The Riemann Integral

50 1. The Riemann Integral
To give Riemann’s definition, we define a tagged partition (P,C) of a compact
interval [a,b] to be a partition
of the interval together with a set
P = {I 1 ,I 2 ,...,I n }
C = {c 1 ,c 2 ,...,c n }
of points such that c k ∈ I k for k = 1,...,n. (Think of c k as a “tag” attached to
I k .)
If f : [a,b] → R, then we define the Riemann sum of f with respect to the
tagged partition (P,C) by
n∑
S(f;P,C) = f(c k )|I k |.
That is, instead of using the supremum or infimum of f on the kth interval in
the sum, we evaluate f at an arbitrary point in the interval. Roughly speaking, a
function is Riemann integrable if its Riemann sums approach the same value as the
partition is refined, independently of how we choose the points c k ∈ I k .
As a measure of the refinement of a partition P = {I 1 ,I 2 ,...,I n }, we define
the mesh (or norm) of P to be the maximum length of its intervals,
k=1
mesh(P) = max
1≤k≤n |I k| = max
1≤k≤n |x k −x k−1 |.
Definition 1.85. A bounded function f : [a,b] → R is Riemann integrable on [a,b]
if there exists a number R ∈ R with the following property: For every ǫ > 0 there
is a δ > 0 such that
|S(f;P,C)−R| < ǫ
for every tagged partition (P,C) of [a,b] with mesh(P) < δ. In that case, R = ∫ b
a f
is the Riemann integral of f on [a,b].
Note that
L(f;P) ≤ S(f;P,C) ≤ U(f;P),
so the Riemann sums are “squeezed” between the upper and lower sums. The
following theorem shows that the Darboux and Riemann definitions lead to the
same notion of the integral, so it’s a matter of convenience which definition we
adopt as our starting point.
Theorem 1.86. A function is Riemann integrable (in the sense of Definition 1.85)
if and only if it is Darboux integrable (in the sense of Definition 1.3). In that case,
the Riemann and Darboux integrals of the function are equal.
Proof. First, suppose that f : [a,b] → R is Riemann integrable with integral R.
Then f must be bounded; otherwise f would be unbounded in some interval I k
of every partition P, and we could make its Riemann sums with respect to P
arbitrarily large by choosing a suitable point c k ∈ I k , contradicting the definition
of R.
Let ǫ > 0. There is a partition P = {I 1 ,I 2 ,...,I n } of [a,b] such that
|S(f;P,C)−R| < ǫ 2