Exercise 6
Miao Weimin, HT070504X, [email protected]
Wong Wei Pin, [email protected]
October 24, 2007
1. (a) Let E be a measurable subset of R2 such that for almost every x ∈ R1 , {y :
(x, y) ∈ E} has R1 -measure zero. Show that E has measure zero, and that for almost
every y ∈ R1 , {x : (x, y) ∈ E} has measure zero.
(b) Let f (x, y) be nonnegative and measurable in R2 . Suppose that for almost every
x ∈ R1 , f (x, y) is finite for almost every y. Show that for almost every y ∈ R1 , f (x, y)
is finite for almost every x.
Proof:(a) XE is a measurable function on R2 since E is a measurable subset of R2 .
R
Then, we have R XE dy = |{y : (x, y) ∈ E}| = 0 for almost every x ∈ R and hence
¢
R ¡R
X
dy
dx = 0. Moreover, from Tonelli’s Theorem, we have
E
R
R
RR
R2
Hence, |E| =
R
R2
XE dxdy =
R ¡R
R
R
¢
XE dy dx = 0.
XE dxdy = 0. Thus, for almost every y ∈ R, {x : (x, y) ∈ E} has
measure zero.
(b) Let E = {(x, y) : f (x, y) = ∞}. Since f (x, y) is measurable, E is a measurable
subset of R2 . Moreover, for almost every x ∈ R, f (x, y) is finite for almost every y,
that is, for almost every x ∈ R, {y : (x, y) ∈ E} has R-measure zero. From (a), we
obtain that for almost every y ∈ R, {x : (x, y) ∈ E} has measure zero. If follows that
f (x, y) is finite for almost every y.
1
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 |.
Proof: Let F (x, y) = f (x), G(x, y) = g(y). Since f (x) and g(y) are measurable in
Rn , it follows that F (x, y) and G(x, y) are measurable in R2n . Hence F (x, y)G(x, y)
is measurable in R2n , that is, h(x, y) = f (x)g(y) is measurable in R2n .
Let f (x) = XE1 (x), g(y) = XE2 (y), h(x, y) = XE1 ×E2 (x, y). By the definition of
Cartesian product, we have h(x, y) = f (x)g(y). From the above, h(x, y) is a measurable function and hence E1 × E2 is measurable. By Tonelli’s Theorem, we have
RR
|E1 × E2 | = E1 ×E2 h(x, y)dxdy
i
R hR
= E2 E1 XE1 (x)XE2 (y)dx dy
hR
i
R
= E2 XE2 (y) E1 XE1 (x)dx dy
hR
i
= |E1 | E2 XE2 (y)dy = |E1 ||E2 |.
3. Let f be measurable on (0, 1). If f (x) − f (y) is integrable over the square 0 ≤ x ≤
1, 0 ≤ y ≤ 1, show that f ∈ L(0, 1).
Proof: Since f is measurable on (0, 1), then f (x) − f (y) is measurable on [0, 1] × [0, 1],
so is |f (x) − f (y)|. [In fact, define F1 (x, y) = f (x), as proved in Exercise 2, we have
F1 (x, y) is measurable on [0, 1] × [0, 1], so is F2 (x, y) = f (y) and f (x) − f (y) =
F1 (x, y) − F2 (x, y).] From Fubini’s Theorem, we have
¸
Z 1 ·Z 1
ZZ
|f (x) − f (y)|dx dy =
|f (x) − f (y)|dxdy < +∞.
0
0
[0,1]×[0,1]
Then we have for almost every y,
Z 1
|f (x) − f (y)|dx < +∞.
0
R1
Moreover, we can choose y such that |f (y)| 6= ∞, and 0 |f (x) − f (y)|dx < +∞.
R1
Clearly 0 |f (y)|dx = |f (y)| < ∞, and hence
Z 1
Z 1
Z 1
|f (x)|dx ≤
|f (x) − f (y)|dx +
|f (y)|dx < +∞.
0
0
0
2
It follows that |f | ∈ L(0, 1). Therefore f ∈ L(0, 1).
4. Let f be measurable and periodic with period 1 : f (t + 1) = f (t). Suppose that
there is a finite c such that
R1
0
|f (a + t) − f (b + t)|dt ≤ c
for all a and b. Show that f ∈ L(0, 1). (Set a = x, b = −x, integrate with respect to x,
and make the change of variable ξ = x + t, η = −x + t.)
Proof: Method 1: Since f is measurable, from Lemma 6.15, we have f (x + t) and
f (−x + t) are both measurable in R2 . Then by letting a = x, b = −x, from Tonelli’s
Theorem, we have
RR
|f (x + t) − f (−x + t)|d(t, x) =
[0,1]2
R 1 hR 1
0
0
i
|f (x + t) − f (−x + t)|dt dx ≤ c < +∞.
Moreover,
consider
à !
à !the transformation
Ã
! ξ = x + t, ηÃ= −x +!t, denoted by T , that is,
ξ
x
x+t
1 1
=T
=
. Thus, T =
and then T −1 [0, 2]2 =
η
t
−x + t
−1 1
convex hull of ((0, 0), (1, 1), (0, 2), (−1, 1)). Since f (t + 1) = f (t), from Exercise 20
in Chapter 5, we obtain
RR
[0,1]2
|f (ξ) − f (η)|d(ξ, η) =
=
=
=
RR
1
|f (ξ) − f (η)|d(ξ, η)
4
[0,2]2
RR
1
|
det
T
|
|f (x + t) − f (−x + t)|d(t, x)
4
T −1 [0,2]2
RR
1
|f (x + t) − f (−x + t)|d(t, x)
2
[0,1]×[0,2]
RR
[0,1]2
|f (x + t) − f (−x + t)|d(t, x)
< ∞.
Thus |f (ξ) − f (η)| is integral over the square [0, 1] × [0, 1], so is f (ξ) − f (η). Then
from Exercise 3, we have f ∈ L(0, 1).
Method 2: Choose a = x, b = 0. Since f is measurable, f (x + t) is measurable, and
so is f (x + t) − f (t). Then we have
Z 1
|f (x + t) − f (t)|dt < c.
0
3
From Tonelli’s Theorem, we have
¸
Z 1 ·Z 1
Z
|f (x + t) − f (t)|dx dt =
0
0
Then,
R1
0
1
·Z
¸
1
Z
1
|f (x + t) − f (t)|dt dx <
0
0
cdx = c.
0
|f (x + t) − f (t)|dx < ∞ for almost every t. Thus
R1
0
|f (x + t)|dx < +∞
for almost every t. Since f is periodic with period 1: f (t + 1) = f (t). Choose one t
R1
R1
and we have 0 |f (x + t)|dx = 0 |f (x)|dx < +∞. Thus, |f | ∈ L(0, 1). Therefore,
f ∈ L(0, 1).
5. (a) If f is nonnegative and measurable on E and ω(t) = |{x ∈ E : f (x) > t}|,
R
R∞
y > 0, use Tonelli’s theorem to prove that E f = 0 ω(y)dy. [By definition of the
R
RR
integral, we have E f = |R(f, E)| =
dxdy. Use the observation in the
R(f,E)
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
R
fp = p
E
R∞
0
y p−1 ω(y)dy
(f ≥ 0, 0 < p < ∞).
Proof: (a) By definition, we have
{x ∈ E : f (x) ≥ y} = {x : (x, y) ∈ R(f, E)}.
Moreover, ω(y) = |{x ∈ E : f (x) > y}| = |{x ∈ E : f (x) ≥ y}| unless y is a point
of discontinuity of ω. Note that ω(y) is a decreasing function on (0, ∞). Thus ω(y)
R∞
R∞
can only have countable discontinuities. Hence we have 0 ω(y)dy = 0 |{x ∈ E :
f (x) ≥ y}|dy. By the definition of integral, we have
R
f = |R(f, E)| =
E
RR
d(x, y)
R(f,E)
i
R ∞ hR
= 0
dx dy
{x:(x,y)∈R(f,E)}
i
R ∞ hR
= 0
dx
dy
{x∈E:f (x)≥y}
R∞
= 0 |{x ∈ E : f (x) ≥ y}|dy
R∞
= 0 ω(y)dy.
This completes the proof.
4
(b) For all 0 < p < ∞, we have ∀x ∈ E, f (x)p =
Z
Z Z f (x)
p
f (x)dx =
p tp−1 dtdx
E
ZE 0
=
p tp−1 d(x, t)
R(f,E)
Z ∞Z
=
p tp−1 dxdt
0
{x∈E : f (x)≥t}
Z ∞
=
w(t)p tp−1 dt
R f (x)
0
p tp−1 dt, thus
by Tonelli’s Theorem
by Tonelli’s Theorem
0
as w(t) = |{x ∈ E : f (x) ≥ t}| a.e.
6. For f ∈ L(R1 ), define the Fourier transform fˆ of f by
fˆ(x) =
R +∞
−∞
f (t)e−ixt dt
(x ∈ R1 ).
(For a complex-valued function F = F0 + iF1 whose real and imaginary parts F0 and
R
R
R
F1 are integrable, we define F = F0 + i F1 .) Show that if f and g belong to
L(R1 ), then
\
(f
∗ g)(x) = fˆ(x)ĝ(x).
Proof: Since f and g belong to L(R), f (t − s)g(s)e−ixt is measurable in R2 . Consider
the measurable function |f (t − s)g(s)e−ixt |. From Tonelli’s Theorem, by considering
the transformation u = t − s, v = s, we have
R
R
−ixt
|f
(t
−
s)g(s)e
|d(t,
s)
=
|f (u)g(v)e−ix(u+v) |d(u, v)
2
R
R2
£R
¤ £R
¤
= R |f (u)|du R |g(v)|dv .
Hence, |f (t − s)g(s)e−ixt | ∈ L(R2 ). Using Fubini’s Theorem, on one hand, we have
¤ −ixt
R
R £R
−ixt
[f
(t
−
s)g(s)e
]
dtds
=
f
(t
−
s)g(s)ds
e dt
2
R
RR R
−ixt
= R (f ∗ g)(t)e dt
\
= (f
∗ g)(x).
On the other hand, consider the transformation u = t − s, v = s, then we have
¤
R
R £
−ixt
−ix(u+v)
dudv
[f
(t
−
s)g(s)e
]
dtds
=
f
(u)g(v)e
2
2
R
¤
¤ £R
£RR
−ixu
du R g(v)e−ixv dv
= R f (u)e
= fˆ(x)ĝ(x).
5
Thus we complete the proof.
7. Let F be a closed subset of R1 and let δ(x) = δ(x, F ) be the corresponding distance
function. If λ > 0 and f is nonnegative and integrable over the complement of F , prove
that the function
R
δ λ (y)f (y)
dy
R1 |x−y|1+λ
is integrable over F , and so is finite a.e. in F. [In case f = X(a,b) , this reduces to (6.17).]
Proof: Let G = R − F . Since δ(y) = 0, y ∈ F , from Tonelli’s Theorem, we have
i
i
hR
i
R hR δλ f (y)
R hR δλ f (y)
R λ
dx
dy dx = F G |x−y|1+λ dy dx = G δ (y)f (y) F |x−y|1+λ dy.
F
R |x−y|1+λ
From the proof of Theorem 6.17 (Marcinkiewicz), we have
R
dx
F |x−y|1+λ
≤ 2λ−1 δ(y)−λ .
Moreover, since f is nonnegative and integrable over G, we have
i
R λ
R
R hR δλ f (y)
−1
−λ
−1
dy
dx
≤
δ
(y)f
(y)[2λ
δ(y)
]dy
=
2λ
f (y)dy < +∞.
1+λ
R |x−y|
F
G
G
Thus we complete the proof.
8. Under the hypotheses of (6.17) and assume that b − a < 1, prove taht the function
M0 (x) =
Rb
a
[log 1/δ(y)]−1 |x − y|−1 dy
is finite a.e. in F .
Proof: Let G = (a, b) − F . Since δ(y) = 0, y ∈ F , from Tonelli’s Theorem, we have
µ
¶
i−1
R
R R h
1
−1
M0 (x)dx = F
log δ(y)
|x − y| dy dx
F
G
i−1 £R
¤
R h
1
−1
|x
−
y|
dx
dy.
= G log δ(y)
F
Fix y ∈ G and note that for any x ∈ F , combining with b − a < 1, we have 1 >
|x − y| ≥ δ(y) > 0. Thus we have
R
F
|x − y|−1 dx ≤
R
δ(y)≤|x−y|<1
|x − y|−1 dx < 2
6
R1
δ(y)
1
t−1 dt = 2 log δ(y)
.
Then we obtain that
R
F
M0 (x) ≤
R h
G
log
1
δ(y)
i−1 h
i
1
2 log δ(y) dy = 2|G| < +∞.
It follows immediately that M0 (x) is finite a.e. in F .
9. Show that Mλ (x, F ) = +∞ if x ∈
/ F, λ > 0.
Proof: Let G = (a, b) − F . Since δ(y) = 0, y ∈ F , from Tonelli’s Theorem, we have
Mλ (x) = Mλ (x; F ) =
R
δ λ (y)
dy.
G |x−y|1+λ
If x ∈
/ F , i.e., F ∈ G, since G is an open set, there exist ε such that the neighborhood
(x − ε, x + ε) ∈ G and inf y∈(x−ε,x+ε) δ(y) = m > 0. Since
have
R
δ λ (y)
dy
G |x−y|1+λ
≥
≥
≥
=
δ λ (y)
|x−y|1+λ
are nonnegative, we
R x+ε
δ λ (y)
dy
x−ε |x−y|1+λ
R
mλ |x−y|<ε |x−y|1 1+λ dy
Rε 1
2mλ 0 t1+λ
dt
λ −1 −λ ε
2m λ t |0
= +∞.
Thus we complete the proof.
10. Let vn be the volume of the unit ball in Rn . Show by using Fubini’s Theorem that
vn = 2vn−1
R1
0
(1 − t2 )(n−1)/2 dt.
(We also observe that the integral can be expressed in term of the Γ-function: Γ(s) =
R ∞ −t s−1
e t dt, s > 0.]
0
Proof: Let Brn (0) = {x ∈ Rn : |x| ≤ r}, then
vn = |B1n (0)| = |{xi ∈ R : x21 + · · · + x2n ≤ 1}|
Z
=
d(x1 , . . . , xn )
Z
{xi ∈R: x21 +···+x2n ≤1}
1 Z
=
−1
Z 1
=
−1
{x22 +···+x2n ≤1−x21 }
n−1
|B√
1−x21
d(x2 , . . . , xn )dx1
(0)|dx1
7
by Tonelli’s Theorem
For all |x1 | ≤ 1, consider invertible linear transformation
n−1
then |B√
1−x21
Tx1 : Rn−1 −→ Rn−1
p
,
x
7−→ x 1 − x21
(0)| = |Tx1 (B1n−1 (0))| = | det(Tx1 )||B1n−1 (0)| = (1 − x21 )
n−1
2
vn−1 . Thus,
from above calculation, we have
Z 1
Z 1
n−1
2 n−1
2
vn =
(1 − x1 ) vn−1 dx1 = 2vn−1
(1 − x21 ) 2 dx1
−1
0
11. Use Fubini’s theorem to prove that
R
2
e−|x| dx = π n/2 .
Rn
R +∞
R +∞ R +∞
2
2
2
[For n = 1, write ( −∞ e−x dx)2 = −∞ −∞ e−x −y dxdy and use polar coordinates.
2
2
2
For n > 1, use the formula e−|x| = e−x1 · · · e−xn and Fubini’s theorem to reduce to the
case n = 1.]
Proof: Let BR (0) = {(x, y) ∈ R2 : x2 + y 2 ≤ R2 }, for R > 0. Since e−x
−x2 −y 2
continous on R2 ⊃ BR (0), e
2 −y 2
is
is Riemann Integrable on BR (0) and by subdividing
this region into small area in the sense of polar coordonate and taking the limit when
these area tends to zero, we have
Z
Z
−x2 −y 2
χBR (0) (x, y)e
d(x, y) =
R2
−x2 −y 2
Since χBN (0) (x, y)e
%e
Z
−x2 −y 2
e
0
R
2
N →∞
e−x
2 −y 2
0
d(x, y) = π
BN (0)
On the other hand, by Tonelli’s Theorem, we have
µZ
¶2
Z
Z Z
−x2 −y 2
−x2 −y 2
−x2
π=
e
d(x, y) =
e e dxdy =
e dx
R2
R
Z
R
R
2
e−x dx =
=⇒
√
π
R
2
2
2
When n > 1, e−|x| = e−x1 · · · e−xn . By applying Tonelli’s Theorem repeatedly, we
have
R
Rn
2
e−|x| dx =
R
2
R
e−x1 dx1 · · ·
Thus we complete the proof.
8
R
2
R
n
e−xn dxn = π 2 .
2
e−r rdrdθ = π(1−e−R )
for N ∈ N,N → ∞. Then by monotone conver-
gence theorem of nonnegative function, we have
Z
Z
−x2 −y 2
e
d(x, y) = lim
R2
Z
d(x, y) =
BR (0)
−x2 −y 2
2π