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
, i.e.,
This function, known as the Dirichlet function, is not Riemann
integrable. To see this, take an arbitrary partition
of the interval
. The supremum
of on any interval
(with non-empty interior) is , whereas its infimum
is . Hence,
the upper Riemann sum of is while the
lower Riemann sum is . 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 cannot exist.
On the other hand, turns out to be
Lebesgue integrable, which we now show. Let us enumerate the rationals
in as . Now cover each with an
open set of size . Hence, the set
is contained in the set . Now it is known that every countable union of
open sets forms a Lebesgue measurable set. Therefore, is
also a Lebesgue measurable set. Consequently, the Lebesgue integral
of this set’s characteristic function, i.e.,
exists.
In fact, the integral of is less than or equal to as the total length of the union of sets is less than or equal to . (There might be overlaps between the .) Now taking the limit it follows that for all ,
Since this holds for all , 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 of length , where the sequence of numbers of
is such that for all but such that
. Observe that is a measurable
set
with measure
(length) less than or equal to . But this
implies that the complement
of , which contains only
irrational numbers, has length at least .
Observe that the characteristic function is
nowhere continuous on , as the rationals lie dense in the
reals. Since, also, has measure at least equal
to 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
is Riemann integrable if and only if is continuous almost
everywhere. Apparently, is bounded
and
discontinuous
on a set with measure larger than . Thus, we may
conclude that is not Riemann integrable.
To prove that 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 , where and , 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 , which is not Riemann integrable. Apparently, sequences of Riemann integrable functions may converge to non-Riemann integrable functions. Interestingly, the sequence of integrals of , i.e. a sequence of reals, has a limit as it is increasing and bounded by . 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 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
to some sequence of functions , we are also allowed to apply
to the function obtained as the limit of . It would be
even nicer if is the same as .
(This is, for instance, useful when it is simple to compute
for each , but difficult to compute , 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
, we use Lebesgue Dominated Convergence Theorem,
which states that when a sequence of Lebesgue
measurable functions is bounded by a Lebesgue integrable function, the
function obtained as the pointwise limit is also Lebesgue
integrable, and . Since,
for all , is bounded and Lebesgue
integrable, is also Lebesgue integrable,
and reversing the (pointwise) limit and the integral is allowed.
1.1 Fubini’s Theorem
Admittedly the function 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 . 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 to be Riemann integrable on . Assume also that the one-dimensional function is Riemann integrable for almost all . Then the function is Riemann integrable and . Note: both conditions are satisfied if is continuous on .
Theorem 1.2.
Lebesgue Case. Assume that is Lebesgue integrable on . Then the function is Lebesgue integrable for almost all on . As a consequence, the function is Lebesgue integrable and .
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 the situation is as shown by the following
Venn Diagram:
In the previous we already discussed the inclusion .
Let us now integrate the function over to show
that some functions exist but are not in . As is not bounded on 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, is not contained in as follows from the fact that the Dirichlet function is not .
Finally, is also not a subset of . Define as on , on ,
on , on , etc. The Riemann integral and the
Lebesgue integral of over are both equal to
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, is not
in .
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 |
Classification | msc 28-00 |