Riemann is the Mann! (But Lebesgue may besgue to differ.)
Leo Livshits
May 2, 2008
1 For finite intervals in R
We have seen in class that every continuous function f : [a, b] −→ R has the property that
for every > 0 there is some δ > 0 such that
mesh(P ) < δ
=⇒ Uf (P ) − Lf (P ) < for partitions P of [a, b]. By the Infrastructure of Integration handout this guarantees that
sup { Lf (P ) | P is a partition of [a, b] } = inf { Uf (P ) | P is a partition of [a, b] } (1)
Rb
with the common value denoted by a f , and furthermore
Z b
f = lim Uf (Pn ).
(2)
[mesh(Pn )] −→ 0 =⇒ lim Lf (Pn ) =
n→∞
a
n→∞
We start by introducing a definition of lower and upper sums (corresponding to a
partition of [a,b]) for functions which are not necessarily continuous on [a, b]. With these
definitions in hand the central question is this: which other functions on [a, b] (besides the
continuous ones) are nice enough so that equation (1) and the implication (2) hold? Before
I tell you about Riemann-Lebesgue theorem which answers that question, let us introduce
the definitions of lower and upper sums as promised.
1 Definition. Given a bounded function g : [a, b] −→ R and a partition
P = {a, x1 , x2 , . . . , xm−1 , b} of [a, b], define
n X
Lg (P ) =
inf g 4i
[xi−1 ,xi ]
i=1
and
Ug (P ) =
n
X
!
sup g 4i
[xi−1 ,xi ]
i=1
where 4i is the length of [xi−1 , xi ].
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2 Observation. The key difference between this new definition and the one used for
continuous functions is that we use inf [xi−1 ,xi ] g instead of min[xi−1 ,xi ] g, and sup[xi−1 ,xi ] g
instead of max[xi−1 ,xi ] g. Extreme Value theorem which guarantees the existence of a minimal/maximal value requires continuity, and when the function has no minimum/maximum
but is bounded, it has excellent substitutes: the inf and the sup!
Of course you already know that when min exists it is the inf, and similarly for max.
3 Definition. A bounded function g : [a, b] −→ R is said to be Riemann-integrable on
[a, b] if
sup { Lg (P ) | P is a partition of [a, b] } = inf { Ug (P ) | P is a partition of [a, b] } .
Rb
In this case the common value is denoted by a g and is called the Riemann integral of the
function.
Ra
When a = b, we say that any function on [a, b] is Riemann-integrable and a f = 0.
4 Observation. Note that to demonstrate that a bounded function g : [a, b] −→ R is
Riemann-integrable on [a, b] all one has to do is show that for any > 0 there exist
partitions P and Q such that 0 ≤ Ug (Q) − Lg (P ) < . This is a consequence of Theorem
1 on Infrastructure of Integration handout.
1 Problem. Show that any increasing bounded function g : [a, b] −→ R is Riemannintegrable. Do the same for decreasing bounded functions.
What about implication (2), you say? Well, believe it or not, it comes “for free”, but
the proof of this remarkable result, which we state below, belongs to a Real Analysis course,
and is too involved to be given here.
5 Theorem. (Riemann’s Theorem) If a bounded function g : [a, b] −→ R is Riemannintegrable on [a, b] then
Z b
[mesh(Pn )] −→ 0 =⇒ lim Lg (Pn ) =
g = lim Ug (Pn ).
n→∞
a
n→∞
2 Riemann sums
6 Definition. Given a partition P = {a, x1 , x2 , . . . , xn−1 , b} of [a, b], suppose one picks an
evaluation point zi in each of the n closed intervals [xi−1 , xi ] generated by the partition.
The object thus obtained is a pointed partition P ∗ of [a, b].
Given a bounded function g : [a, b] −→ R and a pointed partition P ∗ of [a, b], the
Riemann sum Rg (P ∗ ) for g corresponding to P ∗ is defined as follows
∗
Rg (P ) =
n
X
i=1
2
g(zi )4i
7 Observation. The key difference between the definition of a Riemann sum and the
definition of upper/lower sums is that we use g(zi ) instead of sup[xi−1 ,xi ] g (and inf [xi−1 ,xi ] g).
In other words we build rectangles on the subintervals of [a, b] by using values of the function
g at evaluation points as heights, rather than using best upper/lower estimates as heights.
When g is continuous on [a, b], and a partition P is given, the Extreme Value theorem
guarantees the existence of choices of evaluation points which will make the corresponding
Riemann sum equal the upper sum (similarly, the lower sum). Do you see how?
The following corollary to Theorem 8 shows the relevance of Riemann sums.
8 Theorem. (Riemann Sum Theorem) If a bounded function g : [a, b] −→ R is
Riemann-integrable on [a, b] then
[mesh(Pn )] −→ 0 =⇒
lim Rg (Pn∗ )
n→∞
Z
=
b
g.
a
Proof. This is trivial by squeeze theorem and Theorem 8, since clearly
Lg (P ) ≤ Rg (P ∗ ) ≤ Ug (P )
no matter which evaluation points are chosen.
9 Observation. What this theorem states is that once one knows that a given function
is Riemann-integrable, one does not need to bother to use cumbersome upper/lower sums
to approximate the integral. All one has to do is pick any sequence of partitions whose
mesh goes to zero, and pick one’s favorite evaluation points for each partition, and then
the corresponding Riemann sums (are easy to calculate and) converge to the integral.
3 Measure for Measure
A big question remains: “Which bounded functions on [a, b] are Riemann integrable?” The
complete answer to it is given by the following stunning theorem which may be considered
to be a genesis of a whole subject called “Measure Theory”.
10 Theorem. (Riemann-Lebesgue Theorem) A bounded function g : [a, b] −→ R is
Riemann-integrable on [a, b] if and only if the set of discontinuities of g can be “covered”
by a sequence of open intervals with total length as small as desired.
In other words, ∀ > 0 there is a sequence I1 , I2 , I3 , . . . of open intervals such that
1. All points where g is not continuous lie in the set I1 ∪ I2 ∪ I3 ∪ . . .
2. The sum of the lengths of I1 , I2 , I3 , . . . (treated as a limit of an infinite series, of
course!) is less than .
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11 Definition. Subsets of R that can be covered by a sequence of open intervals with
total length as small as desired are said to have measure zero. In particular, empty set has
measure zero.
2 Problem.
1. Show that a subset of a set of measure zero has measure zero also.
2. Show by induction that every finite subset of R has measure zero.
3. Show that every infinite countable subset of R has measure zero. (Recall that an
infinite set is countable if there is a a bijection between it at N; in other words, if
there is a sequence which contains exactly the elements of the set appearing without
repetitions.)
(Hint: if length of Ik is 2αk , what is the sum of lengths of I1 , I2 , I3 , . . .?)
12 Corollary. If a bounded function g : [a, b] −→ R is continuous at every point of [a, b]
with the exception of finitely many points, then it is Riemann-integrable on [a, b]. The same
is true if if the set of discontinuities of g is infinite but countable.
13 Theorem. Changing the values of a bounded function g : [a, b] −→ R at a finite set of
points does not affect Riemann-integrability (or the lack of it) of the function. In the case
when the functionRis Riemann-integrable, such a change does not change the value of the
b
Riemann integral a g itself.
Proof. The proof I know for this theorem is not hard, but is rather technical. I will show
you the idea behind it in class, but will not present it here.
14 Corollary. The concept of Riemann-integrability can be defined for bounded functions
on all finite intervals, including the intervals that are open or half-open.
Proof. The key point is that given such a function h on an interval with endpoints a and
b, we can make it into a bounded function on [a, b] by defining it any way we like at a and
Rb
at b. If by doing it one way we get a Riemann-integrable function g on [a, b] with a g,
then doing it in any other way we get Riemann-integrable functions with exactly the same
value of the integral, by Theorem 13 of course!
15 Example. Recall that the function
 1
 n , if x = m
n in reduced form
f (x) =

0, otherwise
is continuous at every irrational number and zero, and is discontinuous at every non-zero
rational number. Still, this function is “good enough” to be Riemann-integrable to any
[a, b].
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The following theorem is not as trivial as it may seem. In fact it is so non-trivial that
we omit the proof and suggest that you take Real Analysis to see it!
16 Theorem. No interval of positive length has measure zero.
3 Problem.
1. Prove that a union of two sets of measure zero has measure zero.
2. Give an example of aSfamily of sets { Sr | r ∈ R } each of which has measure zero,
such that the union r∈R Sr does not have measure zero.
3. (Extra Credit) Suppose that S1 , S2 , S3 , . . . are sets all of which have measure zero.
Prove that their union also has measure zero. This is usually expressed by saying
that a countable union of sets of measure zero has measure zero.
17 Corollary. The set of irrational numbers in any interval of non-zero length does not
have measure zero.
Proof. The set of rational numbers in an interval J of non-zero length is countable and so
has measure zero. If the set of irrationals in J also had measure zero, so would their union
(by Problem 3) which equals all of J, and does not have measure zero by Theorem 16.
18 Example. Recall that the function
f (x) =
if x ∈ Q
otherwise
1,
0,
is discontinuous at every point of R. Hence it is not Riemann-integrable on any interval of
positive length.
4 For rectangles in R2
Perhaps it should not come as a complete surprise that the results of the first section can be
“boosted” to higher dimensions with little trouble. Here is what one has to do to formulate
the corresponding results in 2-dimensions.
1. Replace bounded closed/open intervals with bounded closed/open rectangles respectively.
19 Definition. Given intervals I and J in R, the set
{ (x, y) | x ∈ I, y ∈ J }
is called a rectangle and is denoted by I × J. Rectangle I × J is bounded/closed/open
if both I and J are bounded/closed/open (respectively).
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2. Lengths of intervals get replaced by areas of rectangles.
3. Functions considered are bounded functions of the form g : K −→ R, where K is a
bounded (most often closed) rectangle in R2 .
RR
4. The integral of such a function is denoted by K g.
The results of Problems 2 and 3 extend to higher dimensions word-for-word. So should
their proofs, after you have performed the terminology switch described above.
The following two-dimensional theorem is new.
20 Theorem. Let f : [a, b] −→ R be a continuous function. Then the graph of f has
measure zero in R2 . Recall that the graph of f is the set { (x, f (x) ) | x ∈ [a, b] }.
Proof. We know that since f is continuous on [a,b], it is Riemann-integrable on [a, b]. In
particular for every > 0 there is a partition P of [a, b] such that Uf (P )−Lf (P ) < . Yet it
is easy to see that Uf (P ) − Lf (P ) is the sum of the areas of finitely many closed rectangles
whose union contains the graph of f (do you see how?). By enlarging each rectangle a bit
in both direction (making sure its area does not double) we can drop off the boundaries
of the rectangles and see that the graph of f can be covered by a union of finitely many
open rectangles with the sum of areas equal at most 2. This is good enough, since was
an unspecified positive number.
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