《實變函數(shù)與泛函分析基礎(程其襄張奠宙著)高等教育出版社課后答案.doc》由會員分享,可在線閱讀,更多相關《實變函數(shù)與泛函分析基礎(程其襄張奠宙著)高等教育出版社課后答案.doc(65頁珍藏版)》請在裝配圖網上搜索。
______________________________________________________________________________________________________________
1.
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
-可編輯修改-
x ∈ (A ∪ (B ∪ C)).
x ∈ A,
x ∈ A ∪ B, x ∈ A ∪ C,
x ∈ (A ∪ B) ∩ (A ∪ C).
x ∈ B ∩ C,
x ∈ A ∪ B
x ∈ A ∪ C,
x ∈ (A ∪ B) ∩ (A ∪ C),
A ∪ (B ∩ C) ? (A ∪ B) ∩ (A ∪ C).
x ∈ (A ∪ B) ∩ (A ∪ C). x ∈ A, x ∈ A ∪ (B ∩ C). x ∈ A,
x ∈ A ∪ B x ∈ A ∪ C, x ∈ B x ∈ C, x ∈ B ∩ C, x ∈ A ∪ (B ∩ C),
(A ∪ B) ∩ (A ∪ C) ? A ∪ (B ∩ C). A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
2.
(1)A ? B = A ? (A ∩ B) = (A ∪ B) ? B;
(2)A ∩ (B ? C) = (A ∩ B) ? (A ∩ C);
(3)(A ? B) ? C = A ? (B ∪ C);
(4)A ? (B ? C) = (A ? B) ∪ (A ∩ C);
(5)(A ? B) ∩ (C ? D) = (A ∩ C) ? (B ∪ D);
(6)A ? (A ? B) = A ∩ B.
(1)A ? (A ∩ B) = A ∩ ?s(A ∩ B) = A ∩ (?sA ∪ ?sB) = (A ∩ ?sA) ∪ (A ∩ ?sB) = A ? B;
(A ∪ B) ? B = (A ∪ B) ∩ ?sB = (A ∩ ?sB) ∪ (B ∩ ?sB) = A ? B;
(2)(A ∩ B) ? (A ∩ C) = (A ∩ B) ∩ ?s(A ∩ C) = (A ∩ B) ∩ (?sA ∪ ?sC) = (A ∩ B ∩ ?sA) ∪ (A ∩
B ∩ ?sC) = A ∩ (B ∩ ?sC) = A ∩ (B ? C);
(3)(A ? B) ? C = (A ∩ ?sB) ∩ ?sC = A ∩ ?s(B ∪ C) = A ? (B ∪ C);
(4)A ? (B ? C) = A ? (B ∩ ?sC) = A ∩ ?s(B ∩ ?sC) = A ∩ (?sB ∪ C) = (A ∩ ?sB) ∪ (A ∩ C) =
(A ? B) ∪ (A ∩ C);
(5)(A ? B) ∩ (C ? D) = (A ∩ ?sB) ∩ (C ∩ ?sD) = (A ∩ C) ∩ ?s(B ∪ D) = (A ∩ C) ? (B ∪ D);
(6)A ? (A ? B) = A ∩ ?s(A ∩ ?sB) = A ∩ (?sA ∪ B) = A ∩ B.
3.
(A ∪ B) ? C = (A ? C) ∪ (B ? C);
A ? (B ∪ C) = (A ? B) ∩ (A ? C).
(A ∪ B) ? C = (A ∪ B) ∩ ?sC = (A ∩ ?sC) ∪ (B ∩ ?sC) = (A ? C) ∪ (B ? C);
(A ? B) ∩ (A ? C) = (A ∩ ?sB) ∩ (A ∩ ?sC) = A ∩ ?sB ∩ ?sC = A ∩ ?s(B ∪ C) = A ? (B ∪ C).
∞ ∞
4.
?s(
Ai) =
?sAi.
i=1
∞
i=1
∞
x ∈ ?s(i=1
Ai),
x ∈ S,
x ∈
i=1
Ai,
i,x ∈ Ai,
x ∈ ?sAi,
1
∞
∞
∞
x ∈
i=1
?sAi.
∞
x ∈
i=1
?sAi,
∞
∞
i,x ∈ ?sAi,
x ∈ S,x ∈ Ai,
x ∈ S,
x ∈
i=1
Ai,
x ∈ ?s(i=1
Ai).
?s(
i=1
Ai) =
i=1
?sAi.
5.
(1) (
α∈Λ
Aα) ? B =
α∈Λ
(Aα ? B);
(2)(
α∈Λ
Aα) ? B =
α∈Λ(Aα ? B).
(1)
α∈Λ
Aα ? B = (α∈Λ
Aα) ∩ ?sB =
α∈Λ
(Aα ∩ ?sB) =
α∈Λ(Aα ? B);
(2)
α∈Λ
Aα ? B = (α∈Λ
Aα) ∩ ?sB =
α∈Λ(Aα ∩ ?sB) =
α∈Λ
(Aα ? B).
n?1
6.
{An}
n
n
B1 = A1, Bn = An ? (
ν=1
Aν), n > 1.
{Bn}
ν=1
Aν =
ν=1
Bν, 1 ≤ n ≤ ∞.
i = j,
i < j.
Bi ? Ai
(1 ≤ i ≤ n).
j?1
Bi ∩ Bj ? Ai ∩ (Aj ?
n=1
n
An) = Ai ∩ Aj ∩ ?sA1 ∩ ?sA2 ∩ · · · ∩ ?sAi ∩ · · · ∩ ?sAj?1 = ?.
n
Bi ? Ai(1 = i = n)
i=1
Bi ?
i=1
Ai.
n
n
x ∈
i=1
Ai,
x ∈ A1,
x ∈ B1 ?
i=1
Bi.
x ∈ A1,
in
x ∈ Ain,
in?1
in?1
n
n
n
x ∈
i=1
Ai
x ∈ Ain.
x ∈ Ain ?
i=1
Ai = Bin ?
i=1
Bi.
i=1
Ai =
i=1
Bi.
7.
A2n?1 = 0, n1 , A2n = (0, n), n = 1, 2, · · · ,
n→∞ An = (0, ∞);
{An}
N
x ∈ (0, ∞),
N,
x < N,
x
n>N
An,
0 < x < n,
x ∈ n→∞ An,
x
n→∞ An ? (0, ∞),
n→∞ An = (0, ∞).
n→∞ An = ?;
x ∈ n→∞ An = ?,
N,
n > N,
x ∈ An.
2n ? 1 > N
x ∈ A2n?1,
0
N,x ∈ An,
x ∈
m=n+1
Am ?
n=1 m=n
Am,
∞
∞
∞
∞
∞
n→∞ An ?
n=1 m=n
Am.
x ∈
n=1 m=n
Am,
n,
x ∈
m=n
Am,
m ≥ n,
x ∈ An,
x ∈ n→∞ An.
lim An =
∞
∞
Am.
n→∞
n=1 m=n
2
lim
x ∈ A2n,
lim
lim
lim
lim
lim
n.
lim
lim
lim
lim
9.
(?1, 1)
(?∞, +∞)
(?∞, ∞)
? : (?1, 1) → (?∞, +∞).
x ∈ (?1, 1), ?(x) = tan π2 x. ?
(?1, 1)
10.
(0, 0, 1)
(x, y, z) ∈ S\(0, 0, 1),
xOy
M
?(x, y, z) =
x y
,
1 ? z 1 ? z
∈ M.
?
S
M
11.
A
A
△z
rz
G
G = {△z|△z
z
},
G
△z
rz,
12.
∞
An
n
n = 1, 2, · · · ,
A =
n=0
An.An
n +1
n
n +1
0
6,An = a,
13.
A
§4
4,A = a.
(
)
§4
A
A
: (x, y, r).
(x, y)
r
x, y
r
0
A = a.
14.
f
(?∞, ∞)
E,
(1)
(2)x ∈ E
x ∈ (?∞, ∞),xlim0+ f(x + △x) = f(x + 0)
f(x + 0) > f(x ? 0).
xlim ? f(x + △x) = f(x ? 0)
(3)
x ∈ E,
15.
x1, x2 ∈ E,
(0, 1)
x1 < x2,
11
[0, 1]
3
(3) E x
S : x2 + y2 + (z ? 12)2 = ( 12 )2
→
→0
f(x1 ? 0) < f(x1 + 0) ≤ f(x2 ? 0) < f(x2 + 0),
(f(x ? 0), f(x + 0)),
?
16.
(0, 1)
[0,1] (0,1)
A
R = {r1, r2, · · · },
?
? ?(1) = r2,
?
A
An.
A = {x1, x2, · · · }, A
An 2n
A =
∞
A.An = {x1, x2, · · · , xn}, An
An, A
A
n=1
A
17.
[0, 1]
c.
[0,1]
B =
A,[0,1]
√2 √2
,
2 3
, ··· ,
√2
n
, ···
{r1, r2, · · · },
? A
?(
?(
√2
)=
2n
2
2n + 1
√2
,
n +1
) = rn,
n = 1, 2, · · ·
n = 1, 2, · · ·
?(x) = x,
x ∈ B.
?
18.
xi
A
[0,1]
A
c
[0,1]
A
c,
c.
A
c.
A = {ax1x2x3···} ,
E∞
Ai R ?i.
ax1x2x3··· ∈ A.?(ax1x2x3···) = (?1(x1), ?2(x2), ?3(x3), · · · ).
?
A
?
?(ax1x2x3···) = ?(ax′1x′2x′3···),
i, ?i(xi) = ?i(x′i).
?i
xi =
xi, ax1x2x3··· = ax1x2x3···.
∞
?
?i(xi) = ai.
ax1x2x3··· ∈ A, ?(ax1x2···) = (?1(x1), ?2(x2), · · · ) =
A E∞ c.
?i
19.
An
c,
n0,
An0
c.
n=1
∞
E∞ = c,
n=1
An = E∞.
An < c, n = 1, 2, · · · .
Pi
E∞
R
x = (x1, x2, · · · , xn, · · · ) ∈ E∞,
Pi(x) = xi.
A?i = Pi(Ai), i = 1, 2, · · · ,
4
? ?(0) = r1,
?
?
?
?
? ?(rn) = rn+2, n = 1, 2, · · ·
?
?
? ?(x) = x, x ∈ ((0, 1)\R),
?
√
xi ∈ Ai, Ai = c, i = 1, 2, · · · .
xi ∈ Ai,
(a1, a2, · · · ),
(a1, a2, a3, · · · ) ∈ E∞, ai ∈ R, i = 1, 2, · · · ,
′ ′ ′ ′
A? < A∞< c, i = 1, 2, · · · .
An.
n=1
ξ ∈
∞
∞
n=1
An,
i,
ξi ∈ R\A?i ,
i,
ξ = (ξi, ξ2, · · · , ξn, · · · ) ∈ E∞.
ξi = Pi(ξ) ∈ Pi(Ai) = A?i ,
ξ ∈ R\A?i
ξ ∈
n=1
An = E∞,
ξ ∈ E∞
i0,
Ai0 = c.
20.
0
1
T ,
T
c.
T = {{ξ1, ξ2, · · · } | ξi = 0 or 1, i = 1, 2, · · · }.
T
x ∈ (0, 1]
(0,1] T
E∞ ?
f((0, 1])
(0,1] 2
ξi 0 1,
T
E∞ ?(T )
f(x) = {ξ1, ξ2, · · · },
A = c.
f
5
ξ ∈
∈ Ai,
i i
? : {ξ1, ξ2, · · · } → {ξ2, ξ3, · · · },
A ≤ E∞ = c,
x = 0.ξ1ξ2 · · · ,
T ≥ (0, 1] = c.
Eo
E′
Eˉ
1.
P0
P0
P0 ∈ E′
P1
U(P, δ)(
E (
P0
P0
P1
)
U(P, δ) (
o
U(P, δ) ? E.
P0
)
P0
P0
P1 ∈ E ∩ U(P0) ? E ∩ U(P, δ)
E.
U(P, δ),
P1 = P ,
P0
P0
U(P0) ? U(P, δ),
P1
P0
P1
P0
E,
P0
′
P1
E,
P0
U(P0)
P0 ∈ Eo,
U(P0) ? E.
P0 ∈ U(P, δ) ? E,
U(P0) ? U(P, δ) ? E,
P0 ∈ Eo.
2.
E1
[0, 1]
E1
R1
E1′ , E1o, Eˉ1.
E1′ = [0, 1],
E1o = ?,
Eˉ1 = [0, 1].
3.
E2 = {(x, y)|x2 + y2 < 1}.
E2
R2
E2′ , E2o, Eˉ2.
E2′ = {(x, y)|x2 + y2 ≤ 1},
E1o = {(x, y)|x2 + y2 < 1},
Eˉ1 = {(x, y)|x2 + y2 ≤ 1}.
4.
E3
R2
y =
0,
x
E3
x = 0,
x =0
E3 E3.
E3′ = E3 ∪ {(0, y)| ? 1 ≤ y ≤ 1},
E3o = ?.
5.
R2
2
E1′ , E1o, Eˉ1
E1′ = {(x, 0)|0 ≤ x ≤ 1, }, E1o = ?, Eˉ1 = E1′ .
6.
F
Fˉ = F.
F
F
F ′ ? F,
Fˉ = F ∪ F ′ = F.
Fˉ = F,
F ′ ? F ∪ F ′ = Fˉ = F,
7.
G
F ? G = F ∩ ?G
F
?G
?F
G ? F = G ∩ ?F
8.
f(x)
E = {x|f(x) ≥ a}
1
a, E = {x|f(x) > a}
P0 ∈ E
),
P0 ∈ E′,
P0 ∈ E .
? sin 1 ,
?
?
′ o
(?∞, ∞)
x0 ∈ E,
(?∞, ∞),|x ? x0| < δ
f(x0) > a.
f(x) > a,
f(x)
x ∈ U(x0, δ)
x ∈ E,
δ > 0,
U(x0, δ) ? E,E
x ∈
xn ∈ E,
xn → x0(n → ∞).
f(xn) ≥ a,
f(x)
f(x0) = n→∞ f(xn) ≥ a,
x0 ∈ E,
9.
E
y0 ∈ F ,
F
1
1
n
1
1
d(x0, F ) = inf d(x0, y) ≥
1
d(x0, F ) <
1
n
).
? =
1
n
? δ > 0,
x ∈ U(x0, ?), d(x0, x) < ?.
d(x, y0) ≤ d(x0, x) + d(x0, y0) < ? + δ = ? +
1
n
? ? =
1
n
.
d(x, F ) = inf d(x, y) ≤ d(x, y0) <
1
x ∈ Gn.
U(x0, ?) ? Gn,
Gn
∞
x ∈
n=1
Gn,
n, x ∈ Gn, d(x, F ) <
1
n → ∞,
d(x, F ) = 0.
∞
F
x ∈ F (
x ∈ F ,
∞
yn ∈ F,
d(x, yn) → 0,
∞
x ∈ F ′ ? F ,
),
n=1
Gn ? F.
Gn ? F, n = 1, 2, · · · ,
n=1
Gn ? F ,
n=1
Gn = F ,F
∞
G
?G
Gn,
?G =
Gn,
n=1
∞
∞
?Gn
G
G = ?(?G) = ?(
n=1
Gn) =
n=1
?Gn,
10.
[0,1]
[0,1]
[0,1]
7
7
(0.7,0.8).
7
······
(0.07, 0.08) (0.17, 0.18)
···
(0.97, 0.98).
[0,1]
n
7
(0.a1a2 · · · an?17, 0.a1a2 · · · an?18),
ai(i = 1, 2, · · · , n ? 1)
n ? 1
0
9
7
{a1, a2, · · · , an?1}
∞
An
n=1
2
lim
Gn = x|d(x, F ) <
,Gn
y ∈ F, d(x0, y) ≥
d(x0, y0) = δ <
(
n.
n,
x0 ∈ Gn, d(x0, F ) <
n,
n,
y∈F
n,
y∈F
n.
[0,1]
7
∞
?
n=1
An ∪ (?∞, 0) ∪ (1, ∞) .
An, (?∞, 0), (1, ∞)
[0,1]
7
11. f(x)
E1 = {x|f(x) ≤ c}
[a, b]
c,
E = {x|f(x) ≥ c}
f(x)
[a, b]
8
E
E1
E
E1
x0 ∈ [a, b]. f(x)
f(xn) ≤ f(x0) ? ?0,
f(x0) < f(x0) + ?0 = c),
x0
E
?0 > 0, xn → x0, f(xn) ≥ f(x0) + ?0
c = f(x0) + ?,
f(x) [a, b]
12.
§2
5:
E = ?, E = Rn,
E
(
?E = ?).
P0 = (x1, x2, · · · , xn) ∈ E, P1 = (y1, · · · , yn) ∈ E.
(1 ? t)x2, · · · , tyn + (1 ? t)xn), 0 ≤ t ≤ 1.t0 = sup{t|Pt ∈ E}.
Pt = (ty1 + (1 ? t)x1, ty2 +
Pt0 ∈ ?E.
t0, tn → t0 Ptn ∈ E, Ptn → Pt0, Pt0 ∈ ?E.
t0 = 1.
Pt ∈ E.
tn, 1 > tn >
Pt0 ∈ ?E. ?E = ?.
t0 = 0,
tn, 0 < tn < t0, tn → t0, Ptn → Pt0, Ptn ∈ E,
P
13.
c.
P
1,
P
1 2
,
3 3
1 2
,
9 9
7 8
,
9 9
······
= (0.1, 0.2),
= (0.01, 0.02),
= (0.21, 0.22),
(
P ),
n
2n?1
(n)
(n)
= (0.a1a2 · · · an?11, 0.a1a2 · · · an?12),
a1, a2, · · · , an?1
1, P
0
2.
[0, 1] ? P
1,
x ∈ P ,
x
x =
a1
3
a2 an
+ 32 + ··· + 3n + ··· ,
an
A ? P .
0
2.
ai
A,
A ? P .
1,
[0, 1] ? P
A ? [0, 1],
A
[0, 1] ? P
3
x0 ∈ E(
xn ∈ E = {x|f(x) ≥ c},
Pt0 ∈ E. t ∈ [0, 1] t0 < t ≤ 1,
Pt0 ∈ E,
Ik , k = 1, 2, · · · , 2n?1
Ik
A
B
φ:
φ : x =
∞
n=1
an
3n
→
∞
n=1
1
2n
·
an
2
,
an = 0
Pˉ ≤ c ,
2,
Pˉ = c.
φ
A
B
1-1
A
c,
A ? P ,
Pˉ ≥ c,
4
1.
E
m?E < +∞.
E
I
E ? I.
m?E ≤ m?I < +∞.
2.
Rp
E = {xi | i = 1, 2, · · · }.
p ?
2i
? > 0,
Ii),
∞
i=1
Ii,
Ii ? E,
xi
∞
i=1
∈ Ii,
|Ii| = ?.
|Ii| =
?
?
m?E = 0.
3.
E
m?E > 0,
m?E
c,
E
E1,
?
[a, b]
a = inf x, b = sup x,
x∈E
△x > 0
E ? [a, b].
Ex = [a, x] ∩ E, a ≤ x ≤ b, f(x) = m?Ex
| f(x + △x) ? f(x) | =| m?Ex+x ? m?Ex |
≤| m?(Ex+△ ? E) |
≤ m?(x, x + △x] = △x.
(x 0 f(x+△x) → f(x),
f(x)
f(x)
[a, b]
△x > 0, △x → 0
f(a) = m?Ea = m?(E ∩ {a}) = 0
f(b) = m?(E ∩ [a, b]) = m?E.
4.
c,c < m?E,
m?E1 = c.
S1, S2, · · · , Sn
x0 ∈ [a, b]
f(x0) = c.
m?Ex0 = m?([a, x0] ∩ E) = c.
, Ei ? Si, i = 1, 2. · · · , n,
m?(E1 ∪ E2 ∪ · · · ∪ En) = m?E1 + m?E2 + · · · + m?En.
T ,
? S1, S2, · · · , Snn
Si) =
i=1 i=1
?
T =
n
i=1
Ei,
3
T ∩ Si = (j=1
n
1,
Ej) ∩ Si =
Ei, T ∩ (i=1
n
Si) =
n
i=1
Ei,
n
m?(
i=1
Ei) = m?(T ∩ (i=1
n
Si)) =
n
i=1
m?(T ∩ Si) =
n
i=1
m?Ei.
5.
m?E = 0,
E
T ,T = (E ∩ T ) ∪ (T ∩ ?E),
m?T ≤ m?(E ∩ T ) + m?(T ∩ ?E).
E ∩ T ? E, m?(E ∩ T ) ≤ m?E = 0.T ∩ ?E ? T, m?(T ∩ ?E) ≤ m?T,
m?(E ∩ T ) + m?(T ∩ ?E) ≤ m?T.
1
xi
2i (
m E1 = c.
x∈E
f x ? →x) → f(x),
E1 = E ∩ [a, x0] ? E.
m (T ∩
m (T ∩ Si).
§2
n
m?T = m?(T ∩ E) + m?(T ∩ ?E),
E
6.
(Cantor)
P
[0, 1]
n?1
3n
, ······ .
1
P
[0, 1]
∞
n=1
2
n?1
3n
,
n
).
m[0, 1] = m(P ∪ ([0, 1] ? P )) = mP + m([0, 1] ? P ).
mP = m[0, 1] ? m([0, 1] ? P ) = 1 ? 1 = 0,
0.
7.
A, B ? Rp
m?B < +∞.
A
m?(A∪B) = mA+m?B ?m?(A∩B).
A
m?(A ∪ B) = m?((A ∪ B) ∩ A) + m?((A ∪ B) ∩ ?A) = mA + m?(B ? A).
m?B = m?(B ∩ A) + m?(B ∩ ?A),
m?B < +∞,
m?(B ∩ ?A) < +∞,
m?(B ? A) = m?B ? m?(A ∩ B),
m?(A ∪ B) = mA + m?B ? m?(A ∩ B).
8. E
m(G ? E) < ?, m(E ? F ) < ?.
? > 0,
G
F ,
F ? E ? G,
∞
∞
mE < ∞
∞
? > 0,
{Ii}, i = 1, 2, · · · ,
∞
i=1
Ii ? E,
∞
i=1
| Ii |< mE + ?.
G =
i=1
Ii,
G
G ? E,
mE ≤ mG ≤
i=1
mIi =
i=1
| Ii |<
mE + ?,
mG ? mE < ?,
m(G ? E) < ?.
∞
mE = ∞
E
E =
n=1
En(mEn < ∞),
∞
En
∞
Gn,
∞
Gn ? En
∞
m(Gn ? En) <
?
G =
i=1
Gn, G
G ? E,
G ? E =
n=1
Gn ?
n=1
En ?
n=1(Gn ? En).
∞
m(G ? E) ≤
n=1
m(Gn ? En) < ?.
E
?E
?> 0
G, G ? ?E,
m(G ? ?E) < ?.
G ? ?E = G ∩ E = E ∩ ?(?G) = E ? ?G,
F = ?G,
F
m(E ? F ) = m(G ? ?E) < ?.
E
9.
E ? Rq,
{An}, {Bn},
An ? E ? Bn
m(Bn?An) → 0(n → ∞),
2
2
3,
9, · · · · · ·
2
= 1(
2n .
∞
∞
i,
n=1
Bn ? Bi,
n=1
Bn ? E ? Bi ? E.
E ? Ai, Bi ? E ? Bi ? Ai,
i,
∞
m?
n=1
Bn ? E
≤ m?(Bi ? E) ≤ m?(Bi ? Ai) = m(Bi ? Ai).
∞
∞
i → ∞,
m(Bi ? Ai) → 0,
m?
n=1
Bn ? E
= 0.
n=1
Bn ? E
Bn
∞
∞
∞
n=1
Bn
E =
n=1
Bn ?
n=1
Bn ? E
10.
A, B ? Rp,
m?(A ∪ B) + m?(A ∩ B) ≤ m?A + m?B.
Gδ
G1
m?A = +∞ m?B = +∞,
G2,
m?A < +∞
mG1 = m?A, mG2 = m?B.
m?B < +∞
m?(A ∪ B) ≤ m(G1 ∪ G2), m?(A ∩ B) ≤ m(G1 ∩ G2).
m?(A ∪ B) + m?(A ∩ B) ≤ m(G1 ∪ G2) + m(G1 ∩ G2) = mG1 + mG2 = m?A + m?B.
11.
E ? Rp.
? > 0,
F ? E,
m?(E ? F ) < ?,
E
∞
n,
Fn ? E,
m?(E ? Fn) <
1
F =
n=1
Fn,
F
F ? E.
n,
m?(E ? F ) = 0,
12.
E ? F
m?(E ? F ) ≤ m?(E ? Fn) <
E = F ∪ (E ? F )
μ
1
n
.
M,
μ ? M,
μ ≤ M.
μ ≥ M,
μ = M.
c
c,
3
G1 ? A, G2 ? B,
n.
1.
f(x)
E
r,
E[f > r]
E[f = r]
f(x)
∞
r,E[f > r]
α,
{rn}
α
E[f > a] =
E[f > rn],
E[f > rn]
E[f > α]
f(x)
E
n=1
E[f >
r,E[f = r]
x ∈ z, f(x) =
2] = z
√
f(x)
3; x ∈ z, f(x) =
f
√2,
E = (?∞, ∞), z (?∞, ∞)
r,E[f = r] = ?
E
2.
f(x), fn(x)(n = 1, 2, · · · )
fn(x)
-7f(x)
∞
k=1
[a, b]
n→∞ E | fn ? f |<
1
k
k
A
| fn(x) ? f(x) |<
1
E
fn
x ∈ A,
1
x ∈ n→∞ E | fn ? f |< k
.
k,
N,
n > N
k
x ∈
∞
k=1
n→∞ E | fn ? f |<
1
k
.
∞
x ∈
k=1 n→∞ E | fn ? f |<
1
k
,
? > 0,
k0,
1
k0
< ?,
x ∈ n→∞ E | fn ? f |<
1
k0
N,
n > N
x ∈ E | fn ? f |<
1
k0
,
| fn(x) ? f(x) |<
1
k0
< ?,
lim fn(x) = f(x),
n→∞
x ∈ A.
3.
{fn}
E
A =
∞
k=1
n→∞ E | fn ? f |<
1
k
.
§1
6, lim fn(x)
n→∞
lim fn(x)
n→∞
E
E[n→∞ fn = +∞]
鏈接地址:http://www.820124.com/p-1588281.html