Riemann Integration
Dung Le1
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Preliminaries
Let f be a bounded function defined on an interval [a, b] (a < b). A partition of [a, b]
is a finite set
P = {x0 , x1 , . . . , xn },
a = x0 < x1 < . . . < xn−1 < xn = b.
We set ∆j = xj − xj−1 (the length of the j th interval [xj−1 , xj ]) and define the mesh
of the partition P as
mesh(P ) = max ∆j .
j
For each j, we also the define
Mj (f, P ) =
sup
f (x),
mj (f, P ) =
xj−1 ≤x≤xj
inf
xj−1 ≤x≤xj
f (x).
We then define the upper and lower sums of f with respect to the partition P by
U (f, P ) =
n
X
Mj (f, P )∆j ,
L(f, P ) =
n
X
mj (f, P )∆j .
j=1
j=1
Furthermore, let T = {tj : tj ∈ [xj−1 , xj ], j = 1, . . . , n}, we define the Riemann
sum
n
I(f, P, T ) =
X
f (tj )∆j .
j=1
From these definitions, we obviously have
L(f, P ) ≤ I(f, P, T ) ≤ U (f, P ).
If P, Q are two partitions of [a, b] and P ⊂ Q, we then say Q is finer than P or Q
S
is a refinement of P . Note that R = P Q is also a partition of [a, b] (by reindexing
the points in R), and R is called a common refinement of P, Q.
1
Department of Applied Mathematics, University of Texas at San Antonio, 6900 North Loop
1604 West, San Antonio, TX 78249. Email: [email protected]
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The definition of Riemann integrals
Before we get to the definition of the integral, we need some easy lemmas.
Lemma 2.1 If R is finer than P then
L(f, P ) ≤ L(f, R) ≤ U (f, R) ≤ U (f, P ).
Proof: Exercise.
The lemma says that while the partition is getting finer, the upper sum will be
smaller and the lower sum will be larger. This lemma has an important consequence
that the lower sums are always less than the upper sums.
Corollary 2.2 Let P, Q be two partitions of [a, b]. Then
L(f, P ) ≤ U (f, Q).
Proof: Consider the common refinement R = P
S
Q and use the above lemma.
We see that the set of (numbers) {L(f, P )} is bounded from above by any upper
sum U (f, Q). Similarly, the set of (numbers) {U (f, P )} is bounded from below by
any lower sum L(f, Q). Hence the following numbers are well defined
L(f ) = sup L(f, P ),
U (f ) = inf U (f, P ).
P
P
By Corollary 2.2, we have L(f ) ≤ U (f ), in general. We then have the following
definition of Riemann integrals.
Definition 2.3 A bounded function f on [a, b] is said to be Riemann integrable iff
L(f ) = U (f ). In this case, the common value of L(f ) and U (f ) is called the integral
of f over [a, b] and it is denoted by
Z
b
f (x)dx.
a
The most elementary test for Riemann integrability is the following
Theorem 2.4 (Riemann’s condition) A bounded function f on [a, b] is Riemann integrable iff for every given ε > 0, ther is a partition P so that U (f, P ) − L(f, P ) < ε.
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Corollary 2.5 Let f : [a, b] → IR be abounded function. If there is a sequence of
partitions Pn such that
lim [U (f, Pn ) − L(f, Pn )] = 0
n→∞
then f is Riemann integrable. Moreover, if Tn is any choice of points tn,j selected
from the subintervals of Pn , then
lim I(f, Pn , Tn ) =
n→∞
Z
b
f (x)dx.
a
The following gives another alternative to check Riemann integrability using the
Riemann sum instead of upper and lower sums.
Theorem 2.6 Let f : [a, b] → IR be abounded function. f is Riemann integrable iff
there is a number I such that for every given ε > 0, there exists δ > 0 so that every
partition P such that mesh(P ) < δ and every choice of T = {tj }nj=1 , we have
|I(f, P, T ) − I| < ε.
In this case,
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Rb
a
f (x)dx = I.
Riemann Integrable functions
The two basic classes of functions that are Riemann integrable are described in the
following theorems.
Theorem 3.1 Every monotone function on [a, b] is Riemann integrable.
Theorem 3.2 Every continuous function on [a, b] is Riemann integrable.
The most general description of Riemann integrable function is given by
Theorem 3.3 A bounded function f on [a, b] is Riemann integrable iff it is continuous almost everywhere.
What does it means by continuous ”almost everywhere”? This is to say that the
set of points where f is discontinuous is negligible or has measure zero. the concept
of measure will be discussed later when we introduce a more powerful theory of
integration: Lebesgue integral. Until then, we will prove these theorems.
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The fundamental theorem of calculus
Our discussion so far is theoretically important, but it is not useful in evaluating
integrals. Fortunately, there is a crucial connection between integrals and derivatives
that makes evaluating integrals is more efficient. keep in mind that this is not the
definition of integrals but a central result connecting differential and integral calculus.
Theorem 4.1 Let f : [a, b] → IR be a bounded Riemann integrable function. We
define
Z x
F (x) =
f (t)dt, x ∈ [a, b].
a
then F is a continuous function. Moreover, if f is continuous at c ∈ [a, b], then F is
differentiable at c and F 0 (c) = f (c).
Before proving the fundamental theorem of calculus, let us recall the so called
second fundamental theorem of calculus, which we all know from calculus courses. The
theorem concerns the concept of antiderivatives from differential calculus.
A function f on [a, b] has an antiderivative F on [a, b] if F 0 (x) = f (x) for all
x ∈ (a, b) and F (x) is continuous on [a, b].
As we all well know that finding antiderivatives explicitly by methods we learned
in calculus courses is not an easy problem (many times, it is even impossible!). The
following says that antiderivatives do exist in general and they can help to compute
Riemann integrals.
Corollary 4.2 Let f be a continuous function on [a, b]. Then f has an antiderivative.
Moreover, if G is any antiderivative of f then
Z
b
a
f (x)dx = G(b) − G(a) = G(x)|ba = [G(x)]ba .
Proof: By Theorem 4.1, F (x) = ax f (t)dt is differentiable and F 0 (x) = f (x) for
all x ∈ (a, b). If G is any other antiderivative then G0 = F 0 and, thus, (G − F )0 = 0
on (a, b). Hence, G − F is a constant on [a, b] (why?) and therefore G(b) − G(a) =
F (b) − F (a). But,
R
F (b) − F (a) =
Z
b
f (x)dx,
a
(why?) which gives the assertion.
We now go back to the proof of Theorem 4.1. We will need the following lemmas,
whose proof are left as exercise.
Lemma 4.3 Let f : [a, b] → IR be Riemann integrable. For any c ∈ [a, b], f : [a, c] →
IR and f : [c, b] → IR are also Riemann integrable. Moreover,
Z
b
a
f (x)dx =
Z
c
f (x)dx +
a
Z
c
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b
f (x)dx.
Lemma 4.4 Let f : [a, b] → IR be Riemann integrable. If m, M are numbers such
that m ≤ f (x) ≤ M for all x ∈ [a, b] then
m(b − a) ≤
Z
b
f (x)dx ≤ M (b − a).
a
We now give the proof of Theorem 4.1. Since f is bounded, there is a number
M such that |f (x)| ≤ M for all x ∈ [a, b]. For any x, y ∈ [a, b], by Lemma 4.3 and
Lemma 4.4, we have
Z
|F (x) − F (y)| = x
f (t)dt −
a
Z
y
a
f (t)dt =
Z
y
x
f (t)dt ≤ M |x − y|.
Thus, F is Lipschitz and therefore continuous.
Now suppose that f is continuous at c. For any given ε > 0, there is δ > 0 such
that |f (t) − f (c)| < ε if t ∈ (c − δ, c + δ). Then for |h| < δ, we compute and have
F (c+h)−F (c)
−
f
(c)
h
Z
1 c+h
1 Z c+h
= f (t)dt −
f (c)dt
h c
h c Z
1 c+h
= (f (t) − f (c))dt ≤ ε.
h c
Thus,
F (c + h) − F (c)
= f (c).
h→0
h
F 0 (c) = lim
This completes our proof of the fundamental theorem.
Exercises:
1. Show that if a function f : [a, b] → IR is Lipschitz with constant K, then for
any partition P of [a, b], we have U (f, P ) − L(f, P ) ≤ Kmesh(P ).
2. Show that if f is Riemann integrable on [a, b], then so is |f |. Is the converse
also true?
3. Prove lemma 2.1.
4. Prove lemma 4.3.
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