x rational; f(x) = 0, x irrational. Show that ƒ is not Riemann integrable on [0,1].
3.Give an example of a function ƒ that is not Riemann integrable but whose absolute value |f| is Riemann integrable.
4.Assume ƒ bounded and a increasing on [a, b]. By Eq. (3) if U(P, ƒ, α) and L(P, f, α) approach a common finite limit 
as |P| → 0, then 
. Prove that, conversely, if 
(finite), then U(P, ƒ, α) and L(P, f, α) also approach 
. Thus, the characterizations of the integral via upper and lower sums and via Riemann–Stieltjes sums are equivalent.
6.2PROPERTIES OF THE INTEGRAL
A number of basic properties of the integral follow directly from the definition. To save space, we adopt the notation ƒ ∈ R(α) to indicate that 
exists.
6.2.1THEOREM
(a)If f1, f2 ∈ R(α), then f1 + f2 ∈ R(α) and
(b)If f ∈ R(α) and c is a constant, then cf ∈ R(α) and
Also, f ∈ R(cα) and 
(c)If f ∈ R(α1), and f ∈ R(α2), then f ∈ R(α1 + α2) and
For the remaining statements, assume α is increasing.
(d)If f1, f2 ∈ R(α) and f1 ≤ f2 on [a, b], then
(e)If f ∈ R(α), then |f| ∈ R(α) and
Thus, if |f| ≤ M on [a, b], we have
(f)Assume a < c < b. If f ∈ R(α) on [a, c] and f ∈ R(α) on [c, b], then f ∈ R(α) on [a, b] and
(g)If f is Riemann integrable on [a, b] and g = f except at finitely many points y1, ..., yr, then g is Riemann integrable on [a, b] and 
(h)If f is piecewise continuous on [a, b], that is, f has only finitely many discontinuities on [a, b], all removable or jumps, then f is Riemann integrable on [a, b].
Proof
(a)
(c)
(d)
(since f1 ≤ f2 and α is increasing, so Δαk ≥ 0). Let |P| → 0 to obtain the desired result.
(e)
hence,
Let |P| → 0 to get 
, provided |f| ∈ R(α). But |f(y)| = |f(y) − f(x) + f(x)| ≤ |f(y) – f(x)| + |f(x)|, and similarly |ƒ(x)| ≤ |f(y) − f(x)| + |ƒ(y)|. Consequently, ||ƒ(y)| − |f(x)|| ≤ |f(y) − f(x)|, and it follows that replacing ƒ by |ƒ| decreases |U(P) − L(P)|. (Let |ƒ(y)| approximate sup |ƒ|, and let |f(x)| approximate inf |f|.) Consequently, |f| ∈ R(α). (See Problem 1.)
(f)We may assume that either ƒ or α is continuous at c; otherwise, 
or 
fails to exist (cf. Section 6.1, Problem 1). Given any sequence of partitions P of [a, b] with |P| → 0, and given ∊ > 0, we wish to refine P by adding the point c, producing a new partition Q with |U(P, f, α) − U(Q, f, α)| < ∊ and |L(P, ƒ, α) − L(Q, ƒ, α)| < ∊. The discussion following Eq. (3) of 6.1.1 shows that this can be accomplished for sufficiently small |P| (in that discussion y is replaced by c). Since Q is formed by a partition of [a, c] followed by a partition of [c, b], U(Q, ƒ, α) and 
as |P|, hence |Q|, approaches 0. It follows that for sufficiently small |P|, U(P, ƒ, α) and L(P, f, α) differ from 
by less than ∊. Since ∊ > 0 is arbitrary, the proof is complete.
(g)If 
and 
, then
(h)The interval [a, b] can be written as [y0, y1] ∪ [у1, у2] ∪ ... ∪ [yr−1, yr], where the discontinuities of ƒ occur at the yi. By (g), ƒ is Riemann integrable on each subinterval, since it becomes continuous when changed at the end points. By (f), ƒ is Riemann integrable on [a, b]. ■
It follows from Theorem 6.2.1(b) and (c) that if ƒ is continuous and α is the difference of two increasing functions, then 
exists. We will see in the next section that if α has a continuous derivative α′ on [a, b] then α can be expressed as α1 − α2, where α1 and α2 are increasing; it follows that ƒ ∈ R(α). (In Chapter 5 we omitted any discussion of the “one-sided” derivative at the end points of a closed interval, but the definition is natural:
These comments allow us to show that a Riemann–Stieltjes integral may often be reduced to a Riemann integral, as follows.
6.2.2THEOREM. If f is continuous on [a, b] and α has a continuous derivative on [a, b], then
Proof. Let P: a = x0 < x1 < ... < xn = b be a partition of [a, b]. By the Mean Value Theorem 5.1.4, α(xk) − α(хк−1) = α′(tk)(xk − Xk−1) for some tk ∈ (xk−1, xk). The points t1, ..., tn yield the following Riemann–Stieltjes sum:
Let |P| → 0 to obtain
as desired. ■
We are now in a position to evaluate Riemann–Stieltjes integrals in practical cases. In the usual physical situation, α = α1 + α2 where α1 has a continuous derivative and α2 is a jump function having jumps of size cn at the points xn, n = 1, 2, ... N ; α2 is constant between jumps (see Fig. 6.2.1). In our discussion we assume α2 is increasing, but the evaluation formula to be obtained is valid for any jump function α2, since α2 is expressible as the difference of two increasing jump functions.
Consider the behavior of α2 at a jump of size Δα2(c) at x = c (see Fig. 6.2.2). Assuming ƒ continuous, we know that 
as |P| → 0, so we are free to choose any convenient sequences of P’s. If we choose P so that [c − h, c + h], h > 0, is one of the subintervals of P, we obtain a contribution of the form
as h → 0, by continuity of ƒ. In general, a jump of size cn at xn contributes f(xn)cn to 
. We have arrived at the following result.
6.2.3Evaluation Formula
Let ƒ be continuous on [a, b], and let α = α1 + α2, where 
is continuous on [a, b] and α2 is a jump function. Assume that the jumps of α2 occur at x1, ..., xN, and let 
; notice that the actual value of α2(x) at x = cn is irrelevant. Then
Figure 6.2.2 Behavior at a Jump
The evaluation formula may be extended to the case where ƒ has a finite number of simple discontinuities as long as ƒ and α are not discontinuous at the same point; when simultaneous discontinuities occur, the Riemann–Stieltjes integral does not exist (see Problem 1 of Section 6.1).
A basic application of Riemann–Stieltjes integrals occurs in probability. The function α is the “distribution function” of a “random variable” X. Intuitively, if α has a jump of size cn at xn, then the probability that X will take the value xn in a particular performance of the random experiment is cn. If 
, then X has a “continuous component” (as well as a discrete component represented by the set of probabilities cn). The term 
represents the probability that X will take a value between x and x + dx, assuming none of the xi, 1 ≤ i ≤ N, belong to [x, x + dx]. The evaluation formula then gives the average value of f(X). For example, if f(x) = x2, then the formula gives the average value of the square of the random variable.
Another basic application is to line integrals. For example, if C is a curve in the plane, parametrized by x = x(t), y = y(t), a ≤ t ≤ b, let s be arc length along the curve, so that s(t0) is the length of the portion of the curve given by (x(t), y(t)), a ≤ t ≤ t0. The line integral ∫C f(x, y) ds is given by the Riemann–Stieltjes integral
In particular, if ƒ is the density of a curved rod, then ∫C f(x, y) ds is the total mass.
In the evaluation formula, the integral 
is evaluated by standard calculus techniques. The following result justifies such a computation.
6.2.4Fundamental Theorem of Calculus
Let ƒ be a continuous real–valued function on [a, b].
(a)If 
, then F′ = f on [a, b].
(b)If G′ = ƒ on [a, b], then 
.
Proof
(a)Let h > 0 (the argument for h <0 is similar). Then
by continuity of ƒ.
(b)By (a), (d/dx)(G(x) − F(x)) = f(x) − f(x) = 0 on [a, b], and hence G − F is constant on [a, b] (use the Mean Value Theorem). Thus, 
■
For an intuitive view of Theorem 6.2.4(a), represent F by the cumulative area A under ƒ between a and x. If we change x by a “small” amount dx, the cumulative area changes by (approximately) f(x) dx (see Fig. 6.2.3). Thus dA = f(x) dx, or dA/dx = f(x), which says that F′ = ƒ.
Figure 6.2.3 Intuitive View of the Fundamental Theorem of Calculus
1.If α is increasing and ƒ is bounded on [a, b], show that ƒ ∈ R(α) iff U(P, f, α) − L(P, f, α) → 0 as |P| → 0.
2.Assume α increasing and ƒ bounded on [a, b]. If a < c < b and ƒ ∈ R(α) on [a, b], show that ƒ ∈ R(α) on both [a, c] and [c, b].
3.Let f(x) = x2, and define α as follows:
Evaluate 
4.Evaluate 
where α(x) = [x] = the largest integer ≤ x.
6.3FUNCTIONS OF BOUNDED VARIATION
We have seen that if ƒ is continuous and a is the difference of two increasing functions, then ƒ ∈ R(α). We now look at the question of when a function α can be expressed as α1 − α2, where α1 and α2 are increasing.
6.3.1Definitions and Comments
Let ƒ: [a, b] → R, and let P: a = x0 < x1 < ... < xn = b be a partition of [a, b]. Define
The variation of ƒ on [a, b] is defined by
If we wish to emphasize the interval [a, b] on which the computation is carried out, we write V (ƒ; [a, b]) instead of simply V(ƒ).
We say that ƒ is of bounded variation (on [a, b]) if V (ƒ) < ∞.
It follows from the definitions that a monotone function is of bounded variation, with V(ƒ) = |f(b) − f(a)|. Also, for arbitrary ƒ we have V(−ƒ ) = V(ƒ). A brief computation shows that
To see this, note that V(f + g, P) ≤ V(f, P) + V(g, P) by the triangle inequality; hence, V(f + g, P) ≤ V(f) + V(g) for any P. Take the sup over P to obtain the desired result. Observe also that since V(−g) = V(g), we have V(f − g) ≤ V(ƒ) + V(g). It follows that if α = f − g, where ƒ and g are increasing, then α is of bounded variation.
We adopt the notation ƒ ∈ BV to indicate that ƒ is of bounded variation. The main result is that if α ∈ BV then α can be expressed as the difference of two increasing functions. It follows that if ƒ is continuous and α ∈ BV, then 
exists. Before proving this, we examine some basic properties and examples.
6.3.2THEOREM
(a)If ƒ ∈ BV, then f is bounded.
(b)If f, g ∈ BV with |f| ≤ A and |g| ≤ B on [a, b], then V(ƒg) ≤ AV(g ) + BV(ƒ); hence, f g ∈ BV.
(c)If f′ exists and is bounded on [a, b], then f ∈ BV.
(d)If f(x) = x sin(1/x), 0 ≤ x ≤ b, ƒ(0) = 0, then f is continuous but not of bounded variation.
(e) If a < c < b, then V (ƒ; [a, b]) = V(ƒ; [a, c]) + V(ƒ ; [c, b]).
Proof
(a)If a < x < b, we have
(b)A typical term in the computation of V (f g, P) is
Since |f(xi)| ≤ A and |g (xi−1)| ≤ B, it follows that
Take the sup over P to obtain the desired result.
(c)By the Mean Value Theorem,
Thus, V(ƒ) ≤ M(b − a) < ∞.
(d)Let yn = 2/nπ, n = 1, 2, ... Then
Let Pn be the partition formed by 0, y2n+1, y2n, ..., y1. Then
(e)If P is any partition of [a, c] and Q is any partition of [c, b], we have V(f, P) + V(f, Q) ≤ V(ƒ; [a, b]), so V(ƒ; [a, c]) + V(f;[c, b]) ≤ V(f;[a, b]). If P0 is any partition of [a, b], we can refine P0 if necessary to obtain a partition Р′ of [a, c] followed by a partition Q′ of [c, b]. The refining process can only increase the variation, by the triangle inequality. Thus,
Take the sup over P0 to finish the argument. ■
We now prove the main result on functions of bounded variation.
6.3.3THEOREM. If f is of bounded variation on [a, b], define
Then F and G are increasing on [a, b], so f is expressible as the difference of two increasing functions.
Proof. Note that F “follows” ƒ, reflecting decreasing portions about the horizontal to obtain increasing portions; see Fig. 6.3.1. It follows from Theorem 6.3.2(e) that F is increasing. To show that G is increasing, let x < y. Then
If ƒ is continuous on [a, b] then so are F and G; see the problems.
Problems for Section 6.3
The purpose of this problem set is to show that, in Theorem 6.3.3, if ƒ is continuous on [a, b] so are F and G.
1.Let c ∈ [a, b), ∊ > 0. Show that there is a δ > 0 and a partition P: c = x0+ < x1 < ... < xn = b such that 0 < x1 − x0 < δ, |f(x1) − ƒ(x0)| < ∊/2, and 
Figure 6.3.1 Expressing a Function of Bounded Variation as the Difference of Two Increasing Functions
2.Show that F(b) − F(c) − ∊/2 < ∊/2 + F(b) − F(x1), and conclude that F is right-continuous at c; i.e.; limx→ c;x > c F(x) = F(c).
3.Let a < c ≤ b, ∊ > 0. Show that there is a δ > 0 and a partition P : a = x0 < x1 < ... < xn−1 < xn = c such that |f(xn) − ƒ(xn−1)| < ∊/2, 0 < xn − xn−1 < δ, and 
4.Show that F(c) − F(a) − ∊/2 < ∊/2 + F(xn−1) − F(a), and conclude that F is left-continuous at c; i.e. limx→c; x<c F(x) = F(c). It follows that F, hence G, is continuous on [a, b].
6.4SOME USEFUL INTEGRATION THEOREMS
We close this chapter with some results that are often useful in computations.
6.4.1Integration by Parts
If 
exists, then so does 
, and
Proof. Consider any Riemann–Stieltjes sum for 
, say
Now
Call this last line A. Then
Since xk−1 ≤ tk ≤ xk, this is a Riemann–Stieltjes sum for 
corresponding to a partition Р′ refining P. It follows that as |P| → 0, 
. Thus, 
, as desired. ■
6.4.2Change of Variable Formula
Let ƒ and h be continuous functions from [a, b] to R, and assume h strictly increasing. Let α be the inverse function of h; in other words, if y = h(x), then x = α(y), a ≤ x ≤ b, h(a) ≤ y ≤ h(b). Then
Proof. Let P: a = x0 < x1 < ... < xn = b, Q: h(a) = y0 < y1 < ... < yn = h(b), where yk = h(xk), 0 ≤ k ≤ n. (Since h is strictly increasing, so is α.) Then
Since h and α are continuous functions defined on closed, bounded (therefore compact) intervals, they are uniformly continuous. (Continuity of a follows from Section 4.2, Problem 5). Thus, if ∊ > 0, there is a δ > 0 such that if |xk − хk−1| < δ then |yk − yk−1| < ∊. It follows that if |P| → 0 then |Q| → 0 (and conversely). Now ƒ ○ α is a composition of continuous functions and is therefore continuous. Since α is monotone, 
exists. Thus, if we let |P| → 0 in (1), we obtain 
, as desired. ■
6.4.3Mean Value Theorem for Integrals
If f is continuous and a is increasing on [a, b], there is a point x0 ∈ [a, b] such that 
Proof. Let M = sup{f(t) : a ≤ t ≤ b} and m = inf {ƒ(t) : a ≤ t ≤ b}. Then 
; hence,
(Note that if α(a) = α(b) then α is constant on [a, b] and the result is trival.) Thus 
for some c ∈ [m, M]. Now if, say, f(x1) = m and f(x2) = M, then by the Intermediate Value Theorem there is a point x0 between x1 and x2 such that f(x0) = c. We conclude that 
. ■
6.4.4Upper Bounds on Integrals
Assume 
exists, and let M = sup{|f(x)| : a ≤ x ≤ b}. Then
where V(α) is the variation of a on [a, b]. Thus, if α is increasing, we have
If α(x) = x, we obtain a bound on the Riemann integral:
Proof. Consider an arbitrary Riemann–Stieltjes sum 
. We have
An integral in which one or both limits of integration are infinite is called improper; it is defined as a limit of ordinary integrals. For example,
In many practical cases, integrals of this type are evaluated in the usual way; for example, 
. For a formal justification, note that 
as b → ∞.
An integral in which the function ƒ is unbounded is also referred to as improper. The theory of improper integrals is best handled within the domain of measure theory. As an example of what might happen, consider the function of Fig. 6.4.1. Here, ƒ is bounded and continuous everywhere, but 
does not exist because 
oscillates between 0 and 1 for n = 0, 1, 2, ... In general, it turns out that if 
exists and is finite—that is, 
approaches a finite limit as a → − ∞ and b → ∞—then 
exists and is finite also.
Figure 6.4.1 Improper Riemann Integral
1.If ƒ and g are continuous and α increasing on [a, b], show that
This is the Cauchy–Schwarz inequality for integrals;, the analogous result for sums is given in Section 3.2, Problem 3.
2.Prove the integral test for convergence of series: if ƒ is continuous, nonnegative, and decreasing, show that 
is finite (that is, 
approaches a finite limit as b → ∞) if and only if 
.
3.Give an example of a function ƒ for which the improper Riemann integral 
is finite, but 
.
4.Give an example of an unbounded function ƒ on (0,1] such that the improper integral 
is
(a)finite.
(b)infinite.
REVIEW PROBLEMS FOR CHAPTER 6
1.Evaluate 
, where ƒ(x) = 1/x and
2.True or false, and explain briefly.
(a)If ƒ is piecewise continuous on [a, b] and α is increasing on [a, b], then 
exists.
(b)Let ƒ be continuous and α increasing on [a, b]. If Q is a refinement of the partition P, then L(Q, f, α) ≥ L(P, ƒ, α).
(c)If ƒ is increasing and a is continuous on [a, b], then 
exists.
3.Let ƒ be continuous on R, and let g and h be differentiable on R. Show that
4.Give an example of a function ƒ : [0, 1] → R such that ƒ is not Riemann integrable on [0,1] but ƒ2 is Riemann integrable on [0,1].
5.Let
If f(x) = 3х, evaluate 
.
6.Let ƒ be a continuous, real-valued function on [a, b], and let 
. Show that F is of bounded variation on [a, b].
7.A student evaluates the integral 
as follows: Make the change of variable y = x2 to get
because 
... is always 0.
But
Where did the student go wrong?
