实分析与复分析部分答案

tinuous.
(b)If f1 and f2 are lower semicontinuous, then f1 + f2 is lower semicon-
tinuous.
∞
(c)If each {fn} is upper semicontinuous, then fn is upper semicontin-
n→∞ X
X
Since f1 ∈ L1(µ), it shows
lim fndµ = f dµ.
n→∞ X
X
3
[counterexample]: Let X = (−∞, +∞), and we define
fn = χEn, En = [n, +∞), n = 1, 2, · · · .
8. Suppose that E is measurable subset of measure space (X, µ) with µ(E) > 0; µ(X − E) > 0, then we can prove that the strict inequality in the Fatou’s lemma can hold.
on X for n = 1, 2, 3, · · ·. Moreover, gn(x) → f1(x) − f (x) as n → ∞, for every
x ∈ X. By the Lebesgue’s monotone convergence theorem, we have
lim gndµ = (f1 − f )dµ.
s with 0 ≤ s ≤ |f | such that
|f |dµ ≤
X
X
sdµ
+
ε. 2
Suppose
that
M
>
0,
satisfying
0
≤
s(x)
≤
M
on
X,
and
let
δ
=
ε 2M
,
we
have
|f |dµ ≤
E
E
sdµ +
ε 2
≤
M m(E) +
ε 2
<
ε
when m(E) < δ.
4
4. Let {an} and {bn} be sequences in [−∞, +∞], and prove the following assertions:
(a)
lim sup(−an) = − lim inf an.
n→∞
n→∞
(b)
lim sup(an + bn) ≤ lim sup an + lim sup bn.
(b) Since Hence
lim sup(−an) = − lim inf an.
n→∞
n→∞
sup(ak + bk) ≤ sup ak + sup bk, n = 1, 2, · · · .
k≥n
k≥n
k≥n
lim sup(ak + bk) ≤ lim [sup ak + sup bk] = lim sup ak + lim sup bk.
n→∞ k≥n
n→∞ k≥n
k≥n
n→∞ k≥n
n→∞ k≥n
By the definations of the upper and the lower limits, that is
lim sup(an + bn) ≤ lim sup an + lim sup bn.
n→∞
n→∞
n→∞
example: we define
an = (−1)n, bn = (−1)n+1, n = 1, 2, · · · .
Then we have
an + bn = 0, n = 1, 2, · · · .
But
lim sup an = lim sup bn = 1.
n→∞
n→∞
(c)Because an ≤ bn for all n, then we have
fn
=
1 n
on
X, n
=
1, 2, · · · .
12. Suppose f ∈ L1(µ). Proved that to each ε > 0 there exists a δ > 0
such that E |f |dµ < ε whenever µ(E) < δ. [proof]: Since
n=1
Hence, ∞ f −1((α, +∞)) = f −1([rn, +∞)). n=1
Since sets f −1([rn, +∞)) are measurable for each n, the set f −1((α, +∞)) is
also measurable. Then f is measurable.
inf (ak) ≤ inf bk, n = 1, 2, · · · .
k≥n
k≥n
By the definations of the lower limits, it follows
lim inf an ≤ lim inf bn.
n→∞
n→∞
5. (a)Supose f : X → [−∞, +∞] and g : X → [−∞, +∞] are measurable. Prove that the sets
the word ”nonnegative” is omitted? Is the truth of the statements affected if
R1 is replaced by a general topological space?
[proof]: (a) Because for every real number α,
n→∞
n→∞
n→∞
provided none of the sums is of the form ∞ − ∞.
(c) If an ≤ bn for all n, then
lim inf an ≤ lim inf bn.
n→∞
n→∞
Show by an example that strict inequality can hold in (b).
|f |dµ = sup sdµ,
X
X
the supremum being taken over all simple measurable functions s such that
0 ≤ s ≤ |f |. Hence, for each ε > 0, there exists a simple measurable function
X
and show that the hypothesis ”µ(X) < ∞”cannot be omitted.
[proof]: There exist Mn > 0, n = 1, 2, · · · such that |fn(x)| ≤ Mn for every x ∈ X, n = 1, 2, · · ·. Since fn → f uniformly on X, there exists M > 0 such that
{x : f (x) < g(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.
∞∞
A=
∞
{x
:
|fn(x)
−
fm(x)|
<
1 r
}.
r=1 k=1 n,m≥k
Therefore, A is measurable.
7. Suppose that fn : X → [0, +∞] is measurable for n = 1, 2, 3, · · · , f1 ≥ f2 ≥ f3 ≥ · · · ≥ 0, fn(x) → f (x) as n → ∞,for every x ∈ Xand f1 ∈ L1(µ).
1}
n
n=1
is also measurable. (b) Let {fn(x)} be the sequence of measurable real-valued functions, A
be the set of points at which the sequence of measurable real-valued functions {fn(x)} converges (to a finite limit). Hence
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chapter one
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 is measurable.
[proof]: (a) Since
sup(−ak) = − inf ak, n = 1, 2, · · · .
k≥n
k≥n
Therefore, let n → ∞, it obtains