周民强《实变函数》解答 第二章 lebesgue测度

周民强《实变函数》解答 第二章 lebesgue测度

Lebesgue

Ex1:

E R1,

q∈(0,1),

∞

(a,b),

{In}:

E∩(a,b)

In,

n=1n=1

∞

m(In)<(b a)q.

m(E)=0.

m(E∩(a,b))≤

∞

m(In)<(b a)q,

I,

n=1

m (E∩I)≤q |I|

E

L

n=1∞

{In}

E

n=1

∞

n=1

m (E∩In)≤q

|In|,

∞

In,

m (E)≤

m (E)≤qm (E)

m E=0,

mE=0.

Ex2:

A1,A2 Rn,A1 A2,A1m(A1)=m (A2)<∞,

A2

A1

Caratheodory

T=A2,

m (A2)=m (A2∩A1)+m (A2∩Ac1)=m(A1)+m(A2\A1)

m(A1)=m (A2)<∞,T Rn,

m (A2\A1)=0.

A1Caratheodory

T∩A2

m (T∩A2)=m (T∩A2∩A1)+m (T∩A2∩Ac1)=m(T∩A1)+m(T∩(A2\A1))=m(T∩A1).

c

m T=m (T∩A1)+m (T∩Ac1)=m(T∩A2)+m(T∩A1)

≥m (T∩A2)+m (T∩Ac2)

m T≤m (T∩A2)+m (T∩Ac2)

m T=m (T∩A2)+m (T∩Ac2)

A2

A1)=0,

A2\A1

Caratheodory

A1

A2

m (A2\

周民强《实变函数》解答 第二章 lebesgue测度

Ex3:

A,B Rn

m(A∪B)+m(A∩B)=m(A)+m(B).

m(A),m(B)

∞,

m(A)<∞,m(B)<

∞.

A,BCaratheodory

m(A∪B)=m((A∪B)∩A)+m((A∪B)∩Ac)=m(A)+m(B\A),

m(B)=m(B∩A)+m(B\A),

(

m(A)<∞,m(B)<∞)

m(A∪B)+m(A∩B)=m(A)+m(B).

Ex4:

F,F [a,b]

F=[a,b],

m(F)=b a?

FF [a,b]m((a,b)\F)>0,

F=[a,b],

(a,b)\F= ,

(a,b)\F

m([a,b])=m(F)+m([a,b]\F)

m(F)=m([a,b]) m([a,b]\F)≤m([a,b]) m((a,b)\F)<b a.F

Ex8:

{Ek}

[0,1]

m(Ek)=1(k=1,2,···), ∞ mEk=1.

k=1

m(Ek)=1,

c

m(Ek)=0(

[0,1]

),

m(

0,

m(

k=1

k

∞

k=1

Ek)=1.

∞

c

Ek)≤

k=1

∞

c

m(Ek)=

Ex9:

Ei(i=1,2,···,k)

[0,1]

m(Ei)>k 1,

i=1

m

i=1

k

Ei

>0.

m

i=1

k

c

Ei

i =1

k

m(Ei)>k 1

k

≤

i=1

k

c

m(Ei)<1,

m(Ei)<1,(1 m(Ei))<1,

i=1 i=1 k mEi>0.([0,1]

k k

i=1

k

c

m(Ei)<1,

).

i=1

周民强《实变函数》解答 第二章 lebesgue测度

Ex14:

E

ε>0,

G1,G2:

G1 E,G2 Ec,

m(G1∩G2)<ε.

ε>0,G1E G1,

F E,m(E\F)<ε/2,G2=Fc,G2

m(G1∩G2)=m(G1\F)≤m(G1\E)+m(E\F)<ε.E

m(G1\E)<ε/2;G2 Ec,

E G,F

F

F E,

1/k,

m(E\F)<ε.εk=1/k,

ε>0,

GFk

Gk

Fk E Gk,m(Gk\Fk)<

m(G\F)=0,

G=

k=1

F E G,

∞

Gk,F=

E

k=1

∞

Fk,

m(G\F)≤1/k

k

ExII(6):

A,B Rn,A∪B

m(A∪B)<∞,

m(A∪B)=m (A)+m (B),

A,B

B

H∩(A∪B)H

H,B

H B,H

A∪B

H A∪B,

Caratheodory

m(A∪B)=m ((A∪B)∩H)+m ((A∪B)∩Hc)

=m (A\H)+m(H)=m (A\H)+m B

m(A∪B)=m (A)+m (B)HCaratheodory

m(A∪B)<∞

m (A\H)=m (A)

m (A)=m (A\H)+m (A∩H)

m (A∩H)=0,A∩HA=((A∪B)\H)∪(A∩H)

A

B

A,B

(

Email:glhe@http://doc.guandang.net,

2007.4.25)