IA Analysis I - The Riemann Integral

7The Riemann Integral

IA Analysis I

7.1 Riemann Integral

Definition (Dissections). Let [

a, b

] be a closed interval. A dissection of [

a, b

] is

a sequence a = x

< x

< ··· < x

= b.

Definition (Upper and lower sums). Given a dissection

, the upper sum and

lower sum are defined by the formulae

(f) =

i=1

− x

i−1

) sup

x∈[x

i−1

]

f(x)

(f) =

i=1

− x

i−1

) inf

x∈[x

i−1

]

f(x)

Sometimes we use the shorthand

= sup

x∈[x

i−1

]

f(x), m

= inf

x∈[x

i−1

−x

]

f(x).

The upper sum is the total area of the red rectangles, while the lower sum is

the total area of the black rectangles:

a x

···

i+1

···

Definition (Refining dissections). If

and

are dissections of [

a, b

], we say

that D

refines D

if every point of D

is a point of D

Lemma. If D

refines D

, then

f ≤ U

f and L

≥ L

Using the picture above, this is because if we cut up the dissections into

smaller pieces, the red rectangles can only get chopped into shorter pieces and

the black rectangles can only get chopped into taller pieces.

Proof.

Let

< x

< ··· < x

. Let

be obtained from

by the

addition of one point z. If z ∈ (x

i−1

, x

), then

f −U

f =

(z − x

i−1

) sup

x∈[x

i−1

,z]

f(x)

− z) sup

x∈[z,x

]

f(x)

− (x

− x

i−1

But

sup

x∈[x

i−1

,z]

(

) and

sup

x∈[z,x

]

(

) are both at most

. So this is at

most M

(z − x

i−1

+ x

− z − (x

− x

i−1

)) = 0. So

f ≤ U

By induction, the result is true whenever D

refines D

A very similar argument shows that L

≥ L

Definition (Least common refinement). If

and

be dissections of [

a, b

Then the least common refinement of

and

is the dissection made out of

the points of D

and D

Corollary. Let D

and D

be two dissections of [a, b]. Then

f ≥ L

Proof.

Let

be the least common refinement (or indeed any common refinement).

Then by lemma above (and by definition),

f ≥ U

f ≥ L

Finally, we can define the integral.

Definition (Upper, lower, and Riemann integral). The upper integral is

f(x) dx = inf

The lower integral is

f(x) dx = sup

If these are equal, then we call their common value the Riemann integral of

and is denoted

f(x) dx.

If this exists, we say f is Riemann integrable.

We will later prove the fundamental theorem of calculus, which says that

integration is the reverse of differentiation. But why don’t we simply define

integration as anti-differentiation, and prove that it is the area of the curve?

There are things that we cannot find (a closed form of) the anti-derivative of,

−x

. In these cases, we wouldn’t want to say the integral doesn’t exist — it

surely does according to this definition!

There is an immediate necessary condition for Riemann integrability — bound-

edness. If

is unbounded above in [

a, b

], then for any dissection

, there must

be some

such that

is unbounded on [

i−1

, x

]. So

∞

. So

∞

Similarly, if

is unbounded below, then

−∞

. So unbounded functions

are not Riemann integrable.

Example. Let

(

) =

on [

a, b

]. Intuitively, we know that the integral is

(

− a

)

2, and we will show this using the definition above. Let

< ··· < x

be a dissection. Then

f =

i=1

− x

i−1

We know that the integral is

−a

. So we put each term of the sum into the

form

−x

i−1

plus some error terms.

i=1



−

i−1

− x

i−1



i=1

− x

i−1

+ (x

− x

i−1

)

− a

) +

i=1

− x

i−1

)

Definition (Mesh). The mesh of a dissection D is max

i+1

− x

Then if the mesh is < δ, then

i=1

− x

i−1

)

≤

i=1

− x

i−1

) =

(b − a).

So by making δ small enough, we can show that for any ε > 0,

x dx <

− a

) + ε.

Similarly,

x dx >

− a

) − ε.

x dx =

− a

Example. Define f : [0, 1] → R by

f(x) =

(

1 x ∈ Q

0 x 6∈ Q

Let

< x

< ··· < x

be a dissection. Then for every

, we have

= 0 (since

there is an irrational in every interval), and

= 1 (since there is a rational in

every interval). So

f =

i=1

− x

i−1

) =

i=1

− x

i−1

) = 1.

Similarly, L

f = 0. Since D was arbitrary, we have

f(x) dx = 1,

f(x) dx = 0.

So f is not Riemann integrable.

Of course, this function is not interesting at all. The whole point of its

existence is to show undergraduates that there are some functions that are not

integrable!

Note that it is important to say that the function is not Riemann integrable.

There are other notions for integration in which this function is integrable. For

example, this function is Lebesgue-integrable.

Using the definition to show integrability is often cumbersome. Most of

the time, we use the Riemann’s integrability criterion, which follows rather

immediately from the definition, but is much nicer to work with.

Proposition (Riemann’s integrability criterion). This is sometimes known as

Cauchy’s integrability criterion.

Let

: [

a, b

]

→ R

. Then

is Riemann integrable if and only if for every

ε > 0, there exists a dissection D such that

− L

< ε.

Proof.

(

⇒

) Suppose that

is integrable. Then (by definition of Riemann

integrability), there exist D

and D

such that

f(x) dx +

and

f(x) dx −

Let D be a common refinement of D

and D

. Then

f −L

f ≤ U

f −L

f < ε.

(⇐) Conversely, if there exists D such that

f −L

f < ε,

then

inf U

f −sup L

f < ε,

which is, by definition, that

f(x) dx −

f(x) dx < ε.

Since ε > 0 is arbitrary, this gives us that

f(x) dx =

f(x) dx.

So f is integrable.

The next big result we want to prove is that integration is linear, ie

(λf(x) + µg(x)) dx = λ

f(x) dx + µ

g(x) dx.

We do this step by step:

Proposition. Let

: [

a, b

]

→ R

be integrable, and

λ ≥

0. Then

λf

is integrable,

and

λf(x) dx = λ

f(x) dx.

Proof. Let D be a dissection of [a, b]. Since

sup

x∈[x

i−1

]

λf(x) = λ sup

x∈[x

i−1

]

f(x),

and similarly for inf, we have

(λf) = λU

(λf) = λL

So if we choose

such that

f − L

f < ε/λ

, then

(

λf

)

−L

(

λf

)

< ε

. So

the result follows from Riemann’s integrability criterion.

Proposition. Let f : [a, b] → R be integrable. Then −f is integrable, and

−f(x) dx = −

f(x) dx.

Proof. Let D be a dissection. Then

sup

x∈[x

i−1

]

−f(x) = − inf

x∈[x

i−1

]

f(x)

inf

x∈[x

i−1

]

−f(x) = − sup

x∈[x

i−1

]

f(x).

Therefore

(−f) =

i=1

− x

i−1

)(−m

) = −L

(f).

Similarly,

(−f) = −U

(−f) − L

(−f) = U

f −L

Hence if

is integrable, then

−f

is integrable by the Riemann integrability

criterion.

Proposition. Let

f, g

: [

a, b

]

→ R

be integrable. Then

is integrable, and

(f(x) + g(x)) dx =

f(x) dx +

g(x) dx.

Proof. Let D be a dissection. Then

(f + g) =

i=1

− x

i−1

) sup

x∈[x

i−1

]

(f(x) + g(x))

≤

i=1

− x

i−1

)

sup

u∈[x

i−1

]

f(u) + sup

v∈[x

i−1

]

g(v)

= U

f + U

Therefore,

(f(x) + g(x)) dx ≤

f(x) dx +

g(x) dx =

f(x) dx +

g(x) dx.

Similarly,

(f(x) + g(x)) dx ≥

f(x) dx +

g(x) dx.

So the upper and lower integrals are equal, and the result follows.

So we now have that

(λf(x) + µg(x)) dx = λ

f(x) dx + µ

g(x) dx.

We will prove more “obvious” results.

Proposition. Let

f, g

: [

a, b

]

→ R

be integrable, and suppose that

(

)

≤ g

(

)

for every x. Then

f(x) dx ≤

g(x) dx.

Proof. Follows immediately from the definition.

Proposition. Let f : [a, b] → R be integrable. Then |f| is integrable.

Proof. Note that we can write

sup

x∈[x

i−1

]

f(x) − inf

x∈[x

i−1

]

f(x) = sup

u,v∈[x

i−1

]

|f(u) − f(v)|.

Similarly,

sup

x∈[x

i−1

]

|f(x)| − inf

x∈[x

i−1

]

|f(x)| = sup

u,v∈[x

i−1

]

||f(u)| − |f(v)||.

For any pair of real numbers,

x, y

, we have that

||x| − |y|| ≤ |x − y|

by the

triangle inequality. Then for any interval u, v ∈ [x

i−1

, x

], we have

||f(u)| − |f(v)|| ≤ |f(u) − f(v)|.

Hence we have

sup

x∈[x

i−1

]

|f(x)| − inf

x∈[x

i−1

]

|f(x)| ≤ sup

x∈[x

i−1

]

f(x) − inf

x∈[x

i−1

]

f(x).

So for any dissection D, we have

(|f|) − L

(|f|) ≤ U

(f) − L

(f).

So the result follows from Riemann’s integrability criterion.

Combining these two propositions, we get that if

|f(x) − g(x)| ≤ C,

for every x ∈ [a, b], then



f(x) dx −

g(x) dx



≤ C(b − a).

Proposition (Additivity property). Let

: [

a, c

]

→ R

be integrable, and let

b ∈

(

a, c

). Then the restrictions of

to [

a, b

] and [

b, c

] are Riemann integrable,

and

f(x) dx +

f(x) dx =

f(x) dx

Similarly, if

is integrable on [

a, b

] and [

b, c

], then it is integrable on [

a, c

] and

the above equation also holds.

Proof.

Let

ε >

0, and let

< x

< ··· < x

be a dissection of

[a, c] such that

(f) ≤

f(x) dx + ε,

and

(f) ≥

f(x) dx − ε.

Let

be the dissection made of

plus the point

. Let

be the dissection of

[

a, b

] made of points of

from

, and

be the dissection of [

b, c

] made of

points of D

from b to c. Then

(f) + U

(f) = U

(f) ≤ U

(f),

and

(f) + L

(f) = L

(f) ≥ L

(f).

Since

(

)

− L

(

)

, and both

(

)

− L

(

) and

(

)

− L

(

)

are non-negative, we have

(

)

− L

(

) and

(

)

− L

(

) are less than

. Since

is arbitrary, it follows that the restrictions of

to [

a, b

] and [

b, c

] are

both Riemann integrable. Furthermore,

f(x) dx +

f(x) dx ≤ U

(f) + U

(f) = U

(f) ≤ U

(f)

≤

f(x) dx + ε.

Similarly,

f(x) dx +

f(x) dx ≥ L

(f) + L

(f) = L

(f) ≥ L

(f)

≥

f(x) dx − ε.

Since ε is arbitrary, it follows that

f(x) dx +

f(x) dx =

f(x) dx.

The other direction is left as an (easy) exercise.

Proposition. Let f, g : [a, b] → R be integrable. Then fg is integrable.

Proof.

Let

be such that

(

)

|, |g

(

)

| ≤ C

for every

x ∈

[

a, b

]. Write

and

for the

sup

and

inf

in [

i−1

, x

]. Now let

be a dissection, and for each

i, let u

and v

be two points in [x

i−1

, x

We will pretend that

and

are the minimum and maximum when we

write the proof, but we cannot assert that they are, since

need not have

maxima and minima. We will then note that since our results hold for arbitrary

and v

, it must hold when f g is at its supremum and infimum.

We find what we pretend is the difference between the upper and lower sum:



i=1



− x

i−1

)(f(v

)g(v

) − f(u

)g(u

)





i=1

− x

i−1

)



f(v

)(g(v

) − g(u

)) + (f(v

) − f(u

))g(u

)





≤

i=1



C(L

− `

) + (M

− m



= C(U

g −L

g + U

f −L

f).

Since u

and v

are arbitrary, it follows that

(fg) − L

(fg) ≤ C(U

f −L

f + U

g −L

g).

Since

is fixed, and we can get

f − L

and

g − L

arbitrary small

(since

and

are integrable), we can get

(

)

−L

(

) arbitrarily small. So

the result follows.

Theorem. Every continuous function

on a closed bounded interval [

a, b

] is

Riemann integrable.

Proof. wlog assume [a, b] = [0, 1].

Suppose the contrary. Let

be non-integrable. This means that there exists

some

such that for every dissection

− L

> ε

. In particular, for every

n, let D

be the dissection 0,

, ··· ,

Since

− L

> ε

, there exists some interval



k+1



in which

sup f −

inf f > ε

. Suppose the supremum and infimum are attained at

and

respectively. Then we have |x

− y

| <

and f(x

) − f(y

) > ε.

By Bolzano Weierstrass, (

) has a convergent subsequence, say (

). Say

→ x

. Since

− y

| <

→

0, we must have

→ x

. By continuity, we

must have

(

)

→ f

(

) and

(

)

→ f

(

), but

(

) and

(

) are always

apart by ε. Contradiction.

With this result, we know that a lot of things are integrable, e.g. e

−x

To prove this, we secretly used the property of uniform continuity:

Definition (Uniform continuity*). Let

A ⊆ R

and let

A → R

. Then

uniformly continuous if

(∀ε)(∃δ > 0)(∀x)(∀y) |x − y| < δ ⇒ |f(x) −f(y)| ≤ ε.

This is different from regular continuity. Regular continuity says that at any

point

, we can find a

that works for this point. Uniform continuity says that

we can find a δ that works for any x.

It is easy to show that a uniformly continuous function is integrable, since by

uniformly continuity, as long as the mesh of a dissection is sufficiently small, the

difference between the upper sum and the lower sum can be arbitrarily small by

uniform continuity. Thus to prove the above theorem, we just have to show that

continuous functions on a closed bounded interval are uniformly continuous.

Theorem (non-examinable). Let

a < b

and let

: [

a, b

]

→ R

be continuous.

Then f is uniformly continuous.

Proof. Suppose that f is not uniformly continuous. Then

(∃ε)(∀δ > 0)(∃x)(∃y) |x − y| < δ and |f(x) − f(y)| ≥ ε.

Therefore, we can find sequences (x

), (y

) such that for every n, we have

− y

| ≤

and |f(x

) − f(y

)| ≥ ε.

Then by Bolzano-Weierstrass theorem, we can find a subsequence (

) converg-

ing to some

. Since

−y

| ≤

→ x

as well. But

(

)

−f

(

)

| ≥ ε

for every

. So

(

) and

(

) cannot both converge to the same limit. So

is not continuous at x.

This proof is very similar to the proof that continuous functions are integrable.

In fact, the proof that continuous functions are integrable is just a fuse of this

proof and the (simple) proof that uniformly continuously functions are integrable.

Theorem. Let f : [a, b] → R be monotone. Then f is Riemann integrable.

Note that monotone functions need not be “nice”. It can even have in-

finitely many discontinuities. For example, if

: [0

→ R

maps

to the

1/(first non-zero digit in the binary expansion of x), with f (0) = 0.

Proof. let ε > 0. Let D be a dissection of mesh less than

f(b)−f (a)

. Then

f −L

f =

i=1

− x

i−1

)(f(x

) − f(x

i−1

))

≤

f(b) − f(a)

i=1

(f(x

) − f(x

i−1

))

= ε.

Pictorially, we see that the difference between the upper and lower sums is

total the area of the red rectangles.

To calculate the total area, we can stack the red areas together to get something

of width

f(b)−f (a)

and height f(b) − f(a). So the total area is just ε.

Lemma. Let

a < b

and let

be a bounded function from [

a, b

]

→ R

that is

continuous on (a, b). Then f is integrable.

An example where this would apply is

sin

. It gets nasty near

= 0, but

its “nastiness” is confined to

= 0 only. So as long as its nastiness is sufficiently

contained, it would still be integrable.

The idea of the proof is to integrate from a point

very near

up to a

point

n−1

very close to

. Since

is bounded, the regions [

a, x

] and [

n−1

, b

]

are small enough to not cause trouble.

Proof.

Let

ε >

0. Suppose that

(

)

| ≤ C

for every

x ∈

[

a, b

]. Let

and

pick

such that

− x

. Also choose

between

and

such that

b − z <

Then

is continuous [

, z

]. Therefore it is integrable on [

, z

]. So we can

find a dissection D

with points x

< x

< ··· < x

n−1

= z such that

f −L

f <

Let D be the dissection a = x

< x

< ··· < x

= b. Then

f −L

f <

· 2C +

· 2C = ε.

So done by Riemann integrability criterion.

Example.

– f (x) =

(

sin

x 6= 0

0 x = 0

defined on [−1, 1] is integrable.

– g(x) =

(

x x ≤ 1

+ 1 x > 1

defined on [0, 1] is integrable.

Corollary. Every piecewise continuous and bounded function on [

a, b

] is inte-

grable.

Proof.

Partition [

a, b

] into intervals

, ··· , I

, on each of which

is (bounded

and) continuous. Hence for every

with end points

j−1

is integrable on

[

j−1

, x

] (which may not equal

, e.g.

could be [

j−1

, x

)). But then by the

additivity property of integration, we get that f is integrable on [a, b]

We defined Riemann integration in a very general way — we allowed arbitrary

dissections, and took the extrema over all possible dissection. Is it possible to

just consider some particular nice dissections instead? Perhaps unsurprisingly,

yes! It’s just that we opt to define it the general way so that we can easily talk

about things like least common refinements.

Lemma. Let

: [

a, b

]

→ R

be Riemann integrable, and for each

, let

the dissection

< x

< ··· < x

, where

i(b−a)

for each

Then

f →

f(x) dx

and

f →

f(x) dx.

Proof.

Let

ε >

0. We need to find an

. The only thing we know is that

Riemann integrable, so we use it:

Since

is integrable, there is a dissection

, say

< u

< ··· < u

, such

that

f −

f(x) dx <

We also know that f is bounded. Let C be such that |f(x)| ≤ C.

For any n, let D

be the least common refinement of D

and D. Then

f ≤ U

Also, the sums

and

are the same, except that at most

of the

subintervals [x

i−1

, x

] are subdivided in D

For each interval that gets chopped up, the upper sum decreases by at most

b−a

· 2C. Therefore

f −U

f ≤

b − a

2C ·m.

Pick n such that 2Cm(b − a)/n <

. Then

f −U

f <

f −

f(x) dx < ε.

This is true whenever

n >

4C(b−a)m

. Since we also have

f ≥

(

) d

therefore

f →

f(x) dx.

The proof for lower sums is similar.

For convenience, we define the following:

Notation. If b > a, we define

f(x) dx = −

f(x) dx.

We now prove that the fundamental theorem of calculus, which says that

integration is the reverse of differentiation.

Theorem (Fundamental theorem of calculus, part 1). Let

: [

a, b

]

→ R

continuous, and for x ∈ [a, b], define

F (x) =

f(t) dt.

Then F is differentiable and F

(x) = f(x) for every x.

Proof.

F (x + h) − F (x)

x+h

f(t) dt

Let

ε >

0. Since

is continuous, at

, then there exists

such that

|y −x| < δ

implies |f(y) − f(x)| < ε.

If |h| < δ, then



x+h

f(t) dt − f(x)



x+h

(f(t) − f(x)) dt



≤

|h|



x+h

|f(t) − f(x)| dt



≤

ε|h|

|h|

= ε.

Corollary. If f is continuously differentiable on [a, b], then

(t) dt = f(b) − f(a).

Proof. Let

g(x) =

(t) dt.

Then

(x) = f

(x) =

(f(x) − f(a)).

Since

(

)

− f

(

) = 0,

(

)

− f

(

) must be a constant function by the mean

value theorem. We also know that

g(a) = 0 = f(a) − f(a)

So we must have

(

) =

(

)

−f

(

) for every

, and in particular, for

Theorem (Fundamental theorem of calculus, part 2). Let

: [

a, b

]

→ R

be a

differentiable function, and suppose that f

is integrable. Then

(t) dt = f(b) − f(a).

Note that this is a stronger result than the corollary above, since it does not

require that f

is continuous.

Proof.

Let

be a dissection

< x

< ··· < x

. We want to make use of this

dissection. So write

f(b) − f(a) =

i=1

(f(x

) − f(x

i−1

)).

For each

, there exists

∈

(

i−1

, x

) such that

(

)

− f

(

i−1j

) = (

−

i−1

) by the mean value theorem. So

f(b) − f(a) =

i=1

− x

i−1

We know that

(

) is somewhere between

sup

x∈[x

i−1

]

(

) and

inf

x∈[x

i−1

]

(

)

by definition. Therefore

≤ f(b) − f(a) ≤ U

Since

is integrable and

was arbitrary,

and

can both get arbitrarily

close to

(t) dt. So

f(b) − f(a) =

(t) dt.

Note that the condition that

is integrable is essential. It is possible to find

a differentiable function whose derivative is not integrable! You will be asked to

find it in the example sheet.

Using the fundamental theorem of calculus, we can easily prove integration

by parts:

Theorem (Integration by parts). Let

f, g

: [

a, b

]

→ R

be integrable such that

everything below exists. Then

f(x)g

(x) dx = f(b)g(b) − f (a)g(a) −

(x)g(x) dx.

Proof. By the fundamental theorem of calculus,

(f(x)g

(x) + f

(x)g(x)) dx =

(fg)

(x) dx = f(b)g(b) − f (a)g(a).

The result follows after rearrangement.

Recall that when we first had Taylor’s theorem, we said it had the Lagrange

form of the remainder. There are many other forms of the remainder term. Here

we will look at the integral form:

Theorem (Taylor’s theorem with the integral form of the remainder). Let

n + 1 times differentiable on [a, b] with with f

(n+1)

continuous. Then

f(b) = f(a) + (b − a)f

(a) +

(b − a)

(2)

(a) + ···

(b − a)

(n)

(a) +

(b − t)

(n+1)

(t) dt.

Proof. Induction on n.

When n = 0, the theorem says

f(b) − f(a) =

(t) dt.

which is true by the fundamental theorem of calculus.

Now observe that

(b − t)

(n+1)

(t) dt =



−(b − t)

n+1

(n + 1)!

(n+1)

(t)



(b − t)

n+1

(n + 1)!

(n+1)

(t) dt

(b − a)

n+1

(n + 1)!

(n+1)

(a) +

(b − t)

n+1

(n + 1)!

(n+2)

(t) dt.

So the result follows by induction.

Note that the form of the integral remainder is rather weird and unexpected.

How could we have come up with it? We might start with the fundamental

theorem of algebra and integrate by parts. The first attempt would be to

integrate 1 to t and differentiate f

(t) to f

(2)

(t). So we have

f(b) = f(a) +

(t) dt

= f(a) + [tf

(t)]

−

(2)

(t) dt

= f(a) + bf

(b) − af

(a) −

(2)

(t) dt

We want something in the form (

b − a

)

(

), so we take that out and see what

we are left with.

= f(a) + (b − a)f

(a) + b(f

(b) − f

(a)) −

(2)

(t) dt

Then we note that f

(b) − f

(a) =

(2)

(t) dt. So we have

= f(a) + (b − a)f

(a) +

(b − t)f

(2)

(t) dt.

Then we can see that the right thing to integrate is (

b −t

) and continue to obtain

the result.

Theorem (Integration by substitution). Let

: [

a, b

]

→ R

be continuous. Let

: [

u, v

]

→ R

be continuously differentiable, and suppose that

(

) =

a, g

(

) =

and f is defined everywhere on g([u, v]) (and still continuous). Then

f(x) dx =

f(g(t))g

(t) dt.

Proof.

By the fundamental theorem of calculus,

has an anti-derivative

defined on g([u, v]). Then

f(g(t))g

(t) dt =

(g(t))g

(t) dt

(F ◦ g)

(t) dt

= F ◦ g(v) − F ◦ g(u)

= F (b) − F (a)

f(x) dx.

We can think of “integration by parts” as what you get by integrating the

product rule, and “integration by substitution” as what you get by integrating

the chain rule.