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 A
1
, A
2
, ···. Without loss of generality,
{A
i
}
∞
i=1
are mutually pairwise disjoint, for otherwise letting
¯
A
1
= A
1
,
¯
A
2
= A
2
− A
1
, · · ·,
¯
A
k
= A
k
−
k−1
[
i=1
A
i
, · · ·
then clearly
¯
A
i
∈ M and
¯
A
i
T
¯
A
j
= ∅ if i 6= j.
Let 2
N
be the collection of all subsets of the natural numbers N, then it is well known that
the cardinal number of 2
N
is 2
N
= [0, 1] = C. (See p25, exercise 3 in the book of Jiang zejian).
For any given Y ∈ 2
N
, since Y is a countable set and M is a σ−algebra,
S
n∈Y
A
n
∈ M. Let
f : 2
N
→ M is given by f(Y ) =
S
n∈Y
A
n
. Then f is well defined and it is one to one, i.e. if
Y, Z ∈ 2
N
, Y 6= Z, then f(Y ) 6= f (Z).
In fact, Y 6= Z implies that there exists an n
0
∈ Y
T
Z
c
⊂ N. Since A
i
T
A
j
= ∅ if i 6= j
and n
0
∈ Z
c
, by the definition of f, A
n
0
T
f(Z) = A
n
0
T
(
S
n∈Z
A
n
) = ∅. So, A
n
0
∈ f(Z)
c
while A
n
0
∈ f(Y ), i.e. f(Y ) 6= f(Z). Therefore f : 2
N
→ M is one-to-one, which implies that
C = 2
N
≤ M ≤ N = C
0
is a contradiction. So, the contradiction shows that M can not has only
countable many members.
We give the proof that 2
N
= [0, 1] = C here. For any given A ⊂ N, set ϕ
A
(n) = {
1, if n ∈ A;
0, if n 6∈ A.
Let g : A → (0, 1) is given by g(A) = 0. ϕ
A
(1)ϕ
A
(2) · ··, it is obvious that g is one-to-one. So,
2
N
≤ [0, 1].
On the other hand, for any x ∈ (0, 1), set A
x
= {r; r ≤ x, r ∈ R}, where R is the set of all
the rational in (0, 1). If x, y ∈ (0, 1) and x 6= y, by Berstein’s theorem, the density of rational in
R, A
x
6= A
y
. So, (0, 1) ≤ R = 2
N
.
So, 2
N
= [0, 1] = C. 2
2, Prove an analogue of Theorem 1.8 for n functions.
Solution: We have the following result.
Theorem 1.8
0
: Let u
1
, u
2
, ..., u
n
be real measurable functions on a measurable space X, let Φ be
a continuous mapping of R
N
onto a topological space Y , and define
h(x) = Φ(u
1
(x), ..., u
n
(x))
for x ∈ X. Then h : X → Y is measurable.
Proof: Since h = Φ ◦ f, where f(x) = (u
1
(x), ..., u
n
(x)) and f maps X into R
n
. Theorem 1.7
shows that it is enough to prove the measurability of f.
1
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ctz19881282014-05-17不错的答案，有了答案，看起书来就没那么费劲了
