实变函数新编——魏勇—第一章答案
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1.
B
,
M ⊂ B, A ∪ B = (A − B ) ∪ (A − B ) ∪ M (A − B ) ∪ M . , , B−M
B = M ∪ (B − M ), M ∩ (B − M ) = ∅, A B−M 2. 3. C [a, b] , , . [a, b] A ∼ R. C [a, b] ≥ A = R = c. r1 , r2 , · · · , rn , · · · . ϕ(f ) = {f (rn )}, ) C [a, b] 4. F , F c. [a, b] F ≥ c. , ϕ . C [a, b] A−B M B A∪B ,
k=1 N =1 n=N
11. 12. 13. . f+ (x0 ) = limx→x+ f (x), 0
.
f− (x0 ) = limx→x− f (x) 0
w(x) = max{|f+ (x) − f (x)|, |f− (x) − f (x)|, |f+ (x) − f− (x)|}, Eε = {x ∈ R1 | w(x) ≥ ε}. ε > 0, Eε , , x0 − xn ), 3 , f− (x0 ), , xn ∈ Eε w(xn ) ≥ ε, |f (xn + an ) − f (xn )| ≥
an → a(n → ∞), n ≥ N,
∞ ∞ 1 |an − a| < k , ∞
k=1 N =1 n=N
1 1 (an − , an + ) = {a}. k k
7.
,
26
,
4
ቤተ መጻሕፍቲ ባይዱ
E1 × E2 × E3 × · · · × En · ·· ≤ c,
E1 × E2 × E3 × · · · × En · ·· 2 8. ( k ξ ∈ [c, x0 ] 0, , s=0 c ∈ (a, b) ) f (ξ ) = k , . c. x0 ∈ (a, b) f (c) = 0, [0, s]( c. f (x0 ) = s, x0 > c , s > 0) ,
B = (A − B ) ∪ M ∪ (B − M ) = [(A − B ) ∪ M ] ∪ (B − M ). [(A − B ) ∪ M ] ∩ (B − M ) = ∅.
A. , E∞ ( [a, b]
A⊂ f ∈ C [a, b].
C [a, b] ≤ E∞ = c.
F. F = c, [a, b] F ∼ [a, b]. t α ∈ [a, b], .
3
9. , 1, 2, · · · .) 10.
E = {r n }, R1 rn = rm + x 0 ,
A = {r n − r m : n = m }. x0 , E ∩ (E + {x0 }) = ∅.
A
x0 = rn − rm (n = m : n, m =
∞
∞
∞
{x| |fn (x)| ≥ k }
ft (x) = F (t, x)(x ∈ [a, b]) , g (x) = F (x, x) + 1 g (x) = fα (x).
F (x, x) + 1 = F (α, x),
x ∈ [a, b].
x=α
1
.
F > c.
2
5. . . 6.
A A
d 2
, . A
. k > 0, N > 0,
f (xn ) → f− (x0 ), limxn →x− 0 Eε
limxn →x− f (xn − an ) → f− (x0 ), 0
1 = {x|w(x) = 0} = ∪∞ n=1 {x|w (x) ≥ n }
ε 2
. xn → x0 , an → 0 ( xn x0 0 < an <
ε 2
|f (xn − an ) − f (xn )| ≥ n (∗). (∗) . f (x) . . x0
ε |f (xn + an ) − f (xn − an )| ≥ 2 ,
xn limxn →x− f (xn + an ) → 0 Eε
B
,
M ⊂ B, A ∪ B = (A − B ) ∪ (A − B ) ∪ M (A − B ) ∪ M . , , B−M
B = M ∪ (B − M ), M ∩ (B − M ) = ∅, A B−M 2. 3. C [a, b] , , . [a, b] A ∼ R. C [a, b] ≥ A = R = c. r1 , r2 , · · · , rn , · · · . ϕ(f ) = {f (rn )}, ) C [a, b] 4. F , F c. [a, b] F ≥ c. , ϕ . C [a, b] A−B M B A∪B ,
k=1 N =1 n=N
11. 12. 13. . f+ (x0 ) = limx→x+ f (x), 0
.
f− (x0 ) = limx→x− f (x) 0
w(x) = max{|f+ (x) − f (x)|, |f− (x) − f (x)|, |f+ (x) − f− (x)|}, Eε = {x ∈ R1 | w(x) ≥ ε}. ε > 0, Eε , , x0 − xn ), 3 , f− (x0 ), , xn ∈ Eε w(xn ) ≥ ε, |f (xn + an ) − f (xn )| ≥
an → a(n → ∞), n ≥ N,
∞ ∞ 1 |an − a| < k , ∞
k=1 N =1 n=N
1 1 (an − , an + ) = {a}. k k
7.
,
26
,
4
ቤተ መጻሕፍቲ ባይዱ
E1 × E2 × E3 × · · · × En · ·· ≤ c,
E1 × E2 × E3 × · · · × En · ·· 2 8. ( k ξ ∈ [c, x0 ] 0, , s=0 c ∈ (a, b) ) f (ξ ) = k , . c. x0 ∈ (a, b) f (c) = 0, [0, s]( c. f (x0 ) = s, x0 > c , s > 0) ,
B = (A − B ) ∪ M ∪ (B − M ) = [(A − B ) ∪ M ] ∪ (B − M ). [(A − B ) ∪ M ] ∩ (B − M ) = ∅.
A. , E∞ ( [a, b]
A⊂ f ∈ C [a, b].
C [a, b] ≤ E∞ = c.
F. F = c, [a, b] F ∼ [a, b]. t α ∈ [a, b], .
3
9. , 1, 2, · · · .) 10.
E = {r n }, R1 rn = rm + x 0 ,
A = {r n − r m : n = m }. x0 , E ∩ (E + {x0 }) = ∅.
A
x0 = rn − rm (n = m : n, m =
∞
∞
∞
{x| |fn (x)| ≥ k }
ft (x) = F (t, x)(x ∈ [a, b]) , g (x) = F (x, x) + 1 g (x) = fα (x).
F (x, x) + 1 = F (α, x),
x ∈ [a, b].
x=α
1
.
F > c.
2
5. . . 6.
A A
d 2
, . A
. k > 0, N > 0,
f (xn ) → f− (x0 ), limxn →x− 0 Eε
limxn →x− f (xn − an ) → f− (x0 ), 0
1 = {x|w(x) = 0} = ∪∞ n=1 {x|w (x) ≥ n }
ε 2
. xn → x0 , an → 0 ( xn x0 0 < an <
ε 2
|f (xn − an ) − f (xn )| ≥ n (∗). (∗) . f (x) . . x0
ε |f (xn + an ) − f (xn − an )| ≥ 2 ,
xn limxn →x− f (xn + an ) → 0 Eε