221 Analysis 2, 2008–09
Suggested solutions to exercise sheet 5
In these questions, the words “measurable” and “integrable” should be taken
to mean “Lebesgue measurable” and “Lebesgue integrable”.
1. Either give an example, or explain why no such example exists, of:
(a) a (Lebesgue) measurable function f : R → R which is not integrable;
(b) an integrable function f : R → R such that −f is not integrable;
(c) an integrable function f : R → R such that f 2 is not integrable;
(d) a convergent sequence of integrable functions fn : R → R such that
lim fn is not integrable.
n→∞
Solution. (a) The constant function f (x) = 1 is a measurable function
R(since it’s equal
R to χR , and χA is measurable iff A is a measurable set). But
|f (x)| dx = χR dm = m(R) = ∞, so f is not integrable.
(b) No such example Rexists. If f is integrable then f is measurable,
so −f
R
is
R measurable. Also, |f (x)| dx < ∞ and | − f | = |f |, so |(−f )(x)| dx =
|f (x)| dx < ∞. So −f is integrable.
(c) Let f (x) = x−1/2 if x ∈ (0, 1), and f (x) = 0 if x 6∈ (0, 1). Then f is measurable [it can be written as a limit of continuous functions] and f χ(1/n,1) ↑ f
as n → ∞, so
Z
Z 1
M CT
FTC
x−1/2 dx = lim 2 − 2 · ( n1 )1/2 = 2 < ∞.
|f (x)| dx = lim
n→∞
n→∞
1/n
So f is integrable. However, f 2 is not integrable since f (x) = x−1 for x ∈
(0, 1) and f 2 χ(1/n,1) ↑ f 2 , as n → ∞, so
Z
2
M CT
|f (x)| dx =
Z
1
lim
n→∞
1/n
FTC
x−1 dx = lim 0 − log( n1 ) = lim log n = ∞.
n→∞
n→∞
(d) Let fn = χ[−n,n]R and let f (x) = 1. Then fn → f as n → ∞ and each fn
is integrable [with fn dm = 2n < ∞], but f is not integrable.
2. Evaluate the following expressions, carefully justifying your working.
Z ∞
Z 1
Z n
1
1 + nx2
sin(n2 x2 )
√
(a)
dx
(b) lim
dx
(c)
lim
dx
n→∞ 0 (1 + x)n
n→∞ −n n2 |x|3/2
1 + x2
−∞
1
1
1 −1
Solution. (a) If x > 1 then 1 + x2 < 4x2 , so √1+x
. So for any
2 > 2x
n ≥ 1,
Z n
Z ∞
Z
1
1
1 n −1
FTC 1
√
√
dx ≥
dx ≥
x dx =
log n.
2 1
2
1 + x2
1 + x2
−∞
1
R∞
1
Since 21 log n → ∞ as n → ∞, this gives −∞ √1+x
2 dx = ∞.
[Alternatively,R you can do this by substituting u = sinh−1 (x) on [−n, n] and
n
−1
1
(n) → ∞ as n → ∞.]
showing that −n √1+x
2 dx = 2 sinh
2
1+nx
(b) Let fn (x) = (1+x)
n χ(0,1) (x). We apply the Dominated Convergence Theorem to the sequence (fn )n≥1 :
• For any x ∈ (0, 1) we have
fn (x) =
1 + nx2
1+n
<
→0
n
(1 + x)
(1 + x)n
as n → ∞, so fn → 0 as n → ∞.
• If x ∈ (0, 1) and n ≥ 1 then x2 < x, so 1 + nx2 < 1 + nx and
1 + nx2
1 + nx2
1 + nx2
=
< 1.
≤
(1 + x)n
1 + nx
1 + nx + n2 x2 + . . .
So fn = |fn | ≤ χ(0,1) , which is integrable and independent of n. So χ(0,1)
will do as our dominating function.
By the Dominated Convergence Theorem,
Z n
Z
Z
Z
1 + nx2
DCT
dm = lim
fn dm =
lim fn dm = 0 dm = 0.
lim
n→∞
n→∞
n→∞ −n (1 + x)n
2 2
x )
(c) Let fn (x) = sin(n
if x ∈ (−n, n)\{0}, and fn (x) = 0 for all other x ∈ R.
n2 |x|3/2
We again apply the Dominated Convergence Theorem:
• For any x ∈ (−n, n) with x 6= 0, using the estimate | sin θ| < |θ| we have
|fn (x)| ≤
1
n2 |x|3/2
→0
as n → ∞. So fn → 0 as n → ∞.
• We have |sin θ| ≤ |θ| for any θ ∈ R, so if x 6= 0 and n ≥ 1 then
sin(n2 x2 ) n 2 x2
2 3/2 ≤ 2 3/2 = |x|1/2 .
n |x|
n |x|
Observe that x 7→ |x|1/2 is integrable near 0, but not near ∞. So we
need another estimate to take care of the growth for large x: | sin θ| ≤ 1
for any θ ∈ R, so
sin(n2 x2 ) 1
2 3/2 ≤ 2 3/2 ≤ |x|−3/2 .
n |x|
n |x|
Now |fn (x)| ≤ |x|1/2 χ(−1,1) (x) + |x|−3/2 χ(−∞,−1]∪[1,∞) (x), which is integrable and independent of n.
2
By the Dominated Convergence Theorem,
Z n
Z
Z
Z
sin(n2 x2 )
DCT
lim
dx = lim
fn dm =
lim fn dm = 0 dm = 0.
n→∞ −n n2 |x|3/2
n→∞
n→∞
3. (a) Let a, b ∈ R with a < b and let f, g : [a, b] → R be continuously differentiable. Use the Fundamental Theorem of Calculus to show that
Z b
Z b
0
f 0 (x)g(x) dx.
f (x)g (x) dx = f (b)g(b) − f (a)g(a) −
a
a
[Hint: what is
Rb
a
(f g)0 (x) dx?]
(b) Show that if f, g : R → R are continuously differentiable functions and
f (x)g(x) → L ∈ R as x → ∞, and f g 0 χ[0,∞) and f 0 gχ[0,∞) are both
integrable, then
Z ∞
Z ∞
0
f (x)g (x) dx = L − f (0)g(0) −
f 0 (x)g(x) dx.
0
0
R∞
(c) Fix a number t ∈ R. Show that 0 e−x cos(xt) dx = (1+t2 )−1 , justifying
your working.
[Hint: call the value of this integral J. Integrate by parts twice (justify
using (b)) to find an equation that J satisfies, and then solve it.]
R∞
(d) For t ∈ R, let I(t) = 0 x−1 e−x sin(xt) dx. Show that I 0 (t) = (1 + t2 )−1 ,
and solve this differential equation to find I(t).
Solution. (a) Since f and g are continuously differentiable, so is f g. By
the Fundamental Theorem of Calculus,
Z b
(f g)0 (x) dx = (f g)(b) − (f g)(a).
a
Moreover, (f g)0 = f g 0 + f 0 g by the product rule, and f g 0 and f 0 g are continuous, so are integrable on [a, b]. So
Z
b
0
Z
(f g) (x) dx =
a
b
Z
0
f (x)g (x) dx +
a
b
f 0 (x)g(x) dx.
a
Equating these two expressions and rearranging gives the result.
(b) If f : R → R and f χ[0,∞) is integrable then by Exercise 4.2,
Z ∞
Z ∞
Z n
Z
f (x) dx =
f (x)χ[0,∞) (x) dx = lim
f (x)χ[0,∞) (x) dx = lim
n→∞
−∞
0
n→∞
−n
For every n ≥ 1, by part (a) we have
Z n
Z
0
f (x)g (x) dx = f (n)g(n) − f (0)g(0) −
0
0
3
n
f 0 (x)g(x) dx.
0
n
f (x) dx.
Since f g 0 and f 0 g are integrable, taking the limit as n → ∞ gives
Z n
Z ∞
0
f 0 (x)g(x) dx
f (x)g (x) dx = lim f (n)g(n) − f (0)g(0) − lim
n→∞
n→∞
0
0
Z ∞
f 0 (x)g(x) dx.
= L − f (0)g(0) −
0
R∞
(c) Let J = 0 e−x cos(xt) dx. First take f (x) = e−x and g(x) = 1t sin(xt):
these are continuously differentiable, and |f g 0 | and |f 0 g| are both dominated
by e−x which is integrable on [0, ∞), so f g 0 χ[0,∞) and f 0 gχ[0,∞) are integrable;
so L = limx→∞ (e−x · 1t sin(xt)) = 0 = f (0)g(0) and g 0 (x) = cos(xt), so by (b)
Z ∞
Z ∞
Z ∞
−x 1
−1
−x
e−x sin(xt) dx.
(−e )· t sin(xt) dx = t
e cos(xt) dx = −
J=
0
0
0
Now take g(x) = − 1t cos(xt). Then g 0 (x) = sin(xt) and all of the integrability
and differentiability conditions of (b) are satisfied again. This time, L = 0
and f (0)g(0) = − 1t , so by (b) we have
Z ∞
Z
−x
1
e sin(xt) dx = 0 − (− t ) − (−e−x ) · (− 1t cos(xt)) dx
0
Z ∞
−1
=t
e−x cos(xt) dx = t−1 (1 − J).
1−
0
So J = t−2 (1 − J), which gives J =
t−2
1+t−2
=
1
.
1+t2
(d) We check that to find I 0 (t), we can interchange the order of differentiation
−x
and integration. Let f (x, t) = e sin(xt)
χ(0,∞) (x). We have:
x
• For all t ∈ R, f (x, t) is an integrable function of x. To see this, observe
that |e−x sin(xt)
| ≤ |e−x xt
| = |t|e−x , which is integrable.
x
x
• For x ∈ R we have
∂
f (x, t)
∂t
−x
• The function g(x) = e
=
e−x
x
· cos(xt) · x = e−x cos(xt);
is integrable and dominates
∂
f (x, t).
∂t
So the hypotheses of the theorem justifying the interchange of integration
and differentiation are satisfied, and by part (c),
Z
Z ∞
Z ∞
d ∞
∂
1
(c)
0
I (t) =
f (x, t) dx =
f (x, t) dx =
e−x cos(xt) dt =
.
dt 0
∂t
1 + t2
0
0
1
So I(t) is an antiderivative of 1+t
2 , so I(t) = arctan(t) + C for some conR∞
stant C. Since I(0) = 0 0 dx = 0, C = 0 so I(t) = arctan(t).
4. Let (an )n≥1 be a sequence in [−∞, ∞]. Prove the following assertions.
(a) lim sup(−an ) = − lim inf an
n→∞
n→∞
(b) an converges as n → ∞ if and only if lim inf an = lim sup an .
n→∞
n→∞
(c) If an → a as n → ∞ then a = lim inf an = lim sup an .
n→∞
4
n→∞
Solution. (a) If S ⊆ [−∞, ∞] and −S = {−x : x ∈ S} then sup(−S) =
− inf S. Applying this to the sets S = {ak : k ≥ n} gives
lim sup(−an ) = lim sup(−ak ) = lim −inf ak = − lim inf ak = − lim inf an .
n→∞
n→∞
n→∞
k≥n
n→∞
k≥n
k≥n
n→∞
(b) and (c) Suppose that an → a as n → ∞.
• If a ∈ R and ε > 0 then there is n0 ≥ 1 with |ak − a| ≤ ε for all k ≥ n0 .
So for n ≥ n0 we have |a − inf ak | ≤ ε and |a − sup ak | ≤ ε. Hence
k≥n
k≥n
inf ak → a and sup ak → a as n → ∞, so a = lim inf an = lim sup an .
k≥n
n→∞
k≥n
n→∞
• If a = ∞ and R ∈ R then there is n0 ≥ 1 with ak ≥ R for all k ≥ n0 .
So for n ≥ n0 we have inf ak ≥ R and sup ak ≥ R. Hence inf ak → ∞
k≥n
k≥n
k≥n
and sup ak → ∞ as n → ∞, so lim inf an = lim sup an = ∞.
n→∞
k≥n
n→∞
• If a = −∞ then −an → ∞ so lim inf (−an ) = lim sup(−an ) = ∞ which,
n→∞
n→∞
by (a), gives lim inf an = lim sup an = −∞.
n→∞
n→∞
Conversely, suppose that lim inf an = lim sup an and let a be this extended
n→∞
n→∞
real number.
• If a ∈ R and ε > 0 then there is n0 ≥ 1 so that |a − inf ak | ≤ ε and
k≥n
|a − sup ak | ≤ ε for all n ≥ n0 . So for all n ≥ n0 we have
k≥n
a − ε ≤ inf ak ≤ an ≤ sup ak ≤ a + ε,
k≥n0
k≥n0
so |a − an | ≤ ε for all n ≥ n0 . So an → a as n → ∞.
• If a = ∞ and R ∈ R then there is n0 ≥ 1 so that inf ak ≥ R for all
k≥n
n ≥ n0 . So for all n ≥ n0 we have R ≤ inf k≥n0 ak ≤ an , so an → a as
n → ∞.
• If a = −∞ and R ∈ R then there is n0 ≥ 1 so that sup ak ≤ R for all
k≥n
n ≥ n0 . So for all n ≥ n0 we have an ≤ supk≥n0 ak ≤ R, so an → −∞
as n → ∞.
5