3Uniform continuity and integration
IB Analysis II
3.2 Applications to Riemann integrability
We can apply the idea of uniform continuity to Riemann integration.
We first quickly recap and summarize things we know about Riemann integrals
IA Analysis I. Let f : [a, b] → R be a bounded function, say m ≤ f(x) ≤ M for
all x ∈ [a, b]. Consider a partition of [a, b]
P = {a
0
, a
1
, ··· , a
n
}.
i.e. a = a
0
< a
1
< a
2
< ··· < a
n
= b. The upper sum is defined as
U(P, f) =
n−1
X
j=0
(a
j+1
− a
j
) sup
[a
j
,a
j+1
]
f.
Similarly, the lower sum is defined as
L(P, f) =
n−1
X
j=0
(a
j+1
− a
j
) inf
[a
j
,a
j+1
]
f.
It is clear from definition that
m(b −a) ≤ L(P, f) ≤ U(P, f) ≤ M(b − a).
Also, if a partition
P
∗
is a refinement of a partition
P
, i.e. it contains all the
points of P and possibly more, then
L(P, f) ≤ L(P
∗
, f) ≤ U(P
∗
, f) ≤ U(P, f).
It thus follows that if P
1
and P
2
are arbitrary partitions, then
L(P
1
, f) ≤ U(P
2
, f),
which we can show by finding a partition that is simultaneously a refinement of
P
1
and P
2
. The importance of this result is that
sup
P
L(P, f) ≤ inf
P
U(P, f),
where we take extrema over all partitions
P
. We define these to be the upper
and lower integrals
I
∗
(f) = inf
P
U(P, f), I
∗
(f) = sup
P
L(P, f).
So we know that
m(b −a) ≤ I
∗
(f) ≤ I
∗
(f) ≤ M(b − a).
Now given any
ε >
0, by definition of the infimum, there is a partition
P
1
such
that
U(P
1
, f) < I
∗
(f) +
ε
2
.
Similarly, there is a partition P
2
such that
L(P
2
, f) > I
∗
(f) −
ε
2
.
So if we let P = P
1
∪ P
2
, then P is a refinement of both P
1
and P
2
. So
U(P, f) < I
∗
(f) +
ε
2
and
L(P, f) > I
∗
(f) −
ε
2
.
Combining these, we know that
0 ≤ I
∗
(f) −I
∗
(f) < U(P, f) − L(P, f) < I
∗
(f) −I
∗
(f) + ε.
We now define what it means to be Riemann integrable.
Definition (Riemann integrability). A bounded function
f
: [
a, b
]
→ R
is
Riemann integrable on [a, b] if I
∗
(f) = I
∗
(f). We write
Z
b
a
f(x) dx = I
∗
(f) = I
∗
(f).
Then the Riemann criterion says
Theorem (Riemann criterion for integrability). A bounded function
f
: [
a, b
]
→
R
is Riemann integrable if and only if for every
ε
, there is a partition
P
such
that
U(P, f) −L(P, f) < ε.
That’s the end of our recap. Now we have a new theorem.
Theorem. If
f
: [
a, b
]
→
[
A, B
] is integrable and
g
: [
A, B
]
→ R
is continuous,
then g ◦f : [a, b] → R is integrable.
Proof.
Let
ε >
0. Since
g
is continuous,
g
is uniformly continuous. So we can find
δ
=
δ
(
ε
)
>
0 such that for any
x, y ∈
[
A, B
], if
|x −y| < δ
then
|g
(
x
)
−g
(
y
)
| < ε
.
Since
f
is integrable, for arbitrary
ε
′
, we can find a partition
P
=
{a
=
a
0
<
a
1
< ··· < a
n
= b} such that
U(P, f) − L(P, f) =
n−1
X
j=0
(a
j+1
− a
j
)
sup
I
j
f − inf
I
j
f
!
< ε
′
. (∗)
Our objective is to make
U
(
P, g ◦ f
)
− L
(
P, g ◦ f
) small. By uniform continuity
of
g
, if
sup
I
j
f − inf
I
j
f
is less than
δ
, then
sup
I
j
g ◦f − inf
I
j
g ◦f
will be less
than ε. We like these sorts of intervals. So we let
J =
(
j : sup
I
j
f − inf
I
j
f < δ
)
,
We now show properly that these intervals are indeed “nice”. For any
j ∈ J
, for
all x, y ∈ I
j
, we must have
|f(x) − f(y)| ≤ sup
z
1
,z
2
∈I
j
(f(z
1
) − f(z
2
)) = sup
I
j
f − inf
I
j
f < δ.
Hence, for each j ∈ J and all x, y ∈ I
j
, we know that
|g ◦f(x) − g ◦f(y)| < ε.
Hence, we must have
sup
I
j
g ◦f(x) − g ◦f(y)
≤ ε.
So
sup
I
j
g ◦f − inf
I
j
g ◦f ≤ ε.
Hence we know that
U(P, g ◦ f) − L(P, g ◦ f) =
n
X
j=0
(a
j+1
− a
j
)
sup
I
j
g ◦f − inf
I
j
g ◦f
!
=
X
j∈J
(a
j+1
− a
j
)
sup
I
j
g ◦f − inf
I
j
g ◦f
!
+
X
j∈J
(a
j+1
− a
j
)
sup
I
j
g ◦f − inf
I
j
g ◦f
!
.
≤ ε(b − a) + 2 sup
[A,B]
|g|
X
j∈J
(a
j+1
− a
j
).
Hence, it suffices here to make
P
j∈J
(
a
j+1
−a
j
) small. From (
∗
), we know that we
must have
X
j∈J
(a
j+1
− a
j
) <
ε
′
δ
,
or else U(P, f) − L(P, f) > ε
′
. So we can bound
U(P, g ◦ f) − L(P, g ◦ f) ≤ ε(b − a) + 2 sup
[A,B]
|g|
ε
′
δ
.
So if we are given an
ε
at the beginning, we can get a
δ
by uniform continuity.
Afterwards, we pick
ε
′
such that
ε
′
=
εδ
. Then we have shown that given any
ε
,
there exists a partition such that
U(P, g ◦ f) − L(P, g ◦ f) <
(b − a) + 2 sup
[A,B]
|g|
!
ε.
Then the claim follows from the Riemann criterion.
As an immediate consequence, we know that any continuous function is
integrable, since we can just let
f
be the identity function, which we can easily
show to be integrable.
Corollary. A continuous function g : [a, b] → R is integrable.
Theorem. Let
f
n
: [
a, b
]
→ R
be bounded and integrable for all
n
. Then if
(
f
n
) converges uniformly to a function
f
: [
a, b
]
→ R
, then
f
is bounded and
integrable.
Proof. Let
c
n
= sup
[a,b]
|f
n
− f |.
Then uniform convergence says that
c
n
→
0. By definition, for each
x
, we have
f
n
(x) − c
n
≤ f(x) ≤ f
n
(x) + c
n
.
Since
f
n
is bounded, this implies that
f
is bounded by
sup |f
n
|
+
c
n
. Also, for
any x, y ∈ [a, b], we know
f(x) − f(y) ≤ (f
n
(x) − f
n
(y)) + 2c
n
.
Hence for any partition P ,
U(P, f) − L(L, f) ≤ U(P, f
n
) − L(P, f
n
) + 2(b − a)c
n
.
So given
ε >
0, first choose
n
such that 2(
b −a
)
c
n
<
ε
2
. Then choose
P
such that
U(P, f
n
) − L(P, f
n
) <
ε
2
. Then for this partition, U (P, f ) − L(P, f ) < ε.
Most of the theory of Riemann integration extends to vector-valued or
complex-valued functions (of a single real variable).
Definition (Riemann integrability of vector-valued function). Let f : [
a, b
]
→ R
n
be a vector-valued function. Write
f(x) = (f
1
(x), f
2
(x), ··· , f
n
(x))
for all
x ∈
[
a, b
]. Then f is Riemann integrable iff
f
j
: [
a, b
]
→ R
is integrable for
all j. The integral is defined as
Z
b
a
f(x) dx =
Z
b
a
f
1
(x) dx, ··· ,
Z
b
a
f
n
(x) dx
!
∈ R
n
.
It is easy to see that most basic properties of integrals of real functions extend
to the vector-valued case. A less obvious fact is the following.
Proposition. If f : [
a, b
]
→ R
n
is integrable, then the function
∥
f
∥
: [
a, b
]
→ R
defined by
∥f∥(x) = ∥f (x)∥ =
v
u
u
t
n
X
j=1
f
2
j
(x).
is integrable, and
Z
b
a
f(x) dx
≤
Z
b
a
∥f∥(x) dx.
This is a rather random result, but we include it here because it will be
helpful at some point in time.
Proof.
The integrability of
∥
f
∥
is clear since squaring and taking square roots
are continuous, and a finite sum of integrable functions is integrable. To show
the inequality, we let
v = (v
1
, ··· , v
n
) =
Z
b
a
f(x) dx.
Then by definition,
v
j
=
Z
b
a
f
j
(x) dx.
If v = 0, then we are done. Otherwise, we have
∥v∥
2
=
n
X
j=1
v
2
j
=
n
X
j=1
v
j
Z
b
a
f
j
(x) dx
=
Z
b
a
n
X
j=1
(v
j
f
j
(x)) dx
=
Z
b
a
v · f(x) dx
Using the Cauchy-Schwarz inequality, we get
≤
Z
b
a
∥v∥∥f ∥(x) dx
= ∥v∥
Z
b
a
∥f∥ dx.
Divide by ∥v ∥ and we are done.