comparison between Lebesgue and Riemann Integration

comparison between Lebesgue and Riemann Integration

The Riemann and

Lebesgue integral

are defined in different ways, with
the latter generally perceived as the more general. The aim of
this article is to clarify this claim by providing a number of,
hopefully simple and convincing, examples and

arguments

. We restrict
this article to a discussion of proper and improper integrals. For

extensions

and even more general

definitions

of the integral we refer
to

http://www.math.vanderbilt.edu/~schectex/ccc/gauge/

:

1

Proper Integrals

The Dirichlet Function

Our first example shows that

functions

exists that are

Lebesgue
integrable

but not

Riemann integrable

.

Consider the

characteristic function

of the

rational numbers

in

[

0

,

1

]

, i.e.,

1

ℚ

⁢

(

x

)

=

{

1

,

if

x

rational

,

0

,

elsewhere

.

This function, known as the

Dirichlet function

, is not Riemann
integrable. To see this, take an arbitrary

partition

of the

interval

[

0

,

1

]

. The

supremum

of

1

ℚ

on any interval
(with non-empty

interior

) is

1

, whereas its

infimum

is

0

. Hence,
the

upper Riemann sum

of

1

ℚ

is

1

while the
lower Riemann sum is

0

. Clearly, the upper and lower Riemann sums

converge

to 1 and 0, respectively, in the limit of the size of largest
interval in the partition going to zero. Obviously, these limits are
not the same. As a

bounded function

is Riemann integrable if and only
if the upper and lower Riemann sums converge to the same number, the
Riemann integral of

1

ℚ

cannot exist.

On the other hand,

1

ℚ

turns out to be
Lebesgue integrable, which we now show. Let us enumerate the rationals
in

[

0

,

1

]

as

{

q

0

,

q

1

,

…

}

. Now cover each

q

i

with an

open set

O

i

of size

ϵ

/

2

i

. Hence, the set

{

q

0

,

q

1

,

…

}

is

contained

in the set

⊂

G

ϵ

:=

∪

i

=

0

∞

O

i

. Now it is known that every

countable

union of
open sets forms a Lebesgue measurable set. Therefore,

G

ϵ

is
also a Lebesgue measurable set. Consequently, the Lebesgue integral
of this set’s characteristic function, i.e.,

1

G

ϵ

⁢

(

x

)

=

{

1

,

if

⁢

x

∈

O

,

0

,

elsewhere

,

exists.

In fact, the integral of

1

G

ϵ

is less than or equal to

ϵ

⁢

∑

i

=

0

∞

2

-

i

=

ϵ

as the total length of
the union of sets

O

i

is less than or equal to

ϵ

. (There
might be overlaps between the

O

i

.) Now taking the limit

ϵ

→

0

it follows that for all

ϵ

,

∫

1

ℚ

⁢

𝑑

x

≤

∫

1

G

ϵ

⁢

𝑑

x

≤

ϵ

Since this holds for all

ϵ

>

0

, the left hand side must be 0.

A ’Worse’ Kind of Dirichlet Function

The above is, arguably, somewhat simple. The only reason that the
Dirichlet function is Lebesgue, but not Riemann,

integrable

, is that
its spikes occur on the rationals, a set of numbers which is, in
comparison to the

irrational numbers

, a very small set. By modifying
the Dirichlet function on

a set of

measure zero

, that is, by
removing its spikes, it becomes the zero function, which is evidently
Riemann integrable. This reasoning might lead us to

conjecture

that
it is possible to turn any Lebesgue integrable function into a Riemann
integrable function by modifying it on a set of measure zero. This
conjecture, however, is false as the next example shows.

Interestingly, the function we seek can be obtained in a more or less
direct way from the previous example. First cover the rationals by
open sets

O

i

of length

y

i

, where the

sequence

of numbers of

{

y

i

}

i

=

0

,

…

is such that

y

i

>

0

for all

i

but such that

∑

i

y

i

=

1

/

2

. Observe that

G

y

:=

∪

i

O

i

is a

measurable
set

with

measure

(length) less than or equal to

1

/

2

. But this
implies that the

complement

G

y

c

of

G

y

, which contains only
irrational numbers, has length at least

1

/

2

.

Observe that the characteristic function

1

G

y

is
nowhere

continuous

on

G

y

c

, as the rationals lie

dense in

the
reals. Since, also,

G

y

c

has measure at least equal
to

1

/

2

it is impossible to remove these discontinuity points by a
mere modification on a countable number of measure zero sets.

Now there is a

theorem

by Lebesgue stating that a bounded function

f

is Riemann integrable if and only if

f

is continuous almost
everywhere. Apparently,

1

G

y

is

bounded

and

discontinuous

on a set with measure larger than

0

. Thus, we may
conclude that

1

G

y

is not Riemann integrable.

To prove that

1

G

y

is Lebesgue integrable follows
easily if we approach the subject of integration from a somewhat more
abstract point of view. This is the topic of the next

section

.

Exchanging Limits and Integrals

Let us first present one further example: the sequence of
characteristic functions of

∪

i

=

1

n

O

i

, where

O

i

,

i

=

1

⁢

…

⁢

n

and

n

<

∞

, are the open sets appearing in the
definition of the ’worse Dirichlet function’. Clearly, any such
function has a finite number of

discontinuities

, hence is Riemann
integrable. However, these functions converge point-wise to

1

G

y

, which is

not

Riemann integrable.
Apparently, sequences of Riemann integrable functions may converge to
non-Riemann integrable functions. Interestingly, the sequence of
integrals of

1

∪

i

=

1

n

O

i

, i.e. a sequence of
reals,

has

a limit as it is

increasing

and bounded by

1

/

2

.
This somewhat disturbing inconsistency is satisfactory resolved
by Lebesgue’s theory of integration.

As a matter of fact, the advantage of Lebesgue integration is perhaps
best appreciated by interpreting this example from a more abstract
(

functional analysis

) point of view. Stated a bit differently, we
might approach the subject not from the bottom up (looking at

individual

functions) but from the top down (looking at classes of
functions). In more detail, suppose we are allowed to apply an

operator

T

to any function that is an element of some

function
space

. It would be nice if this space is

closed under

taking
(point-wise) limits. In other words, besides being allowed to apply

T

to some sequence of functions

f

n

, we are also allowed to apply

T

to the function

f

obtained as the limit of

f

n

. It would be
even nicer if

lim

n

⁡

T

⁢

f

n

is the same as

T

⁢

lim

n

⁡

f

n

=

T

⁢

f

.
(This is, for

instance

, useful when it is simple to compute

T

⁢

f

n

for each

n

, but difficult to compute

T

⁢

f

, while the latter might be
what really

interests

us.)

In the present case, i.e, integration, we perceive the integral of a
function as a (continuous linear) operator. The class of Lebesgue
integrable functions has the desired abstract

properties

(simple
conditions to check whether the exchange of integral and limit is
allowed), whereas the class of Riemann integrable functions does not.

Applying this to the above example, viz. the integration of

1

G

y

, we use

Lebesgue Dominated Convergence Theorem

,
which states that when a sequence

{

f

n

}

of

Lebesgue
measurable

functions is bounded by a Lebesgue integrable function, the
function

f

obtained as the

pointwise

limit

f

n

is also Lebesgue
integrable, and

∫

lim

n

⁡

f

n

=

∫

f

=

lim

n

⁡

∫

f

n

. Since,
for all

n

,

1

∪

i

n

O

i

is bounded and Lebesgue
integrable,

1

G

y

is also Lebesgue integrable,
and reversing the (pointwise) limit and the integral is allowed.

1.1

Fubini’s Theorem

Admittedly the function

1

G

y

is rather artificial.
A really powerful example of the

consequences

of being allowed to
reverse integral and limit is provided by (the proof of)
Fubini’s theorem applied to the

rectangle

Q

=

[

a

,

b

]

×

[

c

,

d

]

.
Compare the following two theorems. See, for instance,

[

Apo69

]

or

[

Lan83

]

for proofs of
the first theorem, and

[

Jon00

]

for
the second theorem.

Theorem 1.1

.

Riemann Case. Assume

f

to be Riemann integrable on

Q

. Assume
also that the one-dimensional function

x

↦

f

⁢

(

x

,

y

)

is Riemann
integrable for almost all

y

∈

[

c

,

d

]

. Then the function

y

↦

∫

a

b

f

⁢

(

x

,

y

)

⁢

𝑑

x

is Riemann integrable and

∫

Q

f

⁢

(

x

,

y

)

⁢

𝑑

x

⁢

𝑑

y

=

∫

c

d

(

∫

a

b

f

⁢

(

x

,

y

)

⁢

𝑑

x

)

⁢

𝑑

y

. Note: both
conditions are satisfied if

f

is continuous on

Q

.

Theorem 1.2

.

Lebesgue Case. Assume that

f

is Lebesgue integrable on

Q

. Then
the function

x

↦

f

⁢

(

x

,

y

)

is Lebesgue integrable for almost all

y

on

[

c

,

d

]

. As a consequence, the function

y

↦

∫

a

b

f

⁢

(

x

,

y

)

⁢

𝑑

x

is Lebesgue integrable and

∫

Q

f

⁢

(

x

,

y

)

⁢

𝑑

x

⁢

𝑑

y

=

∫

c

d

(

∫

a

b

f

⁢

(

x

,

y

)

⁢

𝑑

x

)

⁢

𝑑

y

.

Observe that the second

assumption

in the Riemann case has
turned into a

consequence

in the Lebesgue case. The main
reason behind this

difference

is precisely that the class of Lebesgue
measurable functions is closed under taking limits (under a bounded
condition), whereas the class of Riemann integrable functions is not.

2

Improper Integrals

From the above the reader may conclude that whenever a function is
Riemann integrable, it is Lebesgue integrable. This is true as long we
only include

proper

integrals. If, on the other hand, we also
consider

improper

integrals the statement is no longer valid.
There exist functions whose improper Riemann integral exists, whereas
the Lebesgue integral does not. Concentrating on functions defined on
subsets of

ℝ

n

the situation is as shown by the following

Venn Diagram

:

Figure 1:

R

,

R

⁢

I

, and

L

, are the classes of Riemann, improper
Riemann, and Lebesgue integrable functions, respectively.

In the previous we already discussed the

inclusion

R

⊂

L

.

Let us now

integrate

the function

1

/

x

over

[

0

,

1

]

to show
that some functions exist

R

⁢

I

∪

L

but are not in

R

. As

1

/

x

is not bounded on

[

0

,

1

]

every upper Riemann sum is

infinite

. On
the other hand, both the improper Riemann integral and the Lebesgue
integral exist, and give the same result.

Secondly,

L

is not contained in

R

⁢

I

as follows from the fact that
the Dirichlet function is not

R

⁢

I

.

Finally,

R

⁢

I

is also not a subset of

L

. Define

f

:

[

0

,

∞

)

↦

[

-

1

,

1

]

as

1

on

[

0

,

1

)

,

-

1

/

2

on

[

1

,

2

)

,

1

/

3

on

[

2

,

3

)

,

-

1

/

4

on

[

3

,

4

)

, etc. The Riemann integral and the
Lebesgue integral of

f

over

[

0

,

n

)

are both equal to

∑

k

=

1

n

(

-

1

)

k

+

1

⁢

1

/

k

.

It is well known that this

alternating sum

converges, implying the

existence

of the improper
Riemann integral. However, since a function is Lebesgue integrable if
and only if its

absolute value

is also Lebesgue integrable,

f

is not
in

L

.

References

Apo69

T.M. Apostol.

Calculus

, volume 2.

John Wiley & Sons, 1969.

Jon00

F. Jones.

Lebesgue Integration on

Euclidean Spaces

.

Jones and Bartlett, revised edition edition, 2000.

Lan83

S. Lang.

Undergraduate

Analysis

.

Springer-Verlag, 1983.

Title

comparison between Lebesgue and Riemann Integration

Canonical

name

ComparisonBetweenLebesgueAndRiemannIntegration

Date of creation

2013-03-22 15:29:14

Last modified on

2013-03-22 15:29:14

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Example

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