Problem 1. The strategy is to first integrate |f |, using Tonelli’s theorem. Then if the integral
is finite, we can apply Fubini’s Theorem.
(i) We have
Z
Z 1 Z y 2
Z 1 2
y − x2
x − y2
dx
+
dx
dy
|f | dxdy =
2
2 2
2
2 2
y (x + y )
0
0 (x + y )
Z 1
x y
x 1 −
=
x2 + y 2 0 x2 + y 2 y
0
Z 1
1
1 =
−
dy = ln y 01 − arctan y|10 = ∞
2
y 1+y
0
so we can’t apply Fubini. Moreover
Z 1
Z 1Z 1 2
Z 1
1
x − y2
x 1
π
1
dy
=
−
dxdy
=
−
=
−
arctan
y|
=
−
0
2
2 2
2
2
2
4
0 1+y
0
0 (x + y )
0 x +y 0
and
Z
1
0
Z
0
1
x2 − y 2
dydx =
(x2 + y 2 )2
Z
0
1
y 1
dx =
x2 + y 2 0
Z
0
1
1
π
= arctan y|10 = .
2
1+x
4
(ii) We divide the problem into three cases.
a < 2, a 6= 1. We have
Z
Z 1Z 1
Z 1
Z 1
1 − xy 1−a 1
1
(1 − y)1−a 1 −a
|f | dxdy =
(1 − xy) dxdy =
−
dy
dy =
y(a − 1) 0
a−1 0
y
y
0
0
0
Z
Z 1
1 h 1/2 (1 − y)1−a 1 (1 − y)1−a 1 i
−
dy +
−
dy
a−1 0
y
y
y
y
1/2
moreover the function in the first integral is bounded since
(1 − y)1−a 1 1 − (1 − a)y + O(y 2 ) − 1
−
= lim
=a−1
lim
y→0
y→0
y
y
y
and the denominator in the second interval is ≥ 1/2 so that
Z
|f | dxdy ≤ C < ∞
which allow us to apply Fubini and conclude that all the integrals are equal and
finite.
a = 1. We have
Z
Z 1Z 1
Z 1
Z 1
ln(1 − xy) 1
ln(1 − y
−1
|f | dxdy =
(1 − xy) dxdy =
dy
dy = −
−y
y
0
0
0
0
0
Z 1/2
Z 1
ln(1 − y)
ln(1 − y)
=−
dy −
dy < ∞
y
y
0
1/2
because
ln(1 − y)
−1
= lim
= −1
y→0
y
1−y
which makes the first integral bounded, while for the second we have
Z 1
Z 1
ln(1 − y)
dy ≤ −2
ln(1 − y) dy = ln(2) + 1 < ∞
−
y
1/2
1/2
lim
y→0
which allow us to apply Fubini and conclude that all the integrals are equal and
finite.
1
2
a ≥ 2. Proceeding as in the first case we split the integral in two parts, one in (0, 1/2) the
other in (1/2, 1), and we observe that for the second part
Z 1
Z 1
(1 − y)1−a 1 −
dy ≥
((1 − y)1−a − 1) dy = ∞
y
y
1/2
1/2
so by Tonelli all three integrals are infinite.
(iii) We have
Z Z
Z 1/2 Z 1/2−x Z 1 Z x−1/2 −3
1
1 −3
|f | dxdy =
−x
x−
dydx +
dydx
2
2
0
0
1/2 0
Z 1/2 Z 1
−2
1
1 −2
−x
x−
=
dx +
dx
2
2
0
1/2
−1 1/2 1
1 −1 1
−x
=
− x−
=∞
2
2
1
1/2
so we can’t apply Fubini. Furthermore
Z Z
Z 1/2 Z 1/2−x Z 1 Z x−1/2 1 −3
1 −3
x−
x−
f dydx =
dydx +
dydx = ∞ − ∞
2
2
0
0
1/2 0
is not well defined and
Z Z
Z 1/2 Z
f dxdy =
Z 1/2 Z 1 1 −3
1 −3
dxdy +
x−
dydx
x−
2
2
0
1/2−y
0
0
Z 1/2 Z
i
1 h 1/2 1
1
1
1
=−
−
dy
+
−
dy = −∞
2 0
y2 4
4 (y − 1)2
0
1/2−y Problem 2. By Hölder’s inequality, we have
Z
sup f g ≤ kgkq .
f
So we need only show that
Z
sup
f g ≥ kgkq .
f
For p = ∞, let
f = sgn(g).
(So f (x) = 1 if g(x) is positive, f (x) = −1 if g(x) is negative, and f (x) = 0 if g(x) = 0.) Then
kf k∞ = 1, yet
Z
Z
f g = |g| = kgk1 .
This demonstrates the claim for p = ∞.
For p ∈ (1, ∞), we follow the hint, and note that
Z
Z
p
|fn | = |gn |q → kgkqq ,
by monotone convergence. Likewise,
R
|gn |q ≤ kgkqq for all n. Hence if
φn :=
fn
q/p
kgkq
then
kφn kp ≤ 1.
,
3
Yet
Z
φn g =
Z
1
q/p
|g||gn |
q/p
kgkq
→
=
1
Z
q/p
kgkq
1
q/p
kgkq
Z
|g|1+q/p
|g|q
= kgkq .
= kgkq−q/p
q
This demonstrates the claim for p ∈ (1, ∞).
For p = 1, choose some number N < kgk∞ , and let EN = λ{x : g(x) ≥ N }, and
ψN = 1EN /λ(EN ).
Then kψN k1 = 1, and
Z
ψN g ≥ N.
Hence
Z
ψN g ≥ kgk∞ .
lim
N %kgk∞
This demonstrates the claim for p = 1.
Problem 3. Define A := {(x, t) : 0 ≤ t ≤ f (x)} ⊂ X × R+ . We have
Z
Z Z f (x)
f dµ =
dt dµ
X
X 0
Z Z
1A (x, t) dt dµ.
=
X
R+
As 1A ≥ 0, we may apply Fubini’s theorem; this quantity is equal to
Z
Z Z
µ{x : f (x) > t} dt,
1A (x, t) dx dt =
R+
X
R+
as claimed.
Problem 4. We can write a function f : N × N 3 (n, m) 7→ f (n, m) =: anm R, then the double
integral becomes a double series, that is
Z Z
∞
X
f (n, m) dndm =
anm .
N
N
n,m=1
Now Tonelli’s theorem tells us that if anm ≥ P
0 for every n, m ∈ N, then we can exchange the
series freely. Fubini’s Theorem states the if ∞
n,m=1 |anm | < ∞, then again we can exchange
the series freely.
An example in which neither of these results apply is given by the sequence anm = 1 if n = m,
anm = −1 if n = m + 1 and anm = 0 otherwise. Then one easily sees that
∞
X
anm = 0
n=1
for every m ∈ N, and
∞
X
anm = 0
m=1
if n ≥ 2,
∞
X
m=1
a1m = 1.
4
This implies that
∞ X
∞
X
anm = 0 6= 1 =
m=1 n=1
∞ X
∞
X
anm .
n=1 m=1
This doesn’t contradict Fubini theorem since
∞ X
∞
X
|anm | = ∞.
m=1 n=1
Problem 5. Suppose g = 1[0,1] a.e., and let us try to find a contradiction. Note that because
g is continuous,
Z h
Z h
1
1
1
g(0) = lim
g(x) dx = lim
1[0,1] (x) dx = .
h→0 2h −h
h→0 2h −h
2
But then g(−) = 0 by an analogous argument for all > 0 which is a contradiction.
Problem 6. We write
Z A
Z
−1
x sin x dx =
0
0
A
Z
∞
−xt
sin xe
0
Z
∞
Z
dtdx =
0
A
sin xe−xt dxdt
0
by Fubini’s theorem. Now the inner integral is (1 + t2 )−1 + O(e−At ) so
Z A
Z ∞
1
π
−1
x sin x dx =
dt
=
1 + t2
2
0
0
as by the dominated convergence theorem.