周民强《实变函数》解答 第二章 lebesgue测度

k=1
k=1
0,
m( Ek) = 1.
i pqr Ww% ()0 k=1
Ex 9: Ei (i = 1, 2, · · · , k) [0, 1]
k
m(Ei) > k − 1,
i=1
k
m
Ei
i=1
> 0.
{
k
k
k
k
124¤6 @ ASq q m(Ei) > k−1 k− m(Ei) < 1,
Caratheodory
T = A2,
m∗(A2) = m∗(A2 ∩ A1) + m∗(A2 ∩ Ac1) = m(A1) + m∗(A2 \ A1)
65t QR m(A1) = m∗(A2) < ∞,
m∗(A2 \ A1) = 0.
uWwDP EF c TdagWwv 893x5y T ⊂ Rn,
A1
Caratheodory
T ∩ A2
m∗(T ∩A2) = m∗(T ∩A2∩A1)+m∗(T ∩A2 ∩Ac1) = m∗(T ∩A1)+m∗(T ∩(A2 \A1)) = m∗(T ∩A1).
m∗T = m∗(T ∩ A1) + m∗(T ∩ Ac1) = m∗(T ∩ A2) + m∗(T ∩ Ac1)
r
G1, G2:
m(G1 ∩ G2) < ε.
G1 ⊇ E, G2 ⊇ Ec,
C13v u25W 4wd8a9 W5T{DEg­T v y CGuW v E
ε > 0,
G1
E ⊂ G1,
{ CGu F ⊂ E, m(E \ F) < ε/2, G2 = Fc, G2
T q } daW H, H ⊇ B,H a¹ s B 89 P Caratheodory
65t
A∪B daW
QR
º

H ⊂ A ∪ B,
m(A ∪ B) = m∗((A ∪ B) ∩ H) + m∗((A ∪ B) ∩ Hc)
= m∗(A \ H) + m(H) = m∗(A \ H) + m∗B
XYT
≥ m∗(T ∩ A2) + m∗(T ∩ Ac2)
@
m∗T ≤ m∗(T ∩ A2) + m∗(T ∩ Ac2)

A1

)
=AF 02,4eCQdSdRf aTSAu2gS\shSACS1 CiSpjSa muC∗WwaTraX=thYmeo∗Ad(oT1ry∩d8SAa29 q)+WedkSmrlS∗(smST ∩nSAA2iSc2T)jSdSadSg aSs uSTSgSo
s6
m∗(A2 \
1
Ex 3: $ A,B ⊂ Rn ipqr W ()0
m(A ∪ B) + m(A ∩ B) = m(A) + m(B).
1t2u4wv x m(A), m(B)
∞. A,B Tdag
Pyxyz {∞,
Caratheodory
y| 89
1y} WwP
A∩H
da s Tdag¼½ Fs ¾¿TD A = ((A∪B)\H)∪(A∩H) A
B
gqd Às 4wÁÂÃ ÄÅÆÇÈÉÊÉ ËÌÍÎ (
Email: glhe@zjnu.cn,
A, B 2007.4.25)
3
1325476589@
ABCDEFGHI ∞
m∗(E ∩ (a, b)) ≤ m(In) < (b − a)q,
PI,
n=1
m∗(E ∩ I) ≤ q ∗ |I|
QSR D TSESF USV PSW XSYS`SaSbSTScSdSeSdSfSgSPSW E
L−
{In}
∞
∞
QR P m∗(E ∩ In) ≤ q ∗ |In|,
∞
E ⊂ In,
n=1
m∗(E) ≤
n=1
n=1
m∗(E) ≤ qm∗(E)
QR h m∗E = 0, mE = 0.
ipqr % ()0 i Ex 2: A1,A2 ⊂ Rn,A1 ⊂ A2,A1
m(A1) = m∗(A2) < ∞,
A2
pq13r25s 4765t CdauWwv 893x5y P A1
F s TdagW QR m(A ∪B) = m∗(A) +m∗(B)
m(A ∪ B) < ∞
89 P H
Caratheodory
m∗(A \ H) = m∗(A)
m∗(A) = m∗(A \ H) + m∗(A ∩ H)
65t C@ dauW m∗(A∩H) = 0,
(
m(A) < ∞, m(B) < ∞)
m(A ∪ B) + m(A ∩ B) = m(A) + m(B).
( i r Ex 4:
F ,F ⊂ [a, b] F = [a, b],
m(F ) = b − a?
uWwg 4wP v CuW 65t { q xzG F
m(G1 ∩ G2) = m(G1 \ F ) ≤ m(G1 \ E) + m(E \ F ) < ε.
89 T®¯gWw¡° 89 ± t DE²­ v CGuWw CuWw F ⊂E,
m(E \ F ) < ε.
mSn {SSvSGSHSIS³Se SSHSIS³Se ´ rSW εk =1/k,
i pqr 'W ()0 Ex 8: {Ek} [0,1]
m(Ek) = 1 (k = 1, 2, · · ·),
∞
m
Ek = 1.
k=1
124 6 { ¡S¢S£SuS t y QSR m(Ek) = 1, m(Ekc) = 0( ∞
[0, 1] ),
∞
∞
m( Ekc) ≤ m(Ekc) =
(1−m(Ei)) < 1,
m(Eic) < 1,
i=1
i=1
i=1
i=1
k
k
k
QR ¡¢£u t y m
源自文库
Eic ≤ m(Eic) < 1 ,
m Ei > 0. (
[0, 1] ).
i=1
i=1
i=1
2
()0¥
Ex 14:
r
E pq
¦§¨©ª«
i
4
¬
ε > 0, &
1/k,
∞
∞
µ { DEF ¶Wwh mn G = Gk,F = Fk,
k=1
k=1
s @ da s F ⊂ E ⊂ G, E
ε > 0, Gk m(G \ F ) ≤ 1/k
G Fk
k
m(G1 \ E) < ε/2;
F
G2 ⊇ Ec,
E ⊂ G,F Fk ⊂ E ⊂ Gk,m(Gk \ Fk) <
¢¡¢£ Lebesgue ¤¢¥§¦©¨¢¢
!"# $%&"#' Ex 1: E ⊂ R1,
q ∈ (0, 1),
(a, b),
{In}:
∞
∞
E ∩ (a, b) ⊂ In, m(In) < (b − a)q.
()0
n=1
n=1
m(E) = 0.
m(G \ F ) = 0,
ipqr Ww · Ex II(6): A,B ⊂ Rn,A ∪ B
m(A ∪ B) < ∞,
(» ){ 130y 25HA4 H,∩BT(A¸Bd∪pTaBq}g)saW ¹
m(A ∪ B) = m∗(A) + m∗(B),
F ⊂ [a, b] F = [a, b], (a, b) \ F = ∅, (a, b) \ F
m((a, b) \ F ) > 0,
m([a, b]) = m(F ) + m([a, b] \ F )
QR h x5 T m(F) = m([a, b]) − m([a, b] \ F) ≤ m([a, b]) − m((a, b) \ F) < b − a. u v s F
y~yyy
s QyRSy
m(A) < ∞, m(B) <
m(A ∪ B) = m((A ∪ B) ∩ A) + m((A ∪ B) ∩ Ac) = m(A) + m(B \ A),
d Fs qr m(B) = m(B ∩A) +m(B \A),