Proof: IfP is any partition of [a, b] then
|R(f, P)| ≤ R(|f|, P) . The first assertion follows.
Next, for part (ii),
R(f, P) ≤ R(g, P) . This inequality implies the second assertion.
Another fundamental operation in the theory of the integral is “change of variable” (sometimes called the “u-substitution” in calculus books). We next turn to a careful formulation and proof of this operation. First we need a lemma:
Lemma 7.16 If f is a Riemann integrable function on [a, b] and if φ is a con-tinuous function on a compact interval that contains the range of f then φ◦ f is Riemann integrable.
Proof: Let ǫ > 0. Since φ is a continuous function on a compact set, it is uniformly continuous (Theorem 5.27). Let δ > 0 be selected such that (i) δ < ǫ and (ii) if|x − y| < δ then |φ(x) − φ(y)| < ǫ.
Now the hypothesis that f is Riemann integrable implies that there exists a eδ > 0 such that ifP and P′ are partitions of [a, b] and m(P), m(P′) < eδ then,
144 CHAPTER 7. THE INTEGRAL
by the choice of eδ (and Remark 7.7). Therefore X
ℓ∈B
∆ℓ< δ .
Let M be an upper bound for|φ| (Theorem 5.22). Then the last line does not exceed (b− a) · ǫ. Putting together our estimates, we find that
|R(φ ◦ f, P) − R(φ ◦ f, Q)| < ǫ · (2M + (b − a)) .
By symmetry, an analogous inequality holds forP′. By Lemma 7.9, this is what we needed to prove.
An easier result is that if f is Riemann integrable on an interval [a, b] and if µ : [α, β]→ [a, b] is continuously differentiable, then f ◦µ is Riemann integrable.
Corollary 7.17 If f and g are Riemann integrable on [a, b], then so is the function f· g.
Proof: By Theorem 7.11, f + g is integrable. By the lemma, (f + g)2 = f2+ 2f · g + g2 is integrable. But the lemma also implies that f2 and g2 are
7.2. PROPERTIES OF THE RIEMANN INTEGRAL 145 integrable (here we use the function φ(x) = x2). It results, by subtraction, that 2· f · g is integrable. Hence f · g is integrable.
Theorem 7.18 Let f be an integrable function on an interval [a, b] of positive length. Let ψ be a continuously differentiable function from another interval [α, β] of positive length into [a, b]. Assume that ψ is increasing, one-to-one, and onto. Then
Proof: Since f is integrable, its absolute value is bounded by some number M . Fix ǫ > 0. Since ψ′ is continuous on the compact interval [α, β], it is uniformly continuous (Theorem 5.27). Hence we may choose δ > 0 so small that if|s−t| <
δ then|ψ′(s)− ψ′(t)| < ǫ/ (M · (β − α)). If P = {p0, . . . , pk} is any partition of [a, b] then there is an associated partition eP = {ψ−1(p0), . . . , ψ−1(pk)} of [α, β].
For simplicity denote the points of eP by epj. Let us choose the partitionP so fine that the mesh of eP is less than δ. If tj are points of Ij = [pj−1, pj] then
where we have used the Mean Value Theorem in the last line to find each uj. Our problem at this point is that f◦ ψ and ψ′are evaluated at different points.
So we must do some estimation to correct that problem.
The last displayed line equals Xk error term. Since the points uj and sj are elements of the same interval eIj of length less than δ, we conclude that |ψ′(uj)− ψ′(sj)| < ǫ/(M · |β − α|). Thus the error term in absolute value does not exceed
Xk
146 CHAPTER 7. THE INTEGRAL This shows that every Riemann sum for f on [a, b] with sufficiently small mesh corresponds to a Riemann sum for f (ψ(x))· ψ′(x) on [α, β] plus an error term of size less than ǫ. A similar argument shows that every Riemann sum for f (ψ(x))· ψ′(x) on [α, β] with sufficiently small mesh corresponds to a Riemann sum for f on [a, b] plus an error term of magnitude less than ǫ. The conclusion is then that the integral of f on [a, b] (which exists by hypothesis) and the integral of f (ψ(x))·ψ′(x) on [α, β] (which exists by the corollary to the lemma) agree.
We conclude this section with the very important
Theorem 7.19 (The Fundamental Theorem of Calculus) Let f be an in-tegrable function on the interval [a, b]. For x∈ [a, b] we define
F (x) =
Notice that we rewrote f (x) as the integral with respect to a dummy variable s over an interval of length |t − x| divided by (t − x). Assume for the moment that t > x. Then the last line is dominated by
Rt
A similar estimate holds when t < x (simply reverse the limits of integration).
This shows that
t→xlim
F (t)− F (x) t− x
exists and equals f (x). Thus F′(x) exists and equals f (x).
In the exercises we shall consider how to use the theory of one-sided limits to make the conclusion of the Fundamental Theorem true on the entire interval [a, b]. We conclude with
EXERCISES 147 Corollary 7.20 If f is a continuous function on [a, b] and if G is any continu-ously differentiable function on [a, b] whose derivative equals f on (a, b) then
Z b a
f (x) dx = G(b)− G(a) .
Proof: Define F as in the theorem. Since F and G have the same derivative on (a, b), they differ by a constant (Corollary 6.13). Then
Z b a
f (x) dx = F (b) = F (b)− F (a) = G(b) − G(a)
as desired.
Exercises
1. Imitate the proof of the Fundamental Theorem of Calculus in this section to show that, if f is continuous on [a, b] and if we define
F (x) = Z x
a
f (t) dt ,
then the one-sided derivative F′(a) exists and equals f (a) in the sense that
t→alim+
F (t)− F (a)
t− a = f (a) .
Formulate and prove an analogous statement for the one-sided derivative of F at b.
2. Let f be a bounded function on an unbounded interval of the form [A,∞).
We say that f is integrable on [A,∞) if f is integrable on every compact subinterval of [A,∞) and
B→+∞lim Z B
A
f (x) dx
exists and is finite.
Assume that f is Riemann integrable on [1, N ] for every N > 1 and that f is decreasing. Show that f is Riemann integrable on [1,∞) if and only ifP∞
j=1f (j) is finite.
Suppose that g is nonnegative and integrable on [1,∞). If 0 ≤ |f(x)| ≤ g(x) for x∈ [1, ∞) and f is integrable on compact subintervals of [1, ∞), then prove that f is integrable on [1,∞).
148 CHAPTER 7. THE INTEGRAL 3. Let f be a function on an interval of the form (a, b] such that f is integrable
on compact subintervals of (a, b]. If
ǫ→0lim+ Z b
a+ǫ
f (x) dx
exists and is finite then we say that f is integrable on (a, b]. Prove that, if we restrict attention to bounded f , then in fact this definition gives rise to no new integrable functions. However, there are unbounded functions that can now be integrated. Give an example.
Give an example of a function g that is integrable by the definition in the preceding paragraph but is such that |g| is not integrable.
4. Suppose that f is a continuous, nonnegative function on the interval [0, 1].
Let M be the maximum of f on the interval. Prove that
n→∞lim
5. Let f be a continuously differentiable function on the interval [0, 2π]. Fur-ther assume that f (0) = f (2π) and f′(0) = f′(2π). For n∈ N define
converges. [Hint: Use integration by parts to obtain a favorable estimate on| bf (n)|.] is finite for every x. Is it the case that f is Riemann integrable?
* 8. Suppose that f is a Riemann integrable function on the interval [0, 1]. Let ǫ > 0. Show that there is a polynomial p so that
Z 1
0 |f(x) − p(x)| dx < ǫ . 9. Prove part (b) of Theorem 7.11.