Chapter 5. Integration §1. The Riemann Integral Let a and b be two real numbers with a<b. Then [a, b] is a closed and bounded interval in IR. By a partition P of [a, b] we mean a finite ordered set {t 0 ,t 1 ,...,t n } such that a = t 0 <t 1 < ··· <t n = b. The norm of P is defined by ∥P ∥ := max{t i − t i−1 : i =1, 2,...,n}. Suppose f is a bounded real-valued function on [a, b]. Given a partition {t 0 ,t 1 ,...,t n } of [a, b], for each i =1, 2,...,n, let m i := inf {f (x): t i−1 ≤ x ≤ t i } and M i := sup{f (x): t i−1 ≤ x ≤ t i }. The upper sum U (f,P ) and the lower sum L(f,P ) for the function f and the partition P are defined by U (f,P ) := n ∑ i=1 M i (t i − t i−1 ) and L(f,P ) := n ∑ i=1 m i (t i − t i−1 ). The upper integral U (f ) of f over [a, b] is defined by U (f ) := inf {U (f,P ): P is a partition of [a, b]} and the lower integral L(f ) of f over [a, b] is defined by L(f ) := sup{L(f,P ): P is a partition of [a, b]}. A bounded function f on [a, b] is said to be (Riemann) integrable if L(f )= U (f ). In this case, we write ∫ b a f (x) dx = L(f )= U (f ). By convention we define ∫ a b f (x) dx := − ∫ b a f (x) dx and ∫ a a f (x) dx := 0. A constant function on [a, b] is integrable. Indeed, if f (x)= c for all x ∈ [a, b], then L(f,P )= c(b − a) and U (f,P )= c(b − a) for any partition P of [a, b]. It follows that ∫ b a c dx = c(b − a). 1
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Chapter 5. Integration
§1. The Riemann Integral
Let a and b be two real numbers with a < b. Then [a, b] is a closed and bounded
interval in IR. By a partition P of [a, b] we mean a finite ordered set {t0, t1, . . . , tn} such
that
a = t0 < t1 < · · · < tn = b.
The norm of P is defined by ∥P∥ := max{ti − ti−1 : i = 1, 2, . . . , n}.Suppose f is a bounded real-valued function on [a, b]. Given a partition {t0, t1, . . . , tn}
of [a, b], for each i = 1, 2, . . . , n, let
mi := inf{f(x) : ti−1 ≤ x ≤ ti} and Mi := sup{f(x) : ti−1 ≤ x ≤ ti}.
The upper sum U(f, P ) and the lower sum L(f, P ) for the function f and the partition
P are defined by
U(f, P ) :=n∑
i=1
Mi(ti − ti−1) and L(f, P ) :=n∑
i=1
mi(ti − ti−1).
The upper integral U(f) of f over [a, b] is defined by
U(f) := inf{U(f, P ) : P is a partition of [a, b]}
and the lower integral L(f) of f over [a, b] is defined by
L(f) := sup{L(f, P ) : P is a partition of [a, b]}.
A bounded function f on [a, b] is said to be (Riemann) integrable if L(f) = U(f). In this
case, we write ∫ b
a
f(x) dx = L(f) = U(f).
By convention we define∫ a
b
f(x) dx := −∫ b
a
f(x) dx and
∫ a
a
f(x) dx := 0.
A constant function on [a, b] is integrable. Indeed, if f(x) = c for all x ∈ [a, b], then
L(f, P ) = c(b− a) and U(f, P ) = c(b− a) for any partition P of [a, b]. It follows that∫ b
a
c dx = c(b− a).
1
Let f be a bounded function from [a, b] to IR such that |f(x)| ≤ M for all x ∈ [a, b].
Suppose that P = {t0, t1, . . . , tn} is a partition of [a, b], and that P1 is a partition obtained
from P by adding one more point t∗ ∈ (ti−1, ti) for some i. The lower sums for P and P1
are the same except for the terms involving ti−1 or ti. Let mi := inf{f(x) : ti−1 ≤ x ≤ ti},m′ := inf{f(x) : ti−1 ≤ x ≤ t∗}, and m′′ := inf{f(x) : t∗ ≤ x ≤ ti}. Then
Now suppose that PN is a mesh obtained from P by adding N points. An induction
argument shows that
L(f, PN )− 2MN∥P∥ ≤ L(f, P ) ≤ L(f, PN ). (1)
Similarly we have
U(f, PN ) ≤ U(f, P ) ≤ U(f, PN ) + 2MN∥P∥. (2)
By the definition of L(f) and U(f), for each n ∈ IN there exist partitions P and Q of
[a, b] such that
L(f)− 1/n ≤ L(f, P ) and U(f) + 1/n ≥ U(f,Q).
Consider the partition P ∪Q of [a, b]. Since P ⊆ P ∪Q and Q ⊆ P ∪Q, by (1) and (2) we
get
L(f, P ) ≤ L(f, P ∪Q) ≤ U(f, P ∪Q) ≤ U(f,Q).
It follows that L(f)−1/n ≤ U(f)+1/n for all n ∈ IN. Letting n → ∞ in the last inequality,
we obtain L(f) ≤ U(f).
We are in a position to establish the following criterion for a bounded function to be
integrable.
2
Theorem 1.1. A bounded function f on [a, b] is integrable if and only if for each ε > 0
there exists a partition P of [a, b] such that
U(f, P )− L(f, P ) < ε.
Proof. Suppose that f is integrable on [a, b]. For ε > 0, there exist partitions P1 and P2
such that
L(f, P1) > L(f)− ε
2and U(f, P2) < U(f) +
ε
2.
For P := P1 ∪ P2 we have
L(f)− ε
2< L(f, P1) ≤ L(f, P ) ≤ U(f, P ) ≤ U(f, P2) < U(f) +
ε
2.
Since L(f) = U(f), it follows that U(f, P )− L(f, P ) < ε.
Conversely, suppose that for each ε > 0 there exists a partition P of [a, b] such that
U(f, P )− L(f, P ) < ε. Then U(f, P ) < L(f, P ) + ε. It follows that
U(f) ≤ U(f, P ) < L(f, P ) + ε ≤ L(f) + ε.
Since ε > 0 is arbitrary, we have U(f) ≤ L(f). But L(f) ≤ U(f). Therefore U(f) = L(f);
that is, f is integrable.
Let f be a bounded real-valued function on [a, b] and let P = {t0, t1, . . . , tn} be a
partition of [a, b]. For each i = 1, 2, . . . , n, choose ξi ∈ [xi−1, xi]. The sum
n∑i=1
f(ξi)(ti − ti−1)
is called a Riemann sum of f with respect to the partition P and points {ξ1, . . . , ξn}.
Theorem 1.2. Let f be a bounded real-valued function on [a, b]. Then f is integrable on
[a, b] if and only if there exists a real number I with the following property: For any ε > 0
there exists some δ > 0 such that∣∣∣∣ n∑i=1
f(ξi)(ti − ti−1)− I
∣∣∣∣ ≤ ε (3)
whenever P = {t0, t1, . . . , tn} is a partition of [a, b] with ∥P∥ < δ and ξi ∈ [ti−1, ti] for
i = 1, 2, . . . , n. If this is the case, then∫ b
a
f(x) dx = I.
3
Proof. Let ε be an arbitrary positive number. Suppose that (3) is true for some partition
P = {t0, t1, . . . , tn} of [a, b] and points ξi ∈ [ti−1, ti], i = 1, 2, . . . , n. Then
L(f, P ) = inf
{ n∑i=1
f(ξi)(ti − ti−1) : ξi ∈ [xi−1, xi] for i = 1, 2, . . . , n
}≥ I − ε
and
U(f, P ) = sup
{ n∑i=1
f(ξi)(ti − ti−1) : ξi ∈ [xi−1, xi] for i = 1, 2, . . . , n
}≤ I + ε.
It follows that U(f, P ) − L(f, P ) ≤ 2ε. By Theorem 1.1 we conclude that f is integrable
on [a, b]. Moreover, L(f) = U(f) = I.
Conversely, suppose that f is integrable on [a, b]. Let M := sup{|f(x)| : x ∈ [a, b]}and I := L(f) = U(f). Given an arbitrary ε > 0, there exists a partition Q of [a, b]
such that L(f,Q) > I − ε/2 and U(f,Q) < I + ε/2. Suppose that Q has N points. Let
P = {t0, t1, . . . , tn} be a partition of [a, b] with ∥P∥ < δ. Consider the partition P ∪Q of
[a, b]. By (1) and (2) we have
L(f, P ) ≥ L(f, P ∪Q)− 2MNδ and U(f, P ) ≤ U(f, P ∪Q) + 2MNδ.
But L(f, P ∪ Q) ≥ L(f,Q) > I − ε/2 and U(f, P ∪ Q) ≤ U(f,Q) < I + ε/2. Choose
δ := ε/(4MN). Since ∥P∥ < δ, we deduce from the foregoing inequalities that
I − ε < L(f, P ) ≤ U(f, P ) < I + ε.
Thus, with ξi ∈ [ti−1, ti] for i = 1, 2, . . . , n we obtain
I − ε < L(f, P ) ≤n∑
i=1
f(ξi)(ti − ti−1) ≤ U(f, P ) < I + ε.
This completes the proof.
Theorem 1.3. Let f be a bounded function from a bounded closed interval [a, b] to IR.
If the set of discontinuities of f is finite, then f is integrable on [a, b].
Proof. Let D be the set of discontinuities of f . By our assumption, D is finite. So the
set D ∪ {a, b} can be expressed as {d0, d1, . . . , dN} with a = d0 < d1 < · · · < dN = b. Let
M := sup{|f(x)| : x ∈ [a, b]}. For an arbitrary positive number ε, we choose η > 0 such
4
that η < ε/(8MN) and η < (dj − dj−1)/3 for all j = 1, . . . , N . For j = 0, 1, . . . , N , let
xj := dj − η and yj := dj + η. Then we have
a = d0 < y0 < x1 < d1 < y1 < · · · < xN < dN = b.
Let E be the union of the intervals [d0, y0], [x1, d1], [d1, y1], . . . , [xN−1, dN−1], [dN−1, yN−1],
and [xN , dN ]. There are 2N intervals in total. For j = 1, . . . , N , let Fj := [yj−1, xj ].
Further, let F := ∪Nj=1Fj . The function f is continuous on F , which is a finite union of
bounded closed intervals. Hence f is uniformly continuous on F . There exists some δ > 0
such that |f(x)− f(y)| < ε/(2(b− a)) whenever x, y ∈ F satisfying |x− y| < δ. For each
j ∈ {1, . . . , N}, let Pj be a partition of Fj such that ∥Pj∥ < δ. Let
P := {a, b} ∪D ∪(∪Nj=1Pj
).
The set P can be arranged as {t0, t1, . . . , tn} with a = t0 < t1 < · · · < tn = b. Consider
U(f, P )− L(f, P ) =
n∑i=1
(Mi −mi)(ti − ti−1),
where Mi := sup{f(x) : ti−1 ≤ x ≤ ti} and mi := inf{f(x) : ti−1 ≤ x ≤ ti}. Each interval
[ti−1, ti] is either contained in E or in F , but not in both. Hence
n∑i=1
(Mi −mi)(ti − ti−1) =∑
[ti−1,ti]⊆E
(Mi −mi)(ti − ti−1) +∑
[ti−1,ti]⊆F
(Mi −mi)(ti − ti−1).
There are 2N intervals [ti−1, ti] contained in E. Each interval has length η < ε/(8MN).
Noting that Mi −mi ≤ 2M , we obtain∑[ti−1,ti]⊆E
(Mi −mi)(ti − ti−1) ≤ 2N(2M)η <ε
2.
If [ti−1, ti] ⊆ F , then ti − ti−1 < δ; hence Mi −mi < ε/(2(b− a)). Therefore,∑[ti−1,ti]⊆F
(Mi −mi)(ti − ti−1) ≤ε
2(b− a)
∑[ti−1,ti]⊆F
(ti − ti−1) <ε
2(b− a)(b− a) =
ε
2.
From the above estimates we conclude that U(f, P ) − L(f, P ) < ε. By Theorem 1.1, the
function f is integrable on [a, b].
Example 1. Let [a, b] be a closed interval with a < b, and let f be the function on [a, b]
given by f(x) = x. By Theorem 1.3, f is integrable on [a, b]. Let P = {t0, t1, . . . , tn} be a
partition of [a, b] and choose ξi := (ti−1 + ti)/2 ∈ [ti−1, ti] for i = 1, 2, . . . , n. Then
n∑i=1
f(ξi)(ti−ti−1) =1
2
n∑i=1
(ti+ti−1)(ti−ti−1) =1
2
n∑i=1
(t2i −t2i−1) =1
2(t2n−t20) =
1
2(b2−a2).
5
By Theorem 1.2 we have ∫ b
a
x dx =1
2(b2 − a2).
More generally, for a positive integer k, let fk be the function given by fk(x) = xk for
x ∈ [a, b]. Choose
ξi :=
(tki−1 + tk−1
i−1 ti + · · ·+ tkik + 1
)1/k
, i = 1, 2, . . . , n.
We have ti−1 ≤ ξi ≤ ti for i = 1, 2, . . . , n. Moreover,
n∑i=1
fk(ξi)(ti − ti−1) =1
k + 1
n∑i=1
(tk+1i − tk+1
i−1 ) =1
k + 1(tk+1
n − tk+10 ) =
1
k + 1(bk+1 − ak+1).
By Theorem 1.2 we conclude that
∫ b
a
xk dx =1
k + 1(bk+1 − ak+1).
Example 2. Let g be the function on [0, 1] defined by g(x) := cos(1/x) for 0 < x ≤ 1
and g(0) := 0. The only discontinuity point of g is 0. By Theorem 1.3, g is integrable on
[0, 1]. Note that g is not uniformly continuous on (0, 1). Indeed, let xn := 1/(2nπ) and
yn := 1/(2nπ + π/2) for n ∈ IN. Then limn→∞(xn − yn) = 0. But
Hence g is not uniformly continuous on (0, 1). On the other hand, the function u given
by u(x) := 1/x for 0 < x ≤ 1 and u(0) := 0 is not integrable on [0, 1], even though u is
continuous on (0, 1]. Theorem 1.3 is not applicable to u, because u is unbounded.
Example 3. Let h be the function on [0, 1] defined by h(x) := 1 if x is a rational number
in [0, 1] and h(x) := 0 if x is an irrational number in [0, 1]. Let P = {t0, t1, . . . , tn} be a
partition of [0, 1]. For i = 1, . . . , n we have
mi := inf{h(x) : x ∈ [ti−1, ti]} = 0 and Mi := sup{h(x) : x ∈ [ti−1, ti]} = 1.
Hence L(h, P ) = 0 and U(h, P ) = 1 for every partition P of [0, 1]. Consequently, L(h) = 0
and U(h) = 1. This shows that h is not Riemann integrable on [0, 1].
6
§2. Properties of the Riemann Integral
In this section we establish some basic properties of the Riemann integral.
Theorem 2.1. Let f and g be integrable functions from a bounded closed interval [a, b]
to IR. Then
(1) For any real number c, cf is integrable on [a, b] and∫ b
a(cf)(x) dx = c
∫ b
af(x) dx;
(2) f + g is integrable on [a, b] and∫ b
a(f + g)(x) dx =
∫ b
af(x) dx+
∫ b
ag(x) dx.
Proof. Suppose that f and g are integrable functions on [a, b]. Write I(f) :=∫ b
af(x) dx
and I(g) :=∫ b
ag(x) dx. Let ε be an arbitrary positive number. By Theorem 1.2, there
exists some δ > 0 such that∣∣∣∣ n∑i=1
f(ξi)(ti − ti−1)− I(f)
∣∣∣∣ ≤ ε and
∣∣∣∣ n∑i=1
g(ξi)(ti − ti−1)− I(g)
∣∣∣∣ ≤ ε
whenever P = {t0, t1, . . . , tn} is a partition of [a, b] with ∥P∥ < δ and ξi ∈ [ti−1, ti] for
i = 1, 2, . . . , n. It follows that∣∣∣∣ n∑i=1
(cf)(ξi)(ti − ti−1)− cI(f)
∣∣∣∣ = |c|∣∣∣∣ n∑i=1
f(ξi)(ti − ti−1)− I(f)
∣∣∣∣ ≤ |c|ε.
Hence cf is integrable on [a, b] and∫ b
a(cf)(x) dx = c
∫ b
af(x) dx. Moreover,∣∣∣∣ n∑
i=1
(f + g)(ξi)(ti − ti−1)− [I(f) + I(g)]
∣∣∣∣≤
∣∣∣∣ n∑i=1
f(ξi)(ti − ti−1)− I(f)
∣∣∣∣+ ∣∣∣∣ n∑i=1
g(ξi)(ti − ti−1)− I(g)
∣∣∣∣ ≤ 2ε.
Therefore f + g is integrable on [a, b] and∫ b
a(f + g)(x) dx =
∫ b
af(x) dx+
∫ b
ag(x) dx.
Theorem 2.2. Let f and g be integrable functions on [a, b]. Then fg is an integrable
function on [a, b].
Proof. Let us first show that f2 is integrable on [a, b]. Since f is bounded, there exists
some M > 0 such that |f(x)| ≤ M for all x ∈ [a, b]. It follows that∣∣[f(x)]2 − [f(y)]2∣∣ = |f(x) + f(y)||f(x)− f(y)| ≤ 2M |f(x)− f(y)| for all x, y ∈ [a, b].
We deduce from the above inequality that U(f2, P ) − L(f2, P ) ≤ 2M [U(f, P ) − L(f, P )]
for any partition P of [a, b]. Let ε > 0. Since f is integrable on [a, b], by Theorem 1.1
7
there exists a partition P of [a, b] such that U(f, P ) − L(f, P ) < ε/(2M). Consequently,
U(f2, P )−L(f2, P ) < ε. By Theorem 1.1 again we conclude that f2 is integrable on [a, b].
Note that fg = [(f+g)2− (f−g)2]/4. By Theorem 2.1, f+g and f−g are integrable
on [a, b]. By what has been proved, both (f + g)2 and (f − g)2 are integrable on [a, b].
Using Theorem 2.1 again, we conclude that fg is integrable on [a, b].
Theorem 2.3. Let a, b, c, d be real numbers such that a ≤ c < d ≤ b. If a real-valued
function f is integrable on [a, b], then f |[c,d] is integrable on [c, d].
Proof. Suppose that f is integrable on [a, b]. Let ε be an arbitrary positive number. By
Theorem 1.1, there exists a partition P of [a, b] such that U(f, P )−L(f, P ) < ε. It follows
that U(f, P ∪ {c, d}) − L(f, P ∪ {c, d}) < ε. Let Q := (P ∪ {c, d}) ∩ [c, d]. Then Q is a
partition of [c, d]. We have
U(f |[c,d], Q)− L(f |[c,d], Q) ≤ U(f, P ∪ {c, d})− L(f, P ∪ {c, d}) < ε.
Hence f |[c,d] is integrable on [c, d].
Theorem 2.4. Let f be a bounded real-valued function on [a, b]. If a < c < b, and if f is
integrable on [a, c] and [c, b], then f is integrable on [a, b] and∫ b
a
f(x) dx =
∫ c
a
f(x) dx+
∫ b
c
f(x) dx.
Proof. Suppose that f is integrable on [a, c] and [c, b]. We write I1 :=∫ c
af(x) dx and
I2 :=∫ b
cf(x) dx. Let ε > 0. By Theorem 1.1, there exist a partition P1 = {s0, s1, . . . , sm}
of [a, c] and a partition P2 = {t0, t1, . . . , tn} of [c, b] such that
U(f, P1)− L(f, P1) <ε
2and U(f, P2)− L(f, P2) <
ε
2.
Let P := P1 ∪ P2 = {s0, . . . , sm−1, t0, . . . , tn}. Then P is a partition of [a, b]. We have