Folland: Real Analysis, Chapter 3
Sébastien Picard
Problem 3.3
Let ν be a signed measure on (X, M).
a. L1 (ν) = L1 (|ν|).R
R
b. If f ∈ L1 (ν), | f dν| ≤ R|f |d|ν|.
c. If E ∈ M, |ν|(E) = sup{| E f dν| : |f | ≤ 1}
Solution:
(a)
R
R
1 +
1 −
+
LetRf ∈ L1 (ν). RBy definition,
f
∈
L
(ν
)
∩
L
(ν
).
Therefore
|f
|dν
<
∞
and
|f |dν − < ∞.
R
1
Hence |f |d|ν| = |f |dν + + |f |dν − < ∞
R and so f R∈ L (|ν|).
R
R
1
+
−
Consersely,
suppose
f
∈
L
(|ν|).
Then
|f
|d|ν|
=
|f
|dν
+
|f
|dν
<
∞.
Therefore
|f |dν + <
R
∞ and |f |dν − < ∞. Hence f ∈ L1 (ν + ) ∩ L1 (ν − ) and so f ∈ L1 (ν).
(b)
Let f ∈ L1 (ν). Then
Z
Z
Z
f dν −
Z
Z
+
f dν −
f dν +
≤
Z
Z
+
≤ |f |dν + |f |dν −
Z
= |f |d|ν|
f dν =
+
f dν −
(c)
Let E ∈ M. Then if f is a measurable function such that |f | ≤ 1, we have
Z
Z
Z
f dν ≤
|f |d|ν| ≤
d|ν| = |ν|(E).
E
E
E
R
It follows that sup{| E f dν| : |f | ≤ 1} ≤ |ν|(E).
On the other hand, let X = P ∪ N be a Hahn decomposition for ν. Since ν + (E) = ν(E ∩ P ) and
ν − (E) = −ν(E ∩ N), we can write
1
|ν|(E) = ν + (E) + ν − (E)
= ν(E ∩ P ) − ν(E ∩ N)
Z
Z
=
χP dν −
χN dν
E
E
Z
= (χP − χN )dν
E
Z
=
(χP − χN )dν
E
Therefore |ν|(E) = |
R
E
gdν| where g = χP − χN . Since |g| ≤ 1, we must have
Z
|ν|(E) ≤ sup{| f dν| : |f | ≤ 1}.
E
Problem 3.7
Suppose that ν is a signed measure on (X, M) and E ∈ M.
−
(a) ν + (E) = sup{ν(F
Pn ) : F ∈ M, F ⊂ E} and ν (E) = − inf{ν(F ) :nF ∈ M, F ⊂ E}
(b) |ν|(E) = sup{ 1 |ν(Ej )| : n ∈ N, E1 , . . . , En are disjoint, and ∪1 Ej = E}
Solution:
Let X = P ∪N be a Hahn decomposition for ν. Then ν + (E) = ν(E ∩P ) and ν − (E) = −ν(E ∩N).
(a)
We prove the first statement. Let F ∈ M, F ⊆ E. Then
ν(F ) = ν + (F ) − ν − (F ) ≤ ν + (F ) ≤ ν + (E).
It follows that sup{ν(F ) : F ∈ M, F ⊆ E} ≤ ν + (E).
On the other hand, ν + (E) = ν(E ∩ P ), and E ∩ P ∈ M, E ∩ P ⊆ E. So ν + (E) ≤ sup{ν(F ) :
F ∈ M, F ⊆ E}. Therefore
ν + (E) = sup{ν(F ) : F ∈ M, F ⊆ E}.
We now prove the second statement. Let F ∈ M, F ⊆ E. Then
ν(F ) = ν + (F ) − ν − (F ) ≥ −ν − (F ) ≥ −ν − (E).
Since ν − (E) ≥ −ν(F ) for all F ∈ M, F ⊆ E, then
ν − (E) ≥ sup{−ν(F ) : F ∈ M, F ⊆ E} = − inf{ν(F ) : F ∈ M, F ⊆ E}.
On the other hand, ν − (E) = −ν(E ∩ N), and E ∩ N ∈ M, E ∩ N ⊆ E. So
2
ν − (E) ≤ sup{−ν(F ) : F ∈ M, F ⊆ E} = − inf{ν(F ) : F ∈ M, F ⊆ E}.
Therefore
ν − (E) = − inf{ν(F ) : F ∈ M, F ⊆ E}.
(b)
Let n ∈ N, E1 , . . . , En disjoint measurable sets and ∪n1 Ej = E. Then
n
X
j=1
|ν(Ej )| =
≤
=
n
X
j=1
n
X
j=1
n
X
j=1
|ν + (Ej ) − ν − (Ej )|
|ν + (Ej ) + ν − (Ej )|
|ν|(Ej )
= |ν|(E)
where the last step follows from the fact that |ν| is a measure. Therefore
sup{
n
X
1
|ν(Ej )| : n ∈ N, E1 , . . . , En are disjoint, and ∪n1 Ej = E} ≤ |ν|(E).
On the other hand, we have
|ν|(E) = ν + (E) + ν − (E) = |ν(E ∩ P )| + |ν(E ∩ N)|.
Since E ∩ P and E ∩ N are disjoint and their union is E, then
|ν|(E) ≤ sup{
n
X
1
|ν(Ej )| : n ∈ N, E1 , . . . , En are disjoint, and ∪n1 Ej = E}.
Problem 3.17
1
Let (X, M, µ) be a σ−finite measure space, N a sub-σ-algebra
R of M,R and ν = µ|N . If f ∈ L (µ),
1
there exists g ∈ L (ν) (thus g is N -measurable) such that E f dµ = E gdν for all E ∈ N ; if g ′ is
another such function then g = g ′ ν-a.e. (In probability theory, g is called the conditional expectation
of f on N )
Solution:
Let (X, M, µ) be a σ−finite measure space, N a sub-σ-algebra of M, ν = µ|N , and f ∈ L1 (µ).
3
R
We define λ(E) := E f dµ to be a signed measure on (X, N ). The fact that λ is a signed measure
is explained in the first paragraph on page 86, and follows from the fact that at least one of f + dµ
and f − dµ are finite (indeed, both are finite since f ∈ L1 (µ)).
Let A ∈ N . If ν(A) = 0, then µ(A) = ν(A) = 0, hence λ(A) = 0. It follows that λ << ν.
By the
R Radon-Nikodym theorem, there exists an extended ν-measurable function g such that
λ(E) = E gdν and any two such functions are equal ν-a.e.. It only remains to show that g ∈ L1 (ν).
This is clear since
Z
Z
Z
f dµ ≤ |f |dµ < ∞,
gdν =
hence
Z
gdν =
Z
+
g dν −
Z
g − dν < ∞.
R
R
Since g is an extended ν-measurable function, one of g + dν, g − dν is finite. Since their difference
is finite, they must both be finite. Therefore
Z
Z
Z
+
|g|dν = g dν + g − dν < ∞.
Problem 3.21
Let ν be a complex measure on (X, M). If E ∈ M, define
µ1 (E) = sup{
n
X
µ2 (E) = sup{
∞
X
1
1
|ν(Ej )| : n ∈ N, E1 , . . . , En disjoint, E =
n
[
Ej }
|ν(Ej )| : n ∈ N, E1 , . . . , En disjoint, E =
∞
[
Ej }
µ3 (E) = sup{
Z
E
1
1
f dν : |f | ≤ 1}.
Then µ1 = µ2 = µ3 = |ν|.
S
Solution: First, we show that µ1 ≤ µ2 . Let A1 , . . . , An be disjoint sets such that E = n1 Aj . Define
an infinite collection {Fi }∞
way: for any i ∈ N, let Fi = Ai if 1 ≤ i ≤ n, Fi = ∅ if
i=1 in the followingS
i > n. Then F1 , F2 , . . . are disjoint and E = ∞
1 Fj , hence
n
X
1
|ν(Aj )| =
∞
X
1
|ν(Fj )| ≤ sup{
∞
X
1
|ν(Ej )| : n ∈ N, E1 , . . . , En disjoint, E =
∞
[
1
Ej }.
It follows that µ1 ≤ µ2 .
Next, we show that µ2 ≤ µ3 . Let A1 , A2 , . . . be disjoint sets such that E = ∪Aj . Then
4
∞
X
1
|ν(Aj )| ≤
∞
X
1
|ν|(Aj )
Proposition 3.13a
= |ν|(E)
Z
=
d|ν|
ZE
dν
=
d|ν|
Proposition 3.13b
E d|ν|
Z
dν dν
=
d|ν|
E d|ν| d|ν|
Z
dν
dν
Proposition 3.9a
=
E d|ν|
Z
≤ sup{
f dν : |f | ≤ 1}
E
Hence µ2 ≤ µ3 .
Next, we show µ3 ≤ µ1 . Let f be a measurable function such that |f | ≤ 1. By Theorem 2.10,
there is a sequence {φn } of simple functions such that 0 ≤ |φ1 | ≤ |φ2| ≤ · · · ≤ 1, φn →Rf pointwise.
Since
ν is a complex
measure, we
know νr+ , νr− , νi+ , νi− are finite positive measures, so 1dνr+ < ∞,
R
R
R
−
1dνr− < ∞, 1dνi+ < ∞, and 1dν
i < ∞. By the Dominated Convergence Theorem, for all ǫ > 0
P
there exists a simple function φN = n1 ai χAi where |ai | ≤ 1 and Ai disjoint such that
Z
Z
Z
E
−
f dνr−
Z
+i
f dνi+
Z
− i f dνi−
E
E
ZE
Z
Z E
Z
+
−
+
≤
φN dνr + ǫ −
φN dνr + ǫ + i φN dνi + iǫ + −i φN dνi− + ǫi
E
E
E
E
Z
≤ 4ǫ +
φN dν
f dν =
f dνr+
E
Z X
n
ai χAi )dν
= 4ǫ +
(
E
= 4ǫ +
= 4ǫ +
1
n
X
1
n
X
1
≤ 4ǫ +
n
X
1
E
χAi dν
E
ai ν(Ai ∩ E)
|ν(Ai ∩ E)| ≤ 4ǫ + |ν(
Since A1 ∩ E, . . . , An ∩ E, (
Z
ai
Z
T
f dν ≤ sup{
n
\
1
Aci ∩ E)| +
n
X
1
|ν(Ai ∩ E)|.
Aci ∩ E) are disjoint and their union is equal to E, we have
n
X
1
|ν(Ej )| : n ∈ N, E1 , . . . , En disjoint, E =
5
n
[
1
Ej }.
This proves that µ1 = µ2 = µ3 . To complete the proof we will show that µ3 = |ν|.
For any measurable function f such that |f | ≤ 1, we have
Z
Z
| f dν| ≤
|f |d|ν| ≤ |ν|(E).
E
E
Hence µ3 ≤ |ν|. On the other hand, if we let f = dν/d|ν|, we can redefine f on a set of |ν| measure
zero such that |f | = 1. Then
Z
Z
Z
dν
¯
¯
f
f dν =
d|ν| = |ν|(E).
d|ν| =
E d|ν|
E
E
Therefore
Z
|ν|(E) ≤ sup{
E
f dν : |f | ≤ 1}.
Hence |ν| = µ3 .
Problem 3.22
If f ∈ L1 (Rn ), f 6= 0, there exist C, R > 0 such that Hf (x) ≥ C|x|−n for |x| > R. Hence m({x :
Hf (x) > α}) ≥ C ′ /α when α is small, so the estimate in the maximal theorem is essentially sharp.
Solution:
Since f 6= 0, there exists a R > 1 such that
Z
B(R,0)
|f | > c1 > 0.
If x > |R|, then
1
Hf (x) ≥
m(B(2|x|, x))
1
|f (y)|dy ≥
m(B(2|x|, x))
B(2|x|,x)
Z
Z
B(R,0)
|f (y)|dy >
c1 |x|−n
= C|x|−n ,
2n m(B n )
c1
where B n = {x ∈ Rn : |x| < 1} and C = 2n m(B
n) .
n
If α is small enough such that C/α > R , then for R < |x| < (C/α)1/n we have Hf (x) > α. Hence
m({x : Hf (x) > α}) ≥ m({x : R < |x| < (C/α)
where C ′ = Cm(B n )(1 −
1/n
C
C′
n
}) = m(B )( − R ) = ,
α
α
Rn α
).
C
Problem 3.25
If E is a Borel set in Rn , the density DE (x) of E at x is defined as
m(E ∩ B(r, x))
,
r→0
m(B(r, x))
DE (x) = lim
6
n
whenever the limit exists.
a. Show that DE (x) = 1 for a.e. x ∈ E and DE (x) = 0 for a.e. x ∈ E c .
b. Find examples of E and x such that DE (x) is a given number α ∈ (0, 1), or such that DE (x)
does not exist.
Solution:
(a) By Theorem 3.18, limr→0 Ar χE (x) = χE (x) for a.e. x ∈ Rn . By definition of Ar , this implies
Z
1
m(E ∩ B(r, x))
lim
χE (y)dy = lim
= χE (x)
r→0 m(B(r, x)) B(r,x)
r→0
m(B(r, x))
for a.e. x ∈ Rn . Therefore DE (x) = 1 for a.e. x ∈ E and DE (x) = 0 for a.e. x ∈ E c .
(b) We will find examples in R2 . Let α ∈ (0, 1). Then define
Eα = {(r, θ) ∈ R2 : 0 < r < 1, 0 < θ < 2πα}.
R 2πα R r
Then when 0 < r < 1, we have m(Eα ∩ B(r, 0)) = 0
sdsdθ = παr 2 . Therefore
0
m(E ∩ B(r, 0))
παr 2
=
= α.
m(B(r, 0))
πr 2
So for E = Eα and x = (0, 0), DE (x) = α.
Next, we construct an E such that DE (x) does not exist. Define
1
1
1
1
En = [ n+1 , n ] × [ n+1 , n ]
2
2
2
2
√
S∞
and let E = 0 En . Define rn = 2−n 2, and notice rn → 0 as n → ∞. Since rn is at the top-right
corner of a square, we can sum the area of the squares and compute
m(E ∩ B(0, rn )) =
∞
X
1
41−n
=
.
k
4
3
k=n
Therefore
41−n 22n
2
m(E ∩ B(0, rn ))
=
=
.
m(B(0, rn ))
3(2π)
3π
Hence
2
m(E ∩ B(0, rn ))
=
.
n→∞
m(B(0, rn ))
3π
√
However, if we let r̃n = 3 · 2−n−1 2, then we also have r̃n → 0 as n → ∞. Here, each r̃n is in
the center of a square, so we can get an upper bound on m(E ∩ B(0, r̃n )) by summing the area of all
previous squares and adding half of the area of the square where r̃n is centered.
lim
7
m(E ∩ B(0, r̃n )) ≤
∞
X
1
4−n 4−n
1 1
−n 5
+
=
+
=
4
.
4n 2 k=n+1 4k
2
3
6
This gives an upper bound on the ratio
5
m(E ∩ B(0, r̃n ))
5 22(n+1)
√
=
≤ 4−n
.
2
m(B(0, r̃n ))
6 π(3 2)
27π
Therefore,
m(E ∩ B(0, r̃n ))
m(E ∩ B(0, rn ))
< lim
.
n→∞
n→∞
m(B(0, r̃n ))
m(B(0, rn ))
lim
These limits would be equal if DE (0) existed. Hence the limit as r → 0 of ratio between m(E ∩
B(0, r)) and m(B(0, r) does not exist.
Problem 3.31
Let F (x) = x2 sin(x−1 ) and G(x) = x2 sin(x−2 ) for x 6= 0, and F (0) = G(0) = 0.
a. F and G are differentiable everywhere (including x = 0).
b. F ∈ BV ([−1, 1]), but G ∈
/ BV ([−1, 1]).
Solution:
(a) It is clear that F and G are differentiable at x 6= 0 since they are the product and composition
of differentiable functions. At x = 0 we can use the definition of the derivative:
h2 sin(1/h)
F (0 + h) − F (0)
≤ lim
≤ lim |h| = 0.
h→0
h→0
h→0
h−0
h
′
′
Hence F (0) = 0, and by the same argument G (0) = 0.
|F ′(0)| = lim
(b)
For x 6= 0, we have F ′ (x) = 2x sin(1/x) − cos(1/x). Therefore, on [−1, 1] we have |F ′ (x)| ≤
2(1)(1) + 1 = 3. For any subdivision n ∈ N, −1 = x0 < · · · < xn = 1, we can apply the Mean Value
Theorem to conclude
n
X
1
|F (xj ) − F (xj−1)| ≤
n
X
1
3|xj − xj−1 | = 6.
Therefore F ∈ BV ([−1, 1]).
Next we show that G ∈
/ BV ([−1, 1]). Define xk = (πk +π/2)−1/2 and let k ∈ N, k > 2 and consider
the following partition Pk of [−1, 1] : −1, −x1 , −x2 , . . . , −xk , 0, xk , . . . , x2 , x1 . Then
8
X
Pk
|G(xi ) − G(xi−1 )| ≥
k
X
=
k
X
1
1
sin(πn + π/2) −
sin(π(n − 1) + π/2)
πn + π/2
π(n − 1) + π/2
n=2
n=2
|
1
1
+
|
πn + π/2 π(n − 1) + π/2
k
1
1X
≥
π n=2 n + 1/2
Since the harmonic series P
diverges, if we refine Pk by increasing k, this sum can be made as
large as we like. Hence sup{ n1 |G(xj ) − G(xj−1 )| : n ∈ N, −1 = x0 < · · · < xn = 1} = ∞ and
G∈
/ BV ([−1, 1]).
Problem 3.37
Suppose F : R → C. There is a constant M such that |F (x) − F (y)| ≤ M|x − y| for all x, y ∈ R (that
is, F is Lipschitz continuous) iff F is absolutely continuous and |F ′| ≤ M a.e.
Solution:
Suppose F is Lipschitz continuous. Let
P ǫ > 0, and δ = ǫ/M. Then for any finite set of disjoint
intervals (a1 , b1 ), . . . , (aN , bN ) such that N
1 (bj − aj ) < δ, we have
N
X
1
|F (bj ) − F (aj )| ≤
N
X
1
M|bj − aj | < δM = ǫ.
Hence F is absolutely continuous. Also,
|F (x) − F (y)|
≤ M.
y→x
|x − y|
|F ′(x)| = lim
Conversely, suppose F is absolutely continuous and |F ′| ≤ M a.e. If x, y ∈ R and x > y, by the
Fundamental Theorem of Calculus for Lebesgue Integrals, we have
Z x
Z x
′
|F (x) − F (y)| =
F (t)dt ≤
|F ′(t)|dt ≤ M(x − y).
y
y
It follows that |F (x) − F (y)| ≤ M|x − y| for all x, y ∈ R.
9