Lecture 30
Definition 1.0.1. Riemann sum: For a partition P = {x1 , x2 , · · · , xn }, the Riemann sum
n
X
S(P, f ) is defined as S(P, f ) = f (ξi )(xi − xi−1 ), ξi ∈ [xi−1 , xi ].
i=1
Then it is easy to show the following
Remark 1.0.1. If m = inf f (x), and M = sup f (x). Then
[a,b] [a,b]
m(b − a) ≤ L(P, f ≤ S(P, f ) ≤ U (P, f ) ≤ M (b − a).
In fact, one has the following Darboux theorem:
Theorem 1.0.2. Let f f : [a, b] → IR be a Riemann integrable function. Then for a given > 0,
there exists δ > 0 such that for any partition P with kP k := max |xi − xi−1 | < δ, we have
1≤i≤n
Z b
|S(P, f ) − f (x)dx| < .
a
Corollary 1.0.3. If f ∈ R[a, b], then for any sequence of partitions {Pn } with kPn k → 0, we
Rb Rb
have L(Pn , f ) → a f (x)dx and U (Pn , f ) → a f (x)dx.
Remark 1.0.2. From the above theorem, we note that if there exists a sequence of partition
{Pn } such that kPn k → 0 and U (Pn , f ) − L(Pn , f ) 6→ 0 as n → ∞, then f is not integrable.
Problem 1.0.1. Show that the function f : [0, 1] → IR
1 + x x∈Q
f (x) =
1 − x x 6∈ Q
is not integrable.
Solution: Consider the sequence of partitions Pn = {0, n1 , n2 , ...., nn = 1}. Then
1 1 2 1 n 1
U (Pn , f ) = (1 + ) + (1 + ) + .... + (1 + )
n n n n n n
1
= 1 + 2 (1 + 2 + ... + n)
n
3
→ as n → ∞
2
Now using the fact that infimum of f on [0, n1 ] is 1 − n1 , though it is not achieved, we get
1 1 2 1 n 1 1
L(Pn , f ) = (1 − ) + (1 − ) + .... + (1 − ) → as n → ∞.
n n n n n n 2
1
Hence f is not integrable.
Theorem 1.0.4. Suppose f is a continuous function on [a, b]. Then f ∈ R[a, b].
Proof. Let > 0. By the Theorem on neccessary and sufficient condition for integrability , we
need to show the existence of a partition P such that
Use of uniform continuity
U (P, f ) − L(P, f ) < .
Since f is continuous on [a, b], this implies f is uniformly continuous on [a, b]. Therefore there
exists δ > 0 such that
|x − y| < δ ⇒ |f (x) − f (y)| < . (1.1)
(b − a)
Now choose a partition P such that
sup |xk − xk−1 | < δ. (1.2)
1≤k≤n
00 0 00
As f is continuous on [a, b] there exist x0k , xk ∈ (xk−1 , xk ) such that mk = f (xk ) and Mk = f (xk ).
0 00 00 0
By (1.2), |xk − xk | < δ and hence by (1.1) |f (xk ) − f (xk )| < (b−a) . Thus
n
X
U (P, f − L(P, f ) = (Mk − mk )(xk − xk−1 )
k=1
n
00 0
X
= (f (xk ) − f (xk ))(xk − xk−1 )
k=1
n
X
≤ (xk − xk−1 ) = (b − a) = .
(b − a) (b − a)
k=1
Therefore f ∈ R[a, b].
Integrability and discontinuous functions:
We study the effect of discontinuity on integrability of a function f (x).
Example 1.0.5. Consider the following function f : [0, 1] → R.
1, x 6= 1 Important Step in this type of questions is the
2
f (x) = partitions we choose.
0, 1
x= 2
Clearly U (P, f ) = 1 for any partition P . We notice that L(P, f ) will be less than 1. We
can try to isolate the point x = 21 in a subinterval of small length. Consider the partition
1 1
P = {0, 12 − 2 , 12 + 2 , 1}. Then L(P , f ) = ( − ) + (1 − − ) = 1 − . Therefore, for given
2 2 2 2
> 0 we have U (P , f ) − L(P , f ) = . Hence f is integrable.
2
In fact we have the following theorem.
Theorem 1.0.6. Suppose f : [a, b] → R be a bounded function which has finitely many discon-
tinuities. Then f ∈ R[a, b].
Proof follows by constructing suitable partition with sub-intervals of sufficiently small length
around the discontinuities as observed in the above example.Next we have the following theorem
Theorem 1.0.7. Let f be a monotonically decreasing function on [a, b], then f is integrable.
Proof. Let Pn = {x0 , x1 , · · · , xn } be a partition of [a, b] with xk − xk−1 = b−a
n . Since f is
monotone, it is bounded (exercise!). Also mk = inf f (x) = f (xk ) and Mk = sup f (x) =
[xk−1 ,xk ] [xk−1 ,xk ]
f (xk−1 ). Then
n
X b−a 1
U (Pn , f ) − L(Pn , f ) = [f (xk−1 ) − f (xk )] = (b − a)(f (b) − f (a)) → 0
n n
k=1
as n → ∞. Therefore by the sequential characterization of integrability, f ∈ R[a, b]. ///
1.1 Properties of Definite Integral:
Z b Z b
Property 1: For a constant c ∈ R, cf (x)dx = c f (x)dx.
a a
Property 2: Let f1 , f2 ∈ R[a, b] . Then
Z b Z b Z b
(f1 + f2 )(x)dx = f1 (x)dx + f2 (x)dx.
a a a
Easy to show that for any partition P ,
U (P, f1 + f2 ) ≤ U (P, f1 ) + U (P, f2 ) (1.3)
L(P, f1 + f2 ) ≥ L(P, f1 ) + L(P, f2 ) (1.4)
Since f1 , f2 are integrable, for > 0 there exists P1 , P2 such that
U (P1 , f1 ) − L(P1 , f1 ) <
U (P2 , f2 ) − L(P2 , f2 ) <
Now taking P = P1 ∪ P2 , if necessary, we assume
U (P, f1 ) − L(P, f1 ) < , U (P, f1 ) − L(P, f2 ) < (1.5)
3
Therefore, using (1.3)-(1.5) we get
U (P, f1 + f2 ) − L(P, f1 + f2 ) ≤ U (P, f1 ) + U (P, f2 ) − L(P, f2 ) − L(P, f2 )
< + = 2.
Hence, f1 + f2 is integrable.
Z b n
X
(f1 + f2 )(x)dx = lim S(Pn , f1 + f2 ) = lim (f1 + f2 )(ξk )(xk − xk−1 )
a n→∞ n→∞
k=1
Xn n
X
= lim f1 (ξk )(xk − xk−1 ) + lim f2 (ξk )(xk − xk−1 )
n→∞ n→∞
k=1 k=1
Z b Z b
= f1 (x)dx + f2 (x)dx
a a
Property 3: If f (x) ≤ g(x) on [a, b]. Then
Z b Z b
f (x)dx ≤ g(x)dx.
a a
Z b
First we note that m ≤ f (x) ≤ M implies m(b − a) ≤ f (x) ≤ M (b − a). Then Property 1
Z b Z b Z ab
and f (x) ≤ g(x) imply (g − f ) ≥ 0 or g(x)dx ≥ f (x)dx.
a a Za b Z b
0
Property 4: If f ∈ R[a, b] then |f | ∈ R[a, b] and | f (x)dx| ≤ |f |(x)dx. Let mk =
a a
0
inf |f |(x) and Mk = sup |f |(x). Then we claim
[xk−1 ,xk ] [xk−1 ,xk ]
0 0
Claim: Mk − mk ≥ Mk − mk
Proof of Claim: Note that for x, y ∈ [xi−1 , xi ],
|f |(x) − |f |(y) ≤ |f (x) − f (y)| ≤ Mi (f ) − mi (f ).
Now take supremum over x and infimum over y, to conclude the claim.
Now since f is integrable, there exists partitions {Pn } such that lim U (Pn , f ) − L(Pn , f ) = 0.
n→∞
i.e.,
Xn
lim (Mk − mk )(xk − xk−1 ) = 0.
n→∞
k=1
4
This implies
n
0 0
X
lim (Mk − mk )(xk − xk−1 ) = 0.
n→∞
k=1
Hence |f | is integrable. Note that −|f | ≤ f ≤ |f |. Thus by Property 3 we get
Z b Z b Z b Z b Z b
− |f |(x)dx ≤ f (x)dx ≤ |f |(x)dx =⇒ | f (x)dx| ≤ |f |(x)dx.
a a a a a
