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实分析与复分析第三版详细答案.pdf
Chapter 1 Abstract Integration
1, Does there exist an infinite σ−algebra which has only countably many members?
Solution: No. There exists no such a infinite σ−algebra M which has only countably members.
Proof: Suppose M is a σ−algebra of subsets in a set X which has only infinite countably many
members. So we can pick from M a countably many members A1, A2,···. Without loss of generality,
{Ai}∞
i=1 are mutually pairwise disjoint, for otherwise letting
¯A1 = A1, ¯A2 = A2 − A1, · · ·, ¯Ak = Ak − k−1
Ai, · · ·
i=1
¯Aj = ∅ if i = j.
then clearly ¯Ai ∈ M and ¯Ai
Let 2N be the collection of all subsets of the natural numbers N, then it is well known that
the cardinal number of 2N is 2N = [0, 1] = C. (See p25, exercise 3 in the book of Jiang zejian).
For any given Y ∈ 2N , since Y is a countable set and M is a σ−algebra,
f : 2N → M is given by f(Y ) =
Y, Z ∈ 2N , Y = Z, then f(Y ) = f(Z).
n∈Y An. Then f is well defined and it is one to one, i.e.
n∈Y An ∈ M. Let
if
In fact, Y = Z implies that there exists an n0 ∈ Y
f(Z) = An0
Aj = ∅ if i = j
and n0 ∈ Zc, by the definition of f, An0
n∈Z An) = ∅. So, An0 ∈ f(Z)c
while An0 ∈ f(Y ), i.e. f(Y ) = f(Z). Therefore f : 2N → M is one-to-one, which implies that
C = 2N ≤ M ≤ N = C0 is a contradiction. So, the contradiction shows that M can not has only
countable many members.
Zc ⊂ N. Since Ai
(
We give the proof that 2N = [0, 1] = C here. For any given A ⊂ N, set ϕA(n) = { 1,
0,
if n ∈ A;
if n ∈ A.
Let g : A → (0, 1) is given by g(A) = 0. ϕA(1)ϕA(2) · ··, it is obvious that g is one-to-one. So,
2N ≤ [0, 1].
On the other hand, for any x ∈ (0, 1), set Ax = {r; r ≤ x, r ∈ R}, where R is the set of all
the rational in (0, 1). If x, y ∈ (0, 1) and x = y, by Berstein’s theorem, the density of rational in
R, Ax = Ay. So, (0, 1) ≤ R = 2N .
So, 2N = [0, 1] = C.
2
2, Prove an analogue of Theorem 1.8 for n functions.
Solution: We have the following result.
Theorem 1.8: Let u1, u2, ..., un be real measurable functions on a measurable space X, let Φ be
a continuous mapping of RN onto a topological space Y , and define
h(x) = Φ(u1(x), ..., un(x))
for x ∈ X. Then h : X → Y is measurable.
Proof: Since h = Φ ◦ f, where f(x) = (u1(x),
shows that it is enough to prove the measurability of f.
..., un(x)) and f maps X into Rn. Theorem 1.7
1
If R is any open rectangle in Rn with sides parallel to the coordinate super plane, then
R = I1 × I2 × · · · × In, where Ii = (ai, bi) ⊂ R1 (i = 1, 2, ..., n) are open segments and
f−1(R) = u−1
i (Ii) for i = 1, 2, ..., n, that is x ∈n
u−1
2 (I2)
1 (I1)
· · ·
u−1
n (In).
x ∈n
ui(x) ∈ Ii, then x ∈ u−1
In fact, for any given x ∈ f−1(R), we have f(x) = (u1(x), ..., un(x)) ∈ R = I1×I2×···×In and
i (Ii). On the other hand, for any
i (i), ui(x) ∈ Ii and f(x) = (u1(x), ..., un(x)) ∈ I1 × I2 ×···× In.
i (Ii), we have xi ∈ u−1
So f−1(R) is measurable for each open rectangle interval R ⊂ Rn as ui are measurable on X.
i=1 u−1
i=1 u−1
By Lindelif’s Theorem, every open set V is a countable union of open interval with sides
i=1 f−1(Ri) is measurable as well.
parallel to the coordinate super-plane, V =
Ri, So f−1(V ) =
n
2
3, Prove that if f is a real function on a measurable space X such that {x; f(x) > r} is measurable
for every rational r, then f if measurable.
Proof: For any α ∈ R1, there is a sequence of rational {rn; n = 1, 2, ...} such that r1 > r2 > r3 >
... > rn > ... > α with lim
n→∞ rn = α. So
{x; f(x) > α} =
{x; f(x) ≥ rn}.
+∞
n=1
(b) lim sup
n→∞
(an + bn) = inf
k≥1
= − sup
k≥1
sup{−ak,−ak+1, ...} = inf
k≥1
inf{ak, ak+1, ...} = lim inf
n→∞ an.
[sup{ak + bk, ak+1 + bk+1, ...}]
[sup{ak + bk, ak+1 + bk+1, ...}]
= lim inf
k≥1
{an} + sup
{bn}]
≤ lim inf
k→∞ [sup
n≥k
n≥k
n→∞ [sup{bn}]
n→∞ [sup{an}] + lim inf
n→∞ an + lim sup
n→∞ bn.
= lim inf
= lim sup
2
In fact, for any x ∈ {x; f(x) > α}, there is a n0 such that α < rn0 ≤ f(x), therefore
x ∈ {x; f(x) > α}. On the other hand, rn > α implies that
+∞
n=1
{x; f(x) ≥ rn} ⊂ {x; f(x) > α}.
By the condition, for any given n, {x; f(x) ≥ rn} is measurable, we see that {x; f(x) > α} is
measurable for any α ∈ R1. By Theorem 1.12, f is measurable.
4, Let {an} and {bn} be sequence in [−∞, +∞], prove the following assertions:
(−an) = − lim inf
(a) lim sup
n→∞ an;
n→∞
(an + bn) ≤ lim sup
(b) lim sup
n→∞ an + lim sup
n→∞
(c) If an ≤ bn for all n, then
n→∞ bn provided none of the sums is of the form ∞ − ∞;
n→∞ an ≤ lim inf
lim inf
n→∞ bn
show by an example that strict inequality can hold in (b).
Proof:
(a) lim sup
n→∞
{− inf{−ak,−ak+1, ...}}
(−an) = inf
k≥1
2
{an} ≤ inf
{ak} = lim inf
inf
n≥k
0, lim sup
(c) lim inf
n→∞ an.
k→∞ inf
n≥k
(an + bn) = 0 < 2 = 1 + 1 = lim sup
n→∞ bn = 1, but lim sup
n→∞
n→∞ an = lim
k→inf ty
In (b), the strict inequality can hold. For example, let an = (−1)n, bn = −(−1)n, then an+bn =
n→∞ an = 1, lim sup
n→∞ bn.2
5, (a) Suppose f : X → [−∞,∞] and g : X → [−∞,∞] are measurable. Prove that the sets
{x; f(x) < g(x)}, {x; f(x) = g(x)} are measurable.
(b) Prove that the set of points at which a sequence of measurable real-valued functions converges(to
a finite limit) is measurable.
Proof: (a1) Suppose f, g are measurable on X, then {x; f(x) < g(x)} is measurable. Suppose
{rn} is the set of rational in R1, we claim that
n→∞ an + lim sup
∞
{x; f(x) < g(x)} =
[{x; f(x) < rn}
{x; rn < g(x)}]
In fact,
n=1
x0 ∈ {x; f(x) < g(x)} ⇐⇒ f(x0) < g(x0)
⇐⇒ x0 ∈∞
⇐⇒ ∃rn0, such that f(x0) < rn0 < g(x0)
n=1[{x; f(x) < rn}{x; rn < g(x)}].
X. So, by the claim we have proved above, {x; f(x) < g(x)} is measurable.
For any r ∈ R, {x; f(x) < r} and {x; r < g(x)} are measurable since f, g are measurable on
(a2) By (a1), {x; f(x) = g(x)} = X − [{x; f(x) < g(x)}{x; f(x) > g(x)}] is measurable.
(b) Suppose {fn} is a sequence of real-valued measurable functions on (X,)(a measurable
space), then by Theorem 1.9(c), for any n, m ∈ N, |fn(x) − fm(x)| is a measurable function as | · |
is a continuous function on R1 × R1. So, {x;
|fn(x) − fm(x)| < a} is a measurable set for any
a ∈ R1.
Clearly, the set where {fn(x)} has finite limit is given by
+∞
+∞
A =
k=1
N =1
n,m≥N
{x; |fn(x) − fm(x)| <
}
1
k
by Cauchy’s criterion for convergence.
A is a measurable set because for any n, m ∈ N and k ∈ N, {x; |fn(x) − fm(x)| < 1
and M is a σ−algebra.
k} ∈ M
2
6, Let X be an uncountable set, let M be the collection of all sets E ⊂ X such that either E or
Ec is at most countable, and define µ(E) = 0 in the first case, µ(E) = 1 in the second. Prove that
M is a σ−algebra in X and that µ is a measure on M. Describe the corresponding measurable
functions and their integrals.
Proof: 1.) Since X c = ∅ is at most countable, we see that X ∈ M.
If A ∈ M, then either A or Ac is at most countable, i.e. either Ac or (Ac)c = A is at most countable,
so Ac ∈ M as well.
If An ∈ M for n = 1, 2, 3..., then
n=1 An ∈ M and if
n=1 An is at most countable, we have
n=1 An0
n=1 An is not at most countable, then there exists n0, such that
n=1 An ∈ M, for if
+∞
+∞
+∞
+∞
+∞
3
is not countable by the fact that a countable union of a sequence of countable sets is countable, so
as An0 ∈ M, we must have Ac
n0 is at most countable. hence for
+∞
(
n=1
+∞
n=1
An)c =
n ⊂ Ac
Ac
n0,
+∞
we see that (
n=1 An)c is at most countable.
Thus M is a σ−algebra in X.
2.) By the definition, µ : 2X → [0, +∞]. Suppose Ai ∈ M for i = 1, 2, 3, ..., Ai + Aj = ∅ if
+∞
i = j, then either Ai or Ac
If
n=1 An is countable, then for any n ∈ N, An is also countable.
i is at most countable.
+∞
+∞
µ(
An) = 0 =
µ(An).
n=1
n=1
+∞
+∞
n=1 An is not countable, then there exists n0 ∈ N, A0 is uncountable and Ac
If
countable. Since {An} are mutually disjoint, we have
n0 is at most
n=n0
if
Ai is uncountable, (
and An0 is uncountable, then
n=n0
An − An0 ∈ M.
n=1
Ai =
n=n0
An)c is at most countable, this contradicts to An0 ⊂ (
An is at most countable. So
n=n0
An)c
n=n0
+∞
3.) Since for any given measurable function f(x) ∈ µ(X), X = {x; f(x) = r}{x; f(x) <
r}{x; f(x) > r}=E1
E3 ∈ M for any r ∈ R1. X ∈ M, µ(X) = 1 implies that there exists
µ(An) + µ(An0) =
An) = 1 = 0 + 1 =
+∞
µ(An).
n=n0
E2
µ(
n=1
n=1
a unique uncountable set E of Ei, i ∈ {1, 2, 3}.
In fact, since X is uncountable, there exits at least one of Ei0 is uncountable, then µ(Ei0) = 1
Ei) = 0, and there will be a contradiction if there exist at least two uncountable Ei.
and µ(
If i0 = 1, then f(x) = r a.e. [µ] .
If i0 = 2 or 3, one can take r = sup
µ(E)=0
For the case i0 = 2, let r → −∞, one will get that there exists a r1 ∈ [r, r), such that
|f(x)| and ¯r = inf
|f(x)|.
sup
X\E
inf
X\E
µ(E)=0
i=i0
µ{x; f(x) = r1} = 1, i.e. f(x) = r1 a.e. [µ];
µ{x; f(x) = r2} = 1, i.e. f(x) = r2 a.e. [µ].
For the case i0 = 3, let r → +∞, one will get that there exists a r2 ∈ (r, ¯r], such that
From above all, there exists r ∈ [r, ¯r], such that µ{x; f(x) = r} = 1, i.e. f(x) = r a.e. [µ]. 2
7, Suppose fn : X → [0,∞] is measurable for n = 1, 2, 3,· · ·, f1 ≥ f2 ≥ · · · ≥ 0, fn(x) → f(x) as
n → ∞, for every x ∈ X, and f1 ∈ L1(µ). Prove that then
and show that this conclusion does not follow if the condition “f1 ∈ L1(µ)” is omitted.
lim
n→∞
X
fndµ =
f dµ
X
4
Proof: Since f1 ∈ L1(µ), {x; f1(x) = +∞} is measurable and µ{x; f1(x) = +∞} = 0, i.e.
f(x) < +∞ a.e. [µ]. Since f1 ≥ f2 ≥ ... ≥ fn ≥ 0 and f1 ∈ L1(µ), fn ∈ L1(µ) and f is measurable.
Let A = {x; f1(x) = +∞} and An = {x; fn(x) = +∞}, then µ(An) ≤ µ(A) = 0. Let
X fndµ ≤
gn(x) = f1 − fn if x ∈ Ac and 0 if x ∈ A, then 0 ≤ g1 ≤ g2 ≤ ... ≤ gn ≤ ... and gn is measurable.
X f1dµ < +∞. By
By Fatou’s lemma, 0 ≤
monotone convergence theorem,
X f dµ =
f1dµ − lim
n→∞
X
fndµ = lim
n→∞
gndµ =
X
(f1 − fn)dµ =
f1dµ −
fndµ.
X
X
n→∞ fndµ ≤ lim inf
X lim
n→∞
lim
n→∞ gndµ =
X
X
i.e. − lim
n→∞
Counterexample: fn(x) = χ[n,+∞] and µ is a counting measure on R1(see p17 for the definition
X f dµ, hence lim
n→∞
X fndµ =
X f dµ.
of counting measure), then f1 ≥ f2 ≥ ... ≥ fn ≥ 0.
For any given x ∈ R1, fn(x) : χ[n,+∞](x) = 0 if n ≥ N(x) if N(x) is large enough, so
X
X fndµ = −
fn(x) → f(x) = 0 and
X f dµ = 0.
On the other hand, since
X fndµ = µ([n, +∞]) = +∞,
X f dµ. 2
8, (E is a measurable set in (X, M)) Put fn = χE if n is odd, fn = 1 − χE if n is even. What is
the relevance of this example to Fatou’s lemma?
Solution: Suppose µ(X) = 1, 0 < µ(E) < 1
µ(X) − µ(E) = 1 − µ(E) and
X fndµ = +∞ = 0 =
2. Since f2n = 1 − χE and f2n+1 = χE,
X f2n+1dx = µ(E), then
X f2ndx =
lim
n→∞
lim inf
n→∞ fn =
X
lim inf
n→∞ fn +
E
Ec
lim inf
n→∞ fn =
E
χEdµ < µ(E) = lim inf
n→∞
Ec
fndµ.
X
(1− χE)dµ +
So, the strict inequality in Fatou’s lemma holds sometimes.
9. Suppose µ is a positive measure on X, f : X → [0, +∞] is measurable,
0 < c < ∞, and α is a constant. Prove that
2
X f dµ = c where
lim
n→+∞
X
n log[1 + ( f
n
)α]dµ =
∞,
c,
0,
if 0 < α < 1,
if α = 1,
if 1 < α < ∞.
Hint: If α ≥ 1,then integrands are dominated by αf.If α < 1, Fatou Lemma can be applied.
proof: Let g(x) = αx − n log[1 + ( x
n
g(x) = α − n
α
1 + ( x
n)α],(α ≥ 1). then g(0) = 0 and
n)α−1 1
n)α−1
α( x
n)α = α[1 − ( x
n)α ]
1 + ( x
1 + ( x
− 1)]
n)α [1 + ( x
n)α−1)
If x ≥ n, it is clear that g(x) ≥ α(1 + ( x
≥ 1.
1 + ( x
n)α
)α ≤ ( x
If 0 < x < n,α ≥ 1 implies that ( x
n
n
)α−1(1 − x
n
)α−1( x
n
) ≤ ( x
n
1
1 − x
) ≤ 1
( x
n
,i.e.
=
n
)
n
5
hence
g(x) =
)α−1( x
n
( x
n
1 + ( x
n
− 1) ≥ −1
− 1) ≥ 0
)α−1( x
n
n)α−1( x
1 + ( x
n)α
g(0) = α > 0
α[1 + ( x
n − 1)]
≥ 0
So g(x) ≥ 0, for x ∈ (0, +∞]. We thus have g(x) ≥ 0,i.e. n log(1 + ( f
f dµ = c < +∞. By dominated convergence theorem
n)α) ≤ αf ∈ L1(µ), since
lim
n→+∞
X
n log[1 + ( f
n
)α]dµ =
n→+∞ n log[1 + ( f
lim
n
)α]dµ
X
=
X f dµ = c,
0.
if α = 1,
if 1 < α < ∞.
lim
x→0+
log(1 + ax)
x
= a
This is because when α = 1,we have
for a ≥ 0, and when α > 1,
y
If 0 < α < 1, a > 0 then
lim
x→0+
As
lim
x→0+
log(1 + (ax)α)
x
= lim
x→0+
log(1 + (ax)α)aαxα−1
(ax)α
= lim
x→0+
aαxα−1 lim
y→0+
log(1 + y)
y
= 0
log(1 + (ax)α)
x
aαxα−1 lim
y→0+
= lim
x→0+
log(1 + (ax)α)aαxα−1
= lim
x→0+
log(1 + y)
(ax)α
= 1 · ∞ = ∞
X
lim inf
n→+∞
≥
{x:f (x)>0}
{x:f (x)>0}
=
= +∞
n log[1 + ( f
n
)α]dµ ≥
n→+∞ n log[1 + ( f
lim inf
∞dµ = ∞ · µ{x; f(x) > 0}
n
n→+∞ n log[1 + ( f
lim inf
X
)α]dµ
n
f dµ = c > 0, we see that µ{x : f(x) > 0} > c. Fatou’s Lemma implies that
)α]dµ
2
10. Suppose µ(X) < ∞, {fn} is a sequence of bounded complex measurable functons on X, and
fn → f uniformly on X. Prove that
lim
n→∞
X
fndµ =
f dµ
X
6
and show that the hypothesis µ(X) < ∞ can not be omitted.
Proof: ∀ n,|fn| ≤ Cn < +∞ on X. So when µ(X) < ∞ we have
fndµ ≤ Cnµ(X) < +∞,
which implies that fn ∈ L1(µ). Since fn converges to f uniformly on X, which implies that f is
measurable, then there exits N, s.t. |fN (x) − f(x)| < 1 for x ∈ X. Thus
X
|f| ≤ |fN − f| + |fN| ≤ 1 + CN < +∞
and f ∈ L1(µ). If µ(X) = 0, then clearly 0 =
X f dµ. If
0 < µ(X) < +∞, as fn converges to f uniformly on X, then ∀ ε > 0, ∃ N = N(ε) > 0, s.t. n ≥ N
we have |fn(x) − f(x)| <
X f dµ and limn→∞
for x ∈ X. Thus
X fndµ =
X fndµ =
ε
X
µ(X)
|
X
≤
≤ ε
X
fndµ −
|fn − f|dµ
X
µ(X) µ(X) = ε.
f dµ| = |
(fn − f)dµ|
So limn→∞
X fndµ =
X f dµ.
Now we prove that the condition µ(X) < ∞ can not be omitted. We can construct a example
as follows: set X = {1, 2, ..., N, ...}, µ is the counting measure and fn is defined as: fn =
, ∀
x ∈ X, f ≡ 0. It is obvious that both fn and f are complex measurable and fn → f uniformly on
X. But
1
n
fndµ =
X
then
µ(X) = +∞,
f dµ = 0 · µ(X) = 0
1
n
lim
n→+∞
X
X
fndµ = +∞ = 0 =
f dµ
X
This implies that the condition µ(X) < ∞ can not be omitted.
11. Show that
+∞
+∞
n=1
k=n
Ek
A =
2
(1)
in Theorem 1.41, and hence prove the theorem without any reference to integration.
Solution: Theorem 1.41: Let {Ek} be a sequence of measurable sets in X, such that
+∞
k=1
µ(Ek) < ∞
Then almost all x ∈ X lie in at most finitely many of the sets Ek. If A is the set of all x which lie
in infinitely many Ek, we have to show that µ(A) = 0. We show
A =
Ek
+∞
+∞
n=1
k=n
7
On one hand, if x ∈+∞
n=1
+∞
k=n Ek, then ∀ n, ∃ kn ≥ n, s.t. x ∈ Ek, we may also assume that
k1 < k2 < ... < kn < ...
hence x belongs to infinitely many of the sets Ek(We can first take k1 and then take n2 > k1,and
take k2 > n2, ... ), then x ∈ A.
On the other hand, if x ∈ A,then there are k1 < k2 < ... < kn, s.t.
... x ∈ Ekn.
so x ∈+∞
i=1 Eki,∀ n, ∃ kn1 ≥ n, s.t. x ∈ Ekn, we can get x ∈+∞
+∞
k=n Ek and x ∈+∞
+∞
+∞
x ∈ Ek2,
x ∈ Ek1,
k=n Ek.
n=1
Thus A =
n=1
k=n Ek, ∀ n.
+∞
µ(A) = µ(
+∞
+∞
≤ +∞
n=1
k=n
Ek) ≤ µ(
Ek)
k=n
k=n
µ(Ek) → 0, n → +∞.
+∞
k=n µ(Ek) < ∞.
since
12. Suppose f ∈ L1(µ). Prove that for each ε > 0, there is a δ > 0 such that
µ(E) < δ.
2
E |f|dµ < ε whenever
X |f|dµ < +∞. According to Theorem 1.17 (P15), since |f| is nonnegative
measurable on X, there exist a sequence of infinite valued nonnegative simple functions SN =
s.t. S1 ≤ ... ≤ SN−1 ≤ SN ≤ ..., and SN (x) → |f(x)| for ∀ x ∈ X. By monotone
Proof: f ∈ L1(µ) ⇒
mN
i=1 aN
i χEN
i
convergence theorem , we get
X
X
But
X |f|dµ < +∞, so for ε > 0, ∃ N = N(ε), s.t.
|f|dµ −
i }i=1,...,mN are nonnegative finite real number. Set kN =
, then when
mN
mN
For above fixed SN =
max1≤i≤mN{aN
E ∈ M and µ(E) < δ, we get
i } + 1, then obvious 0 < kN < +∞. Take δ(ε) = δ(ε, N(ε)) =
0 ≤
, {aN
3kN mN
SN dµ <
i=1 aN
i χEN
ε
3
X
X
ε
i
i ∩ E) ≤ mN
kN µ(E)
lim
N→+∞
SN dµ =
|f|dµ
SN dµ =
i µ(EN
aN
E
i=1
≤ mN kN µ(E) <
i=1
mN kN ε
3mN kN
= ε
3
So when µ(E) < δ we have
|f|dµ ≤
E
≤
<
SN dµ +
E
|f|dµ −
(|f| − SN )dµ + ε
3
+ ε
3 < ε.
E
X
ε
3
SN dµ ≤
|f|dµ −
(|f| − SN )dµ + ε
3
SN dµ + ε
3
E
X
E
X
=
8
2
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