The Riemann Integral
Definition Let I = [a, b] ⊂ R. A set P = {x0, x1, . . . , xm} ⊂ I is called a partition of I if a = x0 < x1 < . . . < xm = b.
Remarks
(a) Equivalently, a collection of subintervals
C = {Ij = [xj−1, xj] | 0 ≤ j ≤ m} = {Ij}mj=1 is called a partition of I if
(i)
m
[
j=1
Ij = I,
(ii) (xi−1, xi) ∩ (xj−1, xj) = Int Ii ∩ Int Ij = ∅ for all 1 ≤ i 6= j ≤ m.
(b) In general, if K is a set in Rn, then a collection of setsC = {Kj}mj=1 is called a partition of K if
(i)
m
[
j=1
Kj = K,
(ii) Int Ki∩ Int Kj = ∅ for all 1 ≤ i 6= j ≤ m, where Int Ki and Int Kj denote the interiors of Ki and Kj, respectively.
Example Let K =
n
Y
j=1
[aj, bj] = [a1, b1] × · · · × [an, bn] = I1 × · · · × In be an n-dimensional cell.
For each 1 ≤ j ≤ n, since either
Pj = {aj = xj 0 < xj 1 < . . . < xj m(j) = bj}, or
{Ij i = [xj, i−1, xj, i] | 1 ≤ i ≤ m(j), aj = xj 0 < · · · < xj m(j) = bj} is a partition of Ij, thus either P =
n
Y
j=1
Pj or C = {
n
Y
j=1
[xj, ij−1, xj, ij] | 1 ≤ ij ≤ m(j)} is a partition of K.
Definition (a) Let K =
n
Y
j=1
[aj, bj] be an n-cell and let
P =
n
Y
j=1
Pj and Q =
n
Y
j=1
Qj be partitions of K,
where for each 1 ≤ j ≤ n,
Pj = {aj = xj 0 < xj 1 < . . . < xj l(j)= bj} Qj = {aj = yj 0 < yj 1 < . . . < yj m(j) = bj}
are partions of [aj, bj]. We say that Q is a refinement of P, or Q is finer than P, denoted by P ⊆ Q, if Pj ⊆ Qj for each 1 ≤ j ≤ n.
(b) Let C = {Ii}`i=1 and F = {Kj}mj=1 be partitions of (an n-cell) K. We say that F is a refinement of C , denoted by F ⊃ C , if for each Kj ∈ F there exists a Ii ∈ C such that Kj ⊆ Ii.
Definition
(a) Let P = {x0, x1, . . . , xm} ⊂ I be a partition of the interval I = [a, b]. We definethe norm of the partition P to be
kP k = max
1≤j≤m|xj − xj−1|= the maximum length of {Ij = [xj−1, xj] | 1 ≤ j ≤ m}.
(b) Let P = {Ij}mj=1 be a partition of (an n-cell) K. We define the norm of the partition P to be
kP k = max
1≤j≤mdiam(Ij)= the maximum diameter of {Ij | 1 ≤ j ≤ m}.
Definition Let f : K → R be a bounded function defined on a closed n-cell K and let P = {Kj}mj=1 be a partition of K.
(a) A Riemann sum SP(f, K) corresponding to partition P is a sum of the form SP(f, K) =
m
X
i=1
f (xi)c(Ki),
where xi is any chosen point in Ki and c(Ki) denotes the n-dimensional volume of Ki. (b) A lower sum L(P, f ) corresponding to partition P is a sum of the form
L(P, f ) =
m
X
i=1
inf{f (x) | x ∈ Ki} c(Ki).
(c) An upper sum U (P, f ) corresponding to partition P is a sum of the form U (P, f ) =
m
X
i=1
sup{f (x) | x ∈ Ki} c(Ki).
Remarks Let f : K → R be a bounded function defined on K.
(a) (Monotonicity of lower and upper sums) If I ⊆ J are subsets of K, then inf
J f ≤ inf
I f ≤ sup
I
f ≤ sup
J
f.
This implies that if P, Q are partitions of K satisfying that P ⊆ Q, i.e. Q is finer than P, then
L(P, f ) ≤ L(Q, f ) ≤ SQ(f, K) ≤ U (Q, f ) ≤ U (P, f ).
(b) Given any partitions P1, P2 of K, since P1∪ P2 is a partition finer than P1 or P2, we have L(P1, f ) ≤ L(P1∪ P2, f ) ≤ U (P1∪ P2, f ) ≤ U (P2, f ).
In particular,
(i) L(P, f ) ≤ U (P2, f ) for any partition P of K, and hence the lower integral of f over K defined by R
Kf = sup{L(P, f ) | P is a partition of K} exists.
(ii) L(P1, f ) ≤ U (P, f ) for any partition P of K, and hence theupper integral of f over K defined by R
Kf = inf{U (P, f ) | P is a partition of K} exists.
Hence
L(P, f ) ≤ Z
K
f ≤ SP(f, K) ≤ Z
K
f ≤ U (P, f ) for any partition P of K.
Definition A bounded function f is said to beRiemann integrable on K if Z
K
f = Z
K
f
and the common value, denoted Z
K
f, is called the (Riemann) integral of f on K.
Remarks
(a) (Riemann Criterion for Integrability) f is Riemann integrable over K, i.e.
Z
K
f = Z
K
f, if and only if for each ε > 0, there exists a partition Pε of K such that
|U (Pε, f ) − L(Pε, f )| < ε.
(b) (Cauchy Criterion for Integrability) f is Riemann integrable over K if and only if for each ε > 0, there exists a partition Pε of K such that if P and Q are refinements of Pε and SP(f, K) and SQ(f, K) are any corresponding Riemann sums, then
|SP(f, K) − SQ(f, K)| < ε.
R
Kf + ε 2 R
Kf
U (Pε, f ) L(Pε, f )
U (P, f ) SP(f, K) U (Q, f ) SQ(f, K)
L(P, f )
L(Q, f ) R
Kf R
Kf − ε 2
Example For a < b, let
f (x) =
(a if x ∈ Q ∩ [0, 1], b if x ∈ [0, 1] \ Q.
Then f is not continuous at any x ∈ [a, b] and f is not integrable on [0, 1] since L(P, f ) = a 6= b = U (P, f ) for any partition P of [a, b].
Properties of the Integral (on n-cells)
(a) Suppose that K, K1 and K2 are closed n-cells such that
K = K1∪ K2 and Int(K1) ∩ Int(K2) = ∅.
If f is integrable on K, then f is integrable on K1, and K2 and Z
K
f = Z
K1
f + Z
K2
f.
Proof Given ε > 0, since f is integrable, there is a partition Pε of K such that if P is any refinement of Pε, then |U (P, f, K) − L(P, f, K)| < ε.
Let Pε,i = Pε∩ Ki, for i = 1, 2. Then Pε,i is a partition of Ki, Pε,1∪ Pε,2 is a refinement of Pε such that
ε > U (Pε,1∪ Pε,2, f, K1∪ K2) − L(Pε,1∪ Pε,2, f, K1∪ K2)
= U (Pε,1, f, K1) − L(Pε,1, f, K1) + U (Pε,2, f, K2) − L(Pε,2, f, K2)
≥ U (Pε,i, f, Ki) − L(Pε,i, f, Ki) for each i = 1, 2.
≥ 0.
Thus if Pi is a refinement of Pε,i for i = 1, 2, then P1∪ P2 is a refinement of Pε,1∪ Pε,2 and ε > U (Pε,i, f, Ki) − L(Pε,i, f, Ki) ≥ U (Pi, f, Ki) − L(Pi, f, Ki) ≥ 0.
This implies that f is Riemann integrable on Ki and Li =R
Kif exists, for i = 1, 2, and
L = Z
K
f = sup{L(P, f ) | P is a refinement of Pε}
= sup{L(P, f ) | P is a refinement of Pε,1∪ Pε,2}
= sup{L(P1, f ) + L(P2, f ) | Piis a refinement of Pε,i for i = 1, 2}
≤ L1+ L2
≤ inf{U (P1, f ) + U (P2, f ) | Piis a refinement of Pε,i for i = 1, 2}
= inf{U (P, f ) | P is a refinement of Pε,1∪ Pε,2}
= inf{U (P, f ) | P is a refinement of Pε}
= Z
K
f = L Hence we have L = L1+ L2.
(b) If f and g are integrable on K, then, for any c ∈ R, cf + g is integrable on K, and Z
K
(cf + g) = c Z
K
f + Z
K
g.
Proof Given ε > 0, since f, g are integrable on K, there exists a partition Pε of K such that if P is any refinement of Pε then
|U (P, f ) − L(P, f )| < ε 2(1 + |c|) and
|U (P, g) − L(P, g)| < ε 2. Thus
|U (P, cf + g) − L(P, cf + g)| ≤ |c||U (P, f ) − L(P, f )| + |U (P, g) − L(P, g)|
< |c|ε
2(1 + |c|) + ε 2
< ε
which implies that cf + g is integrable on K. Let {Pj} be a sequence of partitions of K satisfying that Pj ⊂ Pj+1 for all j = 1, 2, . . ., and lim
j→∞kPjk = 0. Since Z
K
f = lim
j→∞SPj(f, K) and Z
K
g = lim
j→∞SPj(g, K), we have
c Z
K
f + Z
K
g = c lim
j→∞SPj(f, K) + lim
j→∞SPj(g, K)
= lim
j→∞SPj(cf, K) + lim
j→∞SPj(g, K)
= lim
j→∞SPj(cf + g, K)
= Z
K
(cf + g).
(c) Suppose that f and g are integrable on K. If f (x) ≤ g(x) for each x ∈ K, then Z
K
f ≤ Z
K
g.
Proof Since −f + g ≥ 0 on K and, by (b), it is integrable on K, we have 0 ≤ L(P, −f + g) ≤
Z
K
(−f + g) = − Z
K
f + Z
K
g, where P is any partition of K. SinceR
Kf ∈ R, by addingR
Kf on both sides of the inequality, we get R
Kf ≤R
Kg.
(d) If f is integrable on K, then |f | is integrable on K, and
Z
K
f
≤ Z
K
|f |.
Proof Given ε > 0, since f is integrable on K, there exists a partition Pε of K such that if P = {Kj}mj=1 is any refinement of Pε then
|U (P, |f |) − L(P, |f |)| = |
m
X
j=1
sup
Kj
|f | − inf
Kj
|f |c(Kj)|
≤ |
m
X
j=1
sup
Kj
f − inf
Kj
fc(Kj)|
= |U (P, f ) − L(P, f )|
< ε.
Thus, |f | is integrable (by Riemann’s Criterion for integrability). Since ±f, |f | are inte- grable, and ±f ≤ |f | on K, we have
± Z
K
f ≤ Z
K
|f | =⇒
Z
K
f
≤ Z
K
|f |.
Examples
(a) Let I be a closed interval in R, and f be a bounded and monotonic function defined on I = [a, b] Then f is integrable on I.
Proof Since f is bounded on I, Z
I
f = sup
P
L(P, f ) and Z
I
f = inf
P U (P, f ) exist.
If Pn is the partition that divides I into 2n equal length subintervals, then
n→∞lim L(Pn, f ) = Z
I
f and lim
n→∞U (Pn, f ) = Z
I
f.
Since
n→∞lim |U (Pn, f ) − L(Pn, f )| = lim
n→∞
f (b) − f (a)
b − a
2n = 0,
we have
Z
I
f = Z
I
f =⇒ f is integrable on I.
(b) Let K be a closed n-cell, and f be a continuous function on K. Then f is integrable on K.
Proof Since f is continuous on (compact set) K, f is uniformly continuous on K.
Hence, for any given ε > 0 there exists δ > 0 such that
if x, y ∈ K and kx − yk < δ then |f (x) − f (y)| < ε c(K) + 1, where c(K) denotes the volume of K. Let P be a partition of K such that
kPεk = max
Kj∈Pε
diam(Kj) = max
Kj∈Pε
sup{kx − yk : x, y ∈ Kj} < δ.
If P is a refinement of Pε then
|U (P, f ) − L(P, f )| < εc(K) c(K) + 1 < ε.
Therefore, f is integrable on K.
Integration and Differentiation
(a) Theorem Let f be integrable on [a, b]. For each x ∈ [a, b], let F (x) =
Z
[a,x]
f = Z x
a
f (t)dt.
Then F is continuous on [a, b]; moreover, F0(x) exists and equals f (x) at every x at which f is continuous.
Remark By the Property (a), f is initegrable on [a, x] for each x ∈ [a, b]. In fact, for each ε > 0, and for each x ∈ [a, b], since f is integrable on [a, b], there exists a partition Pε of [a, b] such that
if P is a refinement of Pε, then |U (P, f, [a, b]) − L(P, f, [a, b])| < ε (∗).
Let
Pε` = Pε∩ [a, x] ∪ {x} and Pεr = {x} ∪ Pε∩ [x, b].
Then Pε`∪ Pεr is a refinement of Pε and if P` is a refinement of Pεl on [a, x], since
|U (P`, f, [a, x]) − L(P`, f, [a, x])| ≤ |U (Pε`, f, [a, x]) − L(Pε`, f, [a, x])
≤ |U (Pε`, f, [a, x]) − L(Pε`, f, [a, x]) + U (Pεr, f, [x, b]) − L(Pεr, f, [x, b])|
= |U (Pε`∪ Pεr, f, [a, b]) − L(Pε`∪ Pεr, f, [a, b])|
< ε by (∗), f is integrable on [a, x], i.e. F (x) exists.
Proof Since f is (Riemann) integrable on [a, b], it is bounded there and c = sup{|f (t)| : t ∈ [a, b]} exists.
Also since
F (y) − F (x) = Z y
x
f (t)dt for x, y ∈ [a, b], and thus
|F (y) − F (x)| ≤ | Z y
x
|f (t)|dt|
≤ c | Z y
x
dt|
= c |y − x|, F is (Lipschitz, and hence) continuous on [a, b].
Suppose that f is continuous at x ∈ [a, b]. This implies that for each ε > 0, there exists a δ > 0 such that
if t ∈ [a, b] and |t − x| < δ then |f (t) − f (x)| < ε.
Thus for each y ∈ [a, b] satisfying that 0 < |y − x| < δ, we have
F (y) − F (x)
y − x − f (x)
=
1 y − x
Z y x
f (t)dt − 1 y − x
Z y x
f (x)dt
≤ 1
|y − x|
Z y x
|f (t) − f (x)|dt
≤ 1
|y − x|
Z y x
εdt
= ε This imples that
y→xlim
F (y) − F (x)
y − x = f (x),
i.e. F0(x) exists and equals f (x) at every x at which f is continuous.
(b) Fundamental Theorem of Calculus Let F be a continuous function on [a, b] that is differentiable except at finitely many points in [a, b], and let f be a function on [a, b] that agrees with F0 at all points where F0 is defined. If f is integrable on [a, b], then
Z b a
f (t) dt = F (b) − F (a).
Proof Suppose {z1, . . . , zm} ⊂ [a, b] is the set of points at which F0 does not exist.
Let
P = {a = x0 < x1 < · · · < xn= b}
be a partition of [a, b] such that {z1, . . . , zm} ⊆ P.
For each 1 ≤ j ≤ n, since F is continuous on [xj−1, xj] and differentiable on (xj−1, xj), by the Mean Value Theorem, there exists tj ∈ (xj−1, xj) such that
F (xj) − F (xj−1) = F0(tj) (xj − xj−1) = f (tj) (xj − xj−1) ∀ 1 ≤ j ≤ n.
This imples that
F (b) − F (a) =
n
X
j=1
F (xj) − F (xj−1) =
n
X
j=1
f (tj) (xj− xj−1).
Thus
L(P, f, [a, b]) ≤ F (b) − F (a) ≤ U (P, f, [a, b])
=⇒ sup
P
L(P, f, [a, b]) ≤ F (b) − F (a) ≤ inf
P U (P, f, [a, b]).
Hence, if f is integrable then sup
P
L(P, f, [a, b]) = inf
P U (P, f, [a, b]) and Z b
a
f (t)dt = F (b) − F (a).
(c) Theorem (Integration by Parts) Suppose F and G are differentiable functions on [a, b].
If F0 = f and G0 = g are (Riemann) integrable functions on [a, b], then Z b
a
F (x)g(x) dx = F (b)G(b) − F (a)G(a) − Z b
a
f (x)G(x) dx
= F (x)G(x)|ba− Z b
a
f (x)G(x) dx.
Integration of Vector-valued Functions
(a) Definition Let f1, . . . , fk be real functions on [a, b] and let f = (f1, . . . , fk) : [a, b] → Rk be the corresponding mapping of [a, b] into Rk. Then f is said to be (Riemann) integrable on [a, b] if and only if each fj, j = 1, . . . , k, is (Riemann) integrable on [a, b]. If f is integrable on [a, b], we define
Z b a
f dx = Z b
a
f1dx, . . . , Z b
a
fkdx.
(b) Theorem Let f and F be vector-valued maps from [a, b] into Rk. If f is integrable on [a, b]
and if F0 = f on [a, b], then
Z b a
f (x) dx = F(b) − F(a).
(c) Theorem Let f be a map from [a, b] into Rk. If f is integrable on [a, b], then
Z b a
f (x) dx
= Z b
a
|f (x)| dx.
Convergence and Differentiation
Theorem (Chapter 4 and 7 Summary) Let fn: [a, b] → C be a sequence of continuously differen- tiable functions defined on I = [a, b]. If fn(x0) converges for some x0 ∈ [a, b] and if fn0 : [a, b] → C converges uniformly on [a, b], then fn converges uniformly on [a, b] to a function f.
Proof For any x ∈ I = [a, b] and for any m, n ∈ N, there exists a y depending on m, n such that [fm(x) − fn(x)] − [fm(x0) − fn(x0)] = [fm0 (y) − fn0(y)](x − x0)
=⇒ kfm− fnkI ≤ |fm(x0) − fn(x0)| + (b − a)kfm0 − fn0kI.
This implies that fn converges uniformly on I to a function f and f is continuous on I.
Definition If {fn} is a sequence of functions defined on a subset D of Rp with values in Rq, the sequence of partial sums (sn) of the series P fn is defined for x in D by
sn(x) =
n
X
j=1
fj(x).
In case the sequence {sn} converges on D to a function f, we say that the infinite series of functionsP fn converges to f on D. If the sequence {sn} converges uniformly on D to a function f, we say that the infinite series of functions P fn converges uniformly to f on D.
Remark (Cauchy Criterion) It is easy to see thatP fk converges uniformly on D if and only if for each ε > 0, there exists M = M (ε) ∈ N such that for any n, m ≥ M and any x ∈ D, we have
ksn(x) − sm(x)k < ε.
Theorem (Term-by-Term Differentiation) For each n ∈ N, let fn be a real-valued function on K = [a, b] which has a derivative fn0 on K. Suppose that
(i) P fn converges at x0 ∈ K, (ii) P fn0 converges uniformly on K,
there exists a real-valued function f on K such that (a) P fn converges uniformly on K to f,
(b) f is differentiable on K and f0 =X
fn0 on K.
Proof Since P fn converges at x0 ∈ K, the partial sum sn of P fn converges at x0 ∈ K. For each x ∈ K and for any m, n ∈ N, by the Mean Value Theorem, the equality
sm(x) − sn(x) = sm(x0) − sn(x0) + (x − x0)(s0m(y) − s0n(y))
holds for some y lying between x and x0. The uniform convergence of P fn0 and the convergence of P fn(x0) lead to the uniform convergence of P fn on K.
Suppose that P fn0 converges uniformly to g on K. For each x, c ∈ K and any m, n ∈ N, by the Mean Value Theorem, the equality
sm(x) − sn(x) = sm(c) − sn(c) + (x − c)(s0m(y) − s0n(y)) holds for some y lying between x and c. We infer that, when x 6= c, then
sm(x) − sm(c)
x − c −sn(x) − sn(c) x − c
≤ ks0m− s0nkK = sup
x∈K
|s0m(x) − s0n(x)|.
Given ε > 0, by the uniform convergence of P fn0, there exists a M (ε) such that if m, n ≥ M (ε) then ks0m− s0nkK < ε. Taking the limit with respect to m, we obtain
f (x) − f (c)
x − c − sn(x) − sn(c) x − c
≤ ε if n ≥ M (ε).
Since g(c) = lim s0n(c), there exists an N (ε) such that if n ≥ N (ε), then
|s0n(c) − g(c)| < ε.
Now let L = max{M (ε), N (ε)}. In view of the existence of s0L(c), there exist δL(ε) > 0 such that if 0 < |x − c| < δL(ε), then
sL(x) − sL(c)
x − c − s0L(c)
< ε.
Therefore, it follows that if 0 < |x − c| < δL(ε), then
f (x) − f (c)
x − c − g(c)
< 3 ε.
This shows that f0(c) exists and equals g(c).
Examples
(a) For each k ∈ N and for each x ∈ [−1, 1], let fk(x) = xk
k2. Then
∞
X
k=1
fk converges uniformly on [−1, 1],
and for any 0 ≤ r < 1, since |x|k−1
k ≤ |x|k−1 and
∞
X
k=1
|x|k−1 converges uniformly on [−r, r],
∞
X
k=0
fk0 =
∞
X
k=1
xk−1
k converges uniformly on any [−r, r].
(b) For each k ≥ 0 and for each x ∈ (−1, 1), let fk(x) = (−1)kxk. Then, for any 0 ≤ r < 1,
∞
X
k=0
fk converges to f (x) = 1
1 + x uniformly on any [−r, r].
(c) For each k ∈ N and for each x ∈ [−1, 1], let fk(x) = (−1)kx2k. Then
∞
X
k=0
fk(x) =
∞
X
k=0
(−1)kx2k converges to 1
1 + x2 uniformly on any [−r, r]
for any 0 ≤ r < 1, and note that it is not convergent at x = ±1.
Since
∞
X
k=0
Z x 0
fk(t) dt =
∞
X
k=0
(−1)k
2k + 1x2k+1 converges uniformly on [−1, 1],
and Z x
0
1
1 + t2 dt = tan−1x, we have
∞
X
k=0
(−1)k
2k + 1x2k+1 = tan−1x uniformly on [−1, 1].
(d) For each k ∈ N and for each x ∈ R, defiletne fk(x) = sin kx
k2 . Since | sin kx|
k2 ≤ 1 k2 and
∞
X
k=1
1
k2 < ∞,
∞
X
k=1
fk =
∞
X
k=1
sin kx k2 converges uniformly on R by the M -test.
However, since
∞
X
k=1
1
k diverges, the criterion or the M -test is not applicable for
∞
X
k=1
fk0 =
∞
X
k=1
cos kx k .
Note that if K = [a, b] ⊂ (0, 2π), then the partial sums
|sn(x)| = |
n
X
k=1
cos kx| =
sin(n +12)x − sin12x 2 sin12x
≤ 1
| sin12x|
are unformly bounded on K. Since the sequence {k1} decreases to zero, sn converges uni- formly on K.
Convergence and Integration
Question Let {fn} be a sequence of integrable functions that converges at every point of a cell K ⊂ Rp to a function f. Is f integrable on K? Suppose that f is integrable on K, is it true that
Z
K
f = lim
n→∞
Z
K
fn?
Theorem Let {fn} be a sequence of integrable functions that converges uniformly on a closed cell K ⊂ Rp to a function f. Then f is integrable and
Z
K
f = lim
n→∞
Z
K
fn.
Proof For each ε > 0, there exists N ∈ N be such that
kfn− f kK = sup{|fn(x) − f (x)| | x ∈ K} < ε for all n ≥ N.
Since fN is integrable, there exists a partition PN of K such that if P, Q are refinements of PN, then
|SP(fN, K) − SQ(fN, K)| < ε for any choice of the intermediate points.
Note that for such partitions P or Q, if we take the same intermediate points for the Riemann sums of f and fN with respect to P or Q, then we have
|SP(f, K) − SP(fN, K)| ≤ kfN − f kKc(K) < εc(K) and
|SQ(f, K) − SQ(fN, K)| ≤ kfN − f kKc(K) < εc(K),
where c(K) is the volume of K. This imples that for any refinements P, Q of PN, the corresponding Riemann sums satisfy
|SP(f, K) − SQ(f, K)|
≤ |SP(f, K) − SP(fN, K)| + |SP(fN, K) − SQ(fN, K)| + |SQ(fN, K) − SQ(f, K)|
≤ ε (1 + 2c(K)).
This implies that f is integrable on K. Since
Z
K
f − Z
K
fn
= Z
K
(f − fn)
≤ kf − fnkKc(K), we have
Z
K
f = lim
n→∞
Z
K
fn.
Theorem If fn is a sequence of continuous function defined on a subset D of Rp with values in Rq and if P fn converges to f uniformly on D, then f is continuous on D.
Theorem (Term-by-Term Integration) For each n ∈ N, let fn be a real-valued integrable function on K = [a, b]. Suppose that the series P fn converges to f uniformly on K.
Then f is integrable on K and
Z
K
f =
∞
X
j=1
Z
K
fn.
Examples
(a) Let Q ∩ [0, 1] = {xn}∞n=1 and fn be a monotone sequence of integrable functions on [0, 1]
defined by
fn(x) =
(1 if x ∈ {x1, x2, . . . , xn}, 0 otherwise.
Then the limit function f is defined by
f (x) = lim
n→∞fn(x) =
(1 if x ∈ Q ∩ [0, 1], 0 if x ∈ [0, 1] \ Q.
Note that f is not integrable on [0, 1] and for each n ∈ N, since sup
x∈[0,1]
|fn(x) − f (x)| = 1,
n→∞lim sup
x∈[0,1]
|fn(x) − f (x)| = 1 6= 0, the convergence of fn to f is not uniform on [0, 1], and
n→∞lim Z 1
0
fn= 0 6= 1 = Z 1
0
f = Z 1
0
n→∞lim fn.
(b) For each n ≥ 1, define (discontinuous) fn and (continuous) f on K = [0, 1] by
fn(x) =
(n if x ∈ (0,n1),
0 otherwise, and f (x) = 0 ∀x ∈ [0, 1].
Then lim
n→∞fn(x) = f (x) for all x ∈ [0, 1] and f is Riemann integrable on K. Since lim
n→∞ sup
x∈[0,1]
|fn(x)−
f (x)| = ∞ 6= 0, the convergence of fn to f is not uniform on [0, 1], and
n→∞lim Z 1
0
fn = 1 6= 0 = Z 1
0
f = Z 1
0
n→∞lim fn.
(c) Let K = [0, 1], and for each n ≥ 2 let fn be (a continuous function) defined by
fn(x) =
n2x if x ∈ [0,1n],
−n2(x −n2) if x ∈ [n1,n2], 0 if x ∈ [n2, 1], and (a continuous function) f be defined by
f (x) = lim
n→∞fn(x) = 0 for all x ∈ K.
Since lim
n→∞ sup
x∈[0,1]
|fn(x) − f (x)| = ∞ 6= 0, the convergence of fn to f is not uniform on [0, 1], f is integrable on K, and
n→∞lim Z 1
0
fn = 1 6= 0 = Z 1
0
f = Z 1
0
n→∞lim fn.
Example Let K = [0, 1], and fn be defined by
fn(x) =
(sin(nπx) if x ∈ [0,1n], 0 if x ∈ (n1, 1].
Note that fn converges to the zero function on [0, 1], and the convergence is not uniform on K.
However,
n→∞lim Z 1
0
fn= lim
n→∞
2
nπ = 0 = Z 1
0
n→∞lim fn.
This example demonstrates that the uniform convergence is not a necessary condition in the theorem.
Bounded Convergence Theorem Let {fn} be a sequence of integrable functions on a closed cell K ⊂ Rp. Suppose that there exists B > 0 such that kfn(x)k ≤ B for all n ∈ N, x ∈ K. If the function f (x) = lim
n→∞fn(x), x ∈ K, exists and is integrable, then Z
K
f = lim
n→∞
Z
K
fn.
Remark This theorem has replaced the uniform convergence of fn by the uniform boundedness of fn and the integrability of f.
Outline of the Proof Since f (x) = lim
n→∞fn(x) for x ∈ K and kfnkK ≤ B for all n ∈ N, there exists M such that
|f (x)| ≤ M and |fn(x)| ≤ M for all x ∈ K and for all n ≥ 1.
Since |f − fn| is integrable on K, there exists a subset A ⊆ K such that c(K \ A) = 0 and |f − fn| converges uniformly to 0 on A. Hence,
Z
K
f = lim
n→∞
Z
K
fn.
Examples Use a suitable convergence theorem to prove the following.
(a) If a > 0, then lim
n→∞
Z a 0
e−nxdx = 0.
Hint Note that f (x) = lim
n→∞e−nx=
(1 if x = 0,
0 if 0 < x ≤ a, is integrable on [0, a] and |e−nx| ≤ 1 for all x ∈ [0, a] and for all n ∈ N.
(b) If 0 < a < 2, then lim
n→∞
Z 2 a
e−nx2dx = 0. What happens if a = 0?
Hint For 0 < a < 2 and for x ∈ [a, 2], note that f (x) = lim
n→∞e−nx2 = 0 is integrable on [a, 2]
and |e−nx2| ≤ e−na2 for all x ∈ [a, 2] and for all n ∈ N, i.e. the convergence is uniform on [a, 2].
For a = 0, note that |e−nx2| ≤ 1 for all x ∈ [0, 2] and for all n ∈ N, i.e. the sequence {fn(x) = e−nx2} is bounded on [0, 2].
(c) If a > 0, then lim
n→∞
Z π a
sin nx
nx dx = 0. What happens if a = 0?
Hint For a > 0 and for x ∈ [a, π], note that f (x) = lim
n→∞
sin nx
nx = 0 is integrable on [a, π]
and |sin nx nx | ≤ 1
na for all x ∈ [a, π] and for all n ∈ N, i.e. the convergence is uniform on [a, π].
For a = 0 and for each n ∈ N, by defining fn(0) = lim
x→0+
sin nx
nx = 1, the function fn(x) =
1 if x = 0,
sin nx
nx if 0 < x ≤ π, is continuous on [0, π]. Also note that for each n ∈ N, since fn(0) = 1 and sin nx ≤ nx =⇒ sin nx
nx ≤ 1 for all x > 0, |fn(x)| ≤ 1 for all x ∈ [0, π]
and for all n ∈ N, i.e. the sequence {fn(x)} is bounded on [0, π].
(d) Let fn(x) = nx
1 + nx for x ∈ [0, 1] and let f (x) =
(0 if x = 0, 1 if x ∈ (0, 1].
Then lim
n→∞fn(x) = f (x) for all x ∈ [0, 1] and that lim
n→∞
Z 1 0
fn(x) dx = Z 1
0
f (x) dx.
Hint Note that |fn(x)| ≤ 1, for all x ∈ [0, 1] and for all n ∈ N, but the convergence is not uniform on [0, 1].
(e) Let hn(x) = nxe−nx2 for x ∈ [0, 1] and let h(x) = 0. Then 0 =
Z 1 0
h(x) dx 6= lim
n→∞
Z 1 0
hn(x) dx = 1 2.
Hint Note that h and every hn are continuous on [0, 1] and limn→∞hn(x) = h(x) for all x ∈ [0, 1], but the convergence is not uniform on [0, 1].
Monotone Convergence Theorem Let {fn} be a monotone sequence of integrable functions on a closed cell K ⊂ Rp. If the function f (x) = lim fn(x), x ∈ K, exists and is integrable, then
Z
K
f = lim
n→∞
Z
K
fn.
Proof Suppose that
f1(x) ≤ f2(x) ≤ · · · ≤ f (x) for all x ∈ K, we have
fn(x) ∈ [f1(x), f (x)] for all n ∈ N
and
kfn(x)k ≤ |f1(x)| + |f (x)|
≤ sup
x∈K
|f1(x)| + sup
x∈K
|f (x)|
= B for all x ∈ K and for all n ∈ N.
By the Bounded Convergence Theorem, we get Z
K
f = lim
n→∞
Z
K
fn.
Remark Note that the convergence theorem may fail if K in not compact.
Example For each n ∈ N, let fn be defined by
fn(x) =
1
x if x ∈ [1, n]
0 if x > n.
Then fn is integrable on [1, ∞), and {fn} is a bounded, monotone sequence that converges uniformly on [1, ∞) to a continuous function f (x) = 1
x. However
n→∞lim Z ∞
1
fn 6=
Z ∞ 1
n→∞lim fn since f is not integrable on [1, ∞).
Example For each n ∈ N, let gn be defined by
gn(x) =
1
n if x ∈ [0, n2] 0 if x > n2.
Then each gn is integrable on [1, ∞), and {gn} is a bounded, monotone sequence that converges [1, ∞) to an integrable function g(x) ≡ 0. However,
n→∞lim Z ∞
1
gn 6=
Z ∞ 1
n→∞lim gn.
