Page 367 - DMTH401_REAL ANALYSIS

Unit 30: Riemann's and Lebesgue

 a = b} is a partition of [a, b], let M = sup{f(x) : a  x  a } and m = inf{f(x) : a  x  a }. The upper Notes
n i i-1 i i i-1 i
and lower Riemann sums with respect to the partition P are defined as U(f,P) = å n i 1 M (a – a )
=
i-1
i
i
and L(f, P) = å n i 1 m (a – a ). A bounded function f: [a, b]   is said to be Riemann integrable if
=
i-1
i
i
for every  > 0 there is a partition P of [a, b] such that U(f,P) – L(f,P) < . The following
characterization says that a Riemann integrable function is not very different from a continuous
function.
Let f : [a, b]   be a bounded function and let Y = {x  [a, b] : f is not continuous at x}. Then, f is
Riemann integrable iff m * L,1 [Y] = 0.
Proof: Let (f,x) be the oscillation of f at x defined earlier.
 : Since Y =  ¥ k 1 Y , where Y = {x  [a, b] : (f, x)  1/k}, it suffices to show m * L,1 [Y ] = 0 for every
=
k
k
k
k  . Fix k   and let  > 0. Let P = {a = a  a    a  a = b} be a partition of [a, b] with U(f,P)
0 1 n-1 n
– L(f,P) < /2k. Let A = (a , a ) and  = {1  i  n : Y  A  Ø}. Note that M – m  1/k for i  .
i i–1 i k i i i
Write Y = Y  Y , where Y = Y  (  i A ) and Y = Y  { a ,…, a }. We have /2k > U(f, P)
²


²
k k k k k i k k 1 n
²
– L(f, P)  å i (M – m )|A |  1/k å i |A | and hence å i | A | < /2. And since Y is a finite
k
i
i
i
i
i
²
m
set, there are finitely many intervals B ,…,B such that Y   |Bj and å m |B | < /2. Thus
=
=
1 m k j 1 j 1 j
Y  [  i A ]  [  m B] and å i |A | + å |B| < . Since  > 0 was arbitrary, m * [Y ] = 0.
m
=
=
k i j 1 j i j 1 j L,1 k
 : Let  > 0 be given. We have to find a partition P of [a, b] such that U(f, P) – L(f, P) < . Let
 = sup{|f(x)| : x  [a, b]} and let  = /[2 + 2(b – a)]. For each x  [a, b]\Y, choose an open
interval A(x)   containing x such that |f(x) – f(z)| <  for every z  [a, b]  A(x) by continuity.
Also choose countably many open intervals B such that Y   ¥ B and  ¥ |B | < . Then
=
=
m m 1 m m 1 m
{A(x) : x  [a,b]\Y}  {B : m  } is an open cover for the compact set [a, b]. Extract a finite
m
subcover {A(x) : 1  j  p}  {B : 1  m  q}. The end points inside [a, b] of these finitely many
j m
intervals determine a partition P = {a = a  a    a  a = b} of [a, b]. Observe that for each
0 1 n-1 n
i  {1,…,n}, we have [a , a ]  A(x ) for some j  {1,…,p}, or [ a , a ]  B for some m  {1,...,q}.
i–1 i j i–1 i m
Let  = {1  i  n : [ a , a ]  A(x ) for some j} and  = {1,…,n}\. Note that M – m < 2 if i  .
i–1 i j i i
Hence U(f, P) – L(f,P)  å i (M – m )(a – a ) + å i (M – m )(a – a )  2 å i (a – a ) + 2 å i
i
i
i
i–1
i
i–1
i
i–1
i
i
(a – a )  2 å n (a – a ) + 2 å q |B | < 2 (b – a) + 2 = .
=
=
i i–1 i 1 i i–1 m 1 m
A corollary is that any bounded function f : [a, b]   with at most countably many points of
discontinuity (in particular, any continuous function) is Riemann integrable. The higher
dimensional generalization can be stated as follows.
d
Let A   be a d-box, let f : A   be a bounded function and let Y be the set {x  A : f is not
continuous at x}. Then, f is Riemann integrable iff m * L,d [Y] = 0.
Example: Let f : [0,1]   be f(0) = 0 and f(x) = sin (1/x) for x = 0. Even though the graph
of f has infinitely many ups and downs (in fact, f is not of bounded variation), f is Riemann
integrable since f is bounded and is discontinuous only at one point, namely 0.
Definition: Let X be a set and A  X. The characteristic function  : X   of the subset A is
A

ì 1, if x A,
defined as  (x) = í
A

î 0, if x X \A.
Example: We discuss an example that illustrates the main drawback of Riemann
integration theory. Write [0,1]   = {r , r ,…}, let f : [0, 1]   be the characteristic function of
1 2 n
{r ,…, r }, and let f : [0,1]   be the characteristic function of [0,1]  . We have 0  f  f   
1 n 1 2
f  1 and the sequence (f ) converges to f pointwise. Each f is Riemann integrable since f is
n n n
discontinuous only at finitely many points. But f is discontinuous at every point of [0,1], and the
Lebesgue outer measure of [0,1] is positive. Hence f is not Riemann integrable by [115]. Thus
even the pointwise limit of a uniformly bounded, monotone sequence of Riemann integrable
functions need not be Riemann integrable.
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