MATH 6337: Homework 8 Solutions
6.1.
(a) Let E be a measurable subset of R2 such that for almost every x ∈ R, {y : (x, y) ∈ E}
has R-measure zero. Show that E has measure zero and that for almost every y ∈ R,
{x : (x, y) ∈ E} has measure zero.
(b) Let f (x, y) be nonnegative and measurable in R2 . Suppose that for almost every
x ∈ R, f (x, y) is finite for almost every y. Show that for almost every y ∈ R, f (x, y)
is finite for almost every x.
Solution.
(a) Consider the indicator function 1E on R2 . This is a nonnegative measurable function,
so by Fubini/Tonelli we have
ˆ ˆ
ˆ ˆ
¨
1E (x, y) dx dy.
1E (x, y) dy dx =
1E =
R2
R
R
R
R
For almost every x ∈ R, we have
ˆ
1E (x, y) dy = |{y : (x, y) ∈ E}| = 0,
R
so all the integrals in the first line above are 0. Since
¨
1E = |E| ,
R2
we have |E| = 0. Moreover, since
ˆ ˆ
ˆ
1E (x, y) dx dy =
|{x : (x, y) ∈ E}| dy
R
R
R
and since |{x : (x, y) ∈ E}| is a nonnegative function of y whose integral over y is
zero, it follows that |{x : (x, y) ∈ E}| = 0 for almost every y.
(b) Let E be the set of points on which f is not finite. This is a measurable subset of
R2 , and Ex = {y : (x, y) ∈ E} has measure zero for almost every x ∈ R, so by part
(a) the result immediately follows.
6.2. If f and g are measurable in Rn , show that the function h(x, y) = f (x)g(y) is measurable
in Rn × Rn . Deduce that if E1 and E2 are measurable subsets of Rn , then their Cartesian
product E1 × E2 = {(x, y) : x ∈ E1 , y ∈ E2 } is measurable in Rn × Rn , and |E1 × E2 | =
|E1 | |E2 |.
Solution. Fix a measurable function f , and let
Gf = {g : Rn → R, is measurable : f (x)g(y) is measurable} .
Let E ⊆ Rn be a measurable set, and let 1E be its indicator function. If h(x, y) = f (x)1E (y),
then

E c ∪ f −1 (a, ∞), a < 0
−1
h (a, ∞) =
E ∩ f −1 (a, ∞), a ≥ 0
is measurable for all a, so h is measurable; hence, Gf contains the indicator functions of measurable subsets of Rn . Moreover, if a, b ∈ R and g1 , g2 ∈ Gf , then f (x) · (ag1 (y) + bg2 (y)) =
af (x)g1 (y)+bf (x)g2 (y), a sum of measurable functions; hence, Gf is closed under linear combinations. Finally, if g1 , g2 , ... is an increasing sequence of functions in Gf which converges to
the function g, then f (x)g1 (y), f (x)g2 (y), ... is an increasing sequence of measurable functions
converging to f (x)g(y). Since a limit of measurable functions is measurable, f (x)g(y) ∈ Gf ;
hence, Gf is closed under increasing limits. Any collection of functions on Rn satisfying these
three properties contains all measurable functions on Rn , so Gf contains all measurable functions. Since this is true for all measurable f , it follows that f (x)g(y) is measurable for all
measurable f and g.∗
Alternatively, we could let F (x, y) = f (x) and G(x, y) = g(y). Then {(x, y) : F (x, y) > a} =
{x : f (x) > a} × Rn and {(x, y) : G(x, y) > a} = Rn × {y : g(y) > a}. These sets are measurable via repeated application of a lemma proved in Chapter 5 that E × R is measurable
for all measurable E ⊆ Rd .†
Given this result, take f = 1E1 and g = 1E2 . Then h(x, y) = 1E1 ×E2 (x, y) = f (x)g(y) is
measurable, so E1 × E2 is measurable. Moreover, via Fubini/Tonelli we have
¨
¨
ˆ
ˆ
|E1 × E2 | =
1E1 ×E2 =
1E1 (x)1E2 (y) dx dy =
1E1 (x) 1E2 (y) dy dx = |E1 | |E2 | .
R2
R2
R
R
∗
This is a common proof technique in analysis called a monotone class argument. It will come up again later
in the section on abstract measure theory.
†
Since the purpose of the problem is to show that the product of two general measurable sets is measurable,
you cannot just cite that result to say that F −1 (a, ∞) and G−1 (a, ∞) are measurable.
6.4. Let f be measurable and periodic with period 1: f (t + 1) = f (t). Suppose that there
is a finite c such that
ˆ 1
|f (a + t) − f (b + t)| dt ≤ c
0
for all a and b. Show that f ∈ L(0, 1). [Hint: Set a = x, b = −x, integrate with respect to
x, and make the change of variables ξ = x + t, η = −x + t.]
Solution.
ˆ
ˆ
1
c≥
1
|f (x + t) − f (−x + t)| dt dx
0
1
=
2
ˆ
0
0
1ˆ ξ
1
|f (ξ) − f (η)| dη dξ +
2
−ξ
ˆ
2
ˆ
1−ξ
|f (ξ) − f (η)| dη dξ.
1
ξ−1
By the periodicity of f , each integral will double in size by integrating η over the entire
interval [−1, 1]. Thus,
ˆ 2ˆ 1
|f (ξ) − f (η)| dη dξ ≤ 2c,
−1
0
so (again by periodicity)
ˆ
1
ˆ
1
c
|f (ξ) − f (η)| dη dξ ≤ .
2
0
0
Thus, |f (ξ) − f (η)| (hence f (ξ) − f (η)) is integrable over the square [0, 1] × [0, 1]. By the
result of Exercise 6.3, f is integrable over (0, 1).‡
‡
Proof of 6.3: ...
6.5.
(a) If f is nonnegative and measurable on E and ω(y) = |{x ∈ E : f (x) > y}|, y > 0,
´
´∞
use Tonelli’s theorem to prove that E f = 0 ω(y) dy. [Hint: By definition of
´
˜
the integral, we have E f = |R(f, E)| = R(f,E) dx dy. Use the observation in
the proof of (6.11) that {x ∈ E : f (x) ≥ y} = {x : (x, y) ∈ R(f, E)}, and recall that
ω(y) = |{x ∈ E : f (x) ≥ y}| unless y is a point of discontinuity of ω.]
(b) Deduce from this special case the general formula
ˆ
ˆ ∞
p
y p−1 ω(y) dy
(f ≥ 0, 0 < p < ∞).
f =p
0
E
Solution.
(a) Since ω is a monotone decreasing function on R, it has only countably many points
of discontinuity, so ω(y) = |{x ∈ E : f (x) ≥ y}| for almost all y. Thus,
ˆ ∞
ˆ ∞
ω(y) dy =
|{x ∈ E : f (x) ≥ y}| dy
0
0
ˆ ∞
|{x : (x, y) ∈ R(f, E)}| dy
=
0
ˆ ∞ˆ
dx dy
=
0
¨
R(f,E)y
ˆ
1 = |R(f, E)| =
=
R(f,E)
f.
E
˜
´∞´
(See Theorem (6.8) on page 90 of the text for why 0 R(f,E)y dx dy = R(f,E) 1.)
(b) Let ωf (y) be the distribution function for f , and let ωf p (y) be the same for f p . Then
we have
ˆ
ˆ ∞
ˆ ∞
p
p
f =
ωf (y) dy =
ωf (y 1/p ) dy.
E
0
0
Make the change of variables u = y . Then du = p1 y (p−1)/p = p1 up−1 , so
ˆ
ˆ ∞
p
up−1 ωf (u) du.
f =p
1/p
E
0
6.6. For f ∈ L(R), define the Fourier transform fb of f by
ˆ +∞
b
f (t)e−ixt dt
(x ∈ R1 ).
f (x) =
−∞
(For a complex-valued function F = F0 + iF1 whose real and imaginary parts F0 and F1 are
´
´
´
integrable, we define F = F0 + i F1 .) Show that if f and g belong to L(R), then
f[
∗ g(x) = fb(x)b
g (x).
Solution.
ˆ ˆ
f[
∗ g(x) =
R
ˆ
ˆ
−ixu
−ixu
f (u − t)g(t) dt e
du =
g(t)
f (u − t)e
du dt,
R
R
R
where the second equality follows from Fubini. Then we have
ˆ
ˆ
ˆ
ˆ
−ixu
−ixt
−ix(u−t)
g(t)
f (u − t)e
du dt =
g(t)e
f (u − t)e
du dt.
R
R
R
R
Changing variables v = u − t, we have
ˆ
ˆ
ˆ
ˆ
−ixv
−ixt
−ix(u−t)
−ixt
f (v)e
dv dt =
g(t)e
f (u − t)e
du dt =
g(t)e
R
R
R
R
ˆ
ˆ
−ixv
−ixt
f (v)e
dv = fb(x)b
g (x).
g(t)e
dt
R
R
6.10. Let vn be the volume of the unit ball in Rn . Show by using Fubini’s theorem that
ˆ 1
(1 − t2 )(n−1)/2 dt.
vn = 2vn−1
0
[We also observe that the integral can be expressed in terms of the Γ-function: Γ(s) =
´ ∞ −t s−1
e t dt, s > 0.]
0
Solution. We proceed by induction, noting a priori that v1 = 2. For n = 2, we have v2 = π,
which jibes with
ˆ 1√
π
2 · v1 ·
1 − t2 dt = 4 · = π.§
4
0
Now suppose the formula holds for the case n − 1. Recall that the n-ball B n is defined by
x21 + · · · + x2n ≤ 1. We have
˙
˙
ˆ 1˙
vn =
1=
dx1 · · · dxn =
dx1 dx2 · · · dxn
Bn
x21 +···+x2n ≤1
−1
x22 +···+x2n ≤1−x21
p
p
Defining yj = xj / 1 − x21 for j = 2, ..., n, and noting that dyj = dxj / 1 − x21 , we make a
change of variables:
ˆ 1˙
ˆ 1
2 (n−1)/2
(1 − x1 )
dx1 dy2 · · · dyn = 2vn−1
(1 − t2 )(n−1)/2 dt.
−1
2 ≤1
y22 +···+yn
0
Since all functions involved are bounded on the compact domain of integration, they are all
integrable, so Fubini/Tonelli justifies the swapping of integrals. The last inequality is justified
by the fact that the integrand is even and that y22 + · · · + yn2 ≤ 1 defines an (n − 1)-ball of
radius 1.
§In
case you’re not familiar with this integral, you can compute it with the trig substitution t = cos θ.
Geometrically, just observe that it is the area under the graph of the portion of the unit circle in the first
quadrant of the plane.
6.11. Use Fubini’s theorem to prove that
ˆ
2
e−|x| dx = π n/2 .
Rn
[Hint: For n = 1, write
´
+∞ −x2
e
−∞
dx
2
2
=
−x21
For n > 1, use the formula e−|x| = e
n = 1.]
´ +∞ ´ +∞
−∞
−∞
−x2n
···e
e−x
2 −y 2
dx dy and use polar coordinates.
and Fubini’s theorem to reduce to the case
Solution. Following the hint, in the case n = 1 we have
ˆ
2 ˆ ˆ
ˆ
ˆ 2π ˆ ∞
−r2
−x2
−x2 −y 2
e r dr dθ = π
e
dx =
e
dx dy =
R
R
0
R
0
∞
e−u du = π,
0
ˆ
so
2
e−x dx = π 1/2 .
R
Assume by induction that the proposition holds true for n − 1; we’ll show the case n.
Write x = (x1 , ..., xn ). Then we have, via induction and Fubini/Tonelli,
!
ˆ
ˆ
ˆ
ˆ Y
n
n
Y
2
2
2
2
e−|x| dx =
e−x1 dx1
e−xj dx2 · · · dxn =
e−xj =
Rn
Rn j=1
R
Rn−1 j=2
π 1/2 π (n−1)/2 = π n/2 .