Hoc Phan Do Do Va Tich Phan

7/30/2019 Hoc Phan Do Do Va Tich Phan

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HC PHN O V TCH PHN

CHNG I. O

$1. I S. -I S1. i sa) nh ngha 1. Cho tp hp . Mt h N cc tp con caX c gi l mt i s cc tp con ca X, nu N tho mn ba iukin sau:

(i) XN ;(ii) A N CXA = X \ AN ;(iii) A1, A2, ... , AnN U

n

k

kA

1=

N .

b)Cc tnh chtCho N l i s cc tp con ca tp hp X. Khi N c cc tnh

cht sau y:

1. N ;2. A1, A2, ... , AnN I

n

kkA

1=

N ;

3. A, B N A \ B N.Chng minh.1. c suy t (i), (ii)

2. c suy t (ii), (iii) v cng thc de Morgan:

I Un

k

n

kkk CAAC

1 1)(

= ==

3. c suy t (ii), tnh cht 2 va chng minh v cng thcA \ B = A CXBNhn xt i s cc tp con ca tp hp X c tnh cht

" khp kn" i vi cc php ton : hp hu hn, giao hu hn, hiucc tp hp v ly phn b ( ngha l : khi ta thc hin cc php tonny trn cc phn t ca N th kt qu s l cc phn t ca N).c)Cc v d1. Cho .t N =

{ }CX,,,.

Khi N l mt i s cc tp con ca X.2. Cho X = { 1, 2, 3, 4, 5, 6, 7 }, A = { 1, 3, 5, 7 }, B = { 2, 4, 6 },


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C = { 1, 2, 4, 7 }, D = { 3, 5, 6 }.

t N = { , X, A, B, C, D }. Hy kim tra xem N c l mt is cc tp con ca X?3. Cho N l mt h khng rng cc tp con ca tp hp X tho mniu kin :Nu A, B N th X \ A N v A B N.Chng minh rng N l mt i s cc tp con ca X.2. - i sa) nh ngha 2. Cho tp hp . Mt h M cc tp con caX c gi l mt - i s cc tp con ca X, nu M tho mnba iu kin sau:

(i) XM ;(ii) A

M CXA = X \ AM ;(iii) A1, A2, ... , An , ... M U

=1kkA

M .

b)Cc tnh chtCho M l mt- i s cc tp con ca tp hp X. Khi M

c cc tnh cht sau y:1. M l mt i s cc tp con ca X;2.

M ;

3. A1, A2, ... , AnM In

kkA

1=

M ;

4. A, B M A \ B M ;

5. A1, A2, ... , An , ... M I

=1kkA

M .

Chng minh.

- Tnh cht 1 c suy t (i), (ii) v (iii) khi tAn+1 = An+2 = ... = .

- Tnh cht 2, 3, 4 c suy t tnh cht 1 va chng minh.- Tnh cht 5 c suy t (ii), (iii) v cng thc de Morgan:

I U

=

==

1 1)(

k kkk CAAC

Nhn xt - i s cc tp con ca tp hp X c tnh cht " khpkn" i vi cc php ton : hp m c, giao m c ca cc tphp, hiu hai tp hp v ly phn b ( ngha l : khi ta thc hin cc


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php ton ny trn cc phn t ca M th kt qu s l cc phn t caM ).c)Cc v d

1. Cho tp hp . H tt c cc tp con ca tp hp X lmt - i s cc tp con ca tp hp X.2. Cho M l mt h khng rng cc tp con ca tp hp X tho mnhai iu kin :

a) A M X \ A M ;b) A1, A, ... , An , ... M I

=1kkA

M .

Chng minh rng M l mt - i s cc tp con ca X.3. Cho M l mt- i s cc tp con ca tp hp X v Z M.t MZl h tt c cc tp hp thuc M v cha trong Z.

Chng minh MZ l mt- i s cc tp con ca tp hp Z.$2. O1. Tp hp s thc khng m m rng

Cho tp hp s thc khng m ),0[ + .Ta b sung cho tp hp ny mt phn t l + , tp hp mi

thu c l ],0[ + . Ta gi y l tp s thc khng m m rng vi

cc quy c v php ton nh sau.a < + vi mi a ),0[ + ;a + (+) = (+) + a = + vi mi a ],0[ + ;a . (+) = (+) . a = + vi mi a ],0( + ;0 . (+) = (+) . 0 = 0

Lu . ng thc a + c = b + c ko theo a = b khi v chkhi+c .

2. Cc khi nimCho M l mt- i s cc tp con ca tp hp X.Xt nh x : M ],0[ + .

nh ngha 1. c gi l nh x cng tnh hu hn, nu c mth hu hn cc tp hp i mt ri nhau A1, A2, ... , An M th

==

=n

kk

n

kk AA

11)()( U

nh ngha 2. c gi l nh x - cng tnh nu c mt hm c cc tp hp i mt ri nhau A1, A2, ... , An , ... M th


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+

=

+

==

11)()(

kk

kk AA U

nh ngha 3. c gi l mt o trn M, nu hai iu kin sauc tho mn:

1. () = 0;2. l - cng tnh.

nh ngha 4. Cp (X, M), trong M l - i s cc tp con catp hp X, cgi l khng gian o c. Mi tp hp A M cgi l mt tp o c.nh ngha 5. B ba (X, M, ), trong M l - i s cc tp conca tp hp X,

l mt o trn M, cgi l khng gian o.

Nu A M th s(A) c gi l o ca tp hp A.nh ngha 5.o c gi l o hu hn nu (X) < + .

o c gi l o - hu hn, nu X = U

=1kkX

, XkM

v (Xk) < + vi mi k.Nhn xt. o hu hn th - hu hn.3. Cc v da) Cho M l mt- i s cc tp con ca tp hp X.

Xt nh x : M ],0[ + xc nh bi (A) = 0 vi mi AM .

Khi l mt o hu hn.b) Cho M l mt- i s cc tp con ca tp hp X.

Xt nh x : M ],0[ + xc nh bi

() = 0 , (A) = + vi mi A M v .

Khi l mt o khng - hu hn.c) Cho M l mt- i s cc tp con ca tp hp X v x0 X.

Xt nh x : M ],0[ + xc nh bi :- Nu A M v x0 A th (A) = 1 ;- Nu A M v x0 A th (A) = 0 .Chng minh rng l mt o hu hn.

Nhn xt. C nhiu cch xy dng o trn cng mt - i scc tp con ca tp hp X, ng vi mi o s c mt khng gian o tng ng vi cc tnh cht khc nhau.


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4. Cc tnh cht ca oCho (X, M, ) l mt khng gian o. Khi ta c cc tnh

cht sau y.1. l cng tnh hu hn.2. Nu A, B M v A B th

(A)

(B) .

Ngoi ra, nu (A) < + th (B \ A) = (B) -(A).3. Nu A1, A2, ... , An , ... M th

+

=

+

=

11)()(

kk

kk AA U

4. Nu A, B M , A B v (B) = 0 th (A) = 0.

5. Nu A, B M v (B) = 0 th(A B) = (A \ B) = (A).

6. Hp ca mt h hu hn cc tp hp c o khng l tphp c o khng:

(Ak ) = 0, k = 1, 2, ... , n 0)(1

==

Un

kkA

7. Hp ca mt hm c cc tp hp c o khng l tphp c o khng:

(Ak ) = 0, k = 1, 2, ... 0)(1

=+

=U

kkA

8. Nu l o - hu hn thi) X = U

=1kkY

, trong cc tp hp Yk i mt ri nhau,

YkM v (Yk) < + vi mi k;ii) A = U

=1kkA

, trong cc tp hp Ak i mt ri nhau,

AkM v (Ak) < + vi mi A M v mi k.9. Nu { An } , n N, l dy n iu tng cc tp hp o c,

ngha l A1 A2 ... An ... , th

U+

= +=

1)()( lim

nn

nn AA

10. Nu { An } , n N, l dy n iu gim cc tp hp oc, ngha l A1 A2 ... An ... , v (A1) < + th


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)(lim)(1

nnn

n AA +

+

==I

5.o rng tp con ca mt tp o c cha chc l tp hp o

c, ngha l nu A

M

, B

A th c th B M

.nh ngha 6.o c gi l o nu mi tp con ca tpc o khng u l tp o c.

Nhn xt. Nu l o khng th ta c th thc trin thnh mt o nhnh l di y.nh l. Gi s (X, M, ) l mt khng gian o.

Gi M' l h tt c cc tp hp A c dngA = B C (1)

trong B M, C D, D M, (D) = 0.Vi mi tp hp A c dng (1), t ' l nh x sao cho

'(A) = (B) (2)Khi :

i) (X, M', ') l mt khng gian o;ii) ' l o .

nh ngha 7. M' c gi l b sung Lebesgue ca - i sM v' c gi l thc trin Lebesgue ca o .

6. Thc trin nh x- cng tnhthnh onh l (Hahn). Cho N l mt i s cc tp con ca tp hp X vm : N ],0[ + l mt nh x- cng tnh. Khi tn ti mt- i sM cha N v mt o : M ],0[ + sao cho(A) = m(A) vi mi A N . Ngoi ra, nu m l - hu hn th xc nh mt cch duy nht.nh ngha 8.o c gi l thc trin ca m ti s N ln- i sM.

$3. O LEBERGUE TRN 1. Khong trong nh ngha 1. Cc tp hp sau y c gi l cc khong trong :(a, b), [a, b], (a, b], [a, b), (- , a), (- , a], (a, +), [a, +)(- , +).


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rng giao ca hai khong bt k trong cng l khongtrong hoc l tp hp rng.

nh ngha 2. Nu l khong trong c hai u mt l a, b(- a b +) th ta gi s = b - a l di ca .2. i s cc tp con ca Xt h N cc tp hp P l hp ca hu hn cc khong trong khng giao nhau:

N = Un

ijii jiIIIPP

1)(,/

=== (1)

Trn N xt nh x m : N],0[

+ xc nh bi

=

=n

iiIPm

1)(

nu P c biu din nh trong (1).

nh l 1. N l mt i s cc tp con ca .Chng minh.

Ta kim tra ba iu kin ca nh ngha i s.

(i) Ta c = (- , +) ( hp ca mt khong) nn hin nhinN .

(ii) Gi s P N th P l hp ca hu hn khong khng giaonhau. Khi d thy \ P cng l hp ca hu hn khongkhng giao nhau. Vy \ P N.

(iii) Gi s P, Q N, ta cn chng minh P Q N.

Trc ht ta chng minh PQ N.Tht vy, v P, Q u l hp ca hu hn khong khng giao

nhau nn ta c biu din:

IU )'(,, '1

iiIIIP iin

ii ==

=

IU

)'(,, '1 jjJJJQ jjk

j j==

=

Khi


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I U U IU U

U II I U

k

j

n

i

jij

k

j

n

i

i

k

jj

k

jj

JIJI

JPJPQP

1 11 1

11

)(])[(

)()(

= == =

==

==

===

Th m I ijji LJ = ( i = 1, 2, ... , n ; j = 1, 2, ... , k) l cckhong khng giao nhau i mt nn PQ N.

By gi ta chng minh P Q N khi P, Q N .Thy vy, ta c P, Q N nn theo (ii) \ P N , \ Q N. Khi, theo phn va chng minh, ( \ P) ( \ Q) N , hay \ (P Q) N, li theo (ii) suy ra P Q N .Vy, N l i s cc tp con ca , nh l c chng minh.nh l 2.nh x m nh x - cng tnh.Chng minh.

Gi s Q = U

=1kkP

, trong cc tp hp Pk i mt ri nhau,

Q, Pk N (Q v Pku l hp ca hu hn khong khng giao nhau).Ta cn chng minh

+

=

=1

)()(k

kPmQm

Khng mt tnh tng qut ta c th xem Q v mi Pk chl mtkhong trong .Trc ht ta chng minh cho trng hp Q l khong hu hn.Khi cc Pk cng l khong hu hn.

Gi s Q l khong hu hn c hai u mt l a, b , cn Pk chai u mt l ak, bk .

- Vi mi n = 1, 2, ... , lun tn ti hu hn cc khong i( i = 1, 2, ... , ni ) sao cho

U UUin

ii

n

kk IPQ

11)()(

===

trong cc Pk , i ri nhau.Khi

===

+=n

kk

n

ii

n

kk PIPQ

i

111

Cho n + , ta c


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+

=

1kkPQ (2)

- Cho > 0 tu sao cho 2ab


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2)(21

+ +

=kkk abab

hay

41 + +

=kkPQ

Cho 0, ta c

+

=

1kkPQ (3)

T (2) , (3) suy ra

+

==

1k kPQ

hay

+

==

1)()(

kkPmQm

By gi ta chng minh cho trng hp Q l khong v hn.Khi +=Q .

R rng ta lun c th biu din Q dng

+==+

= +U

121 lim...,,

nn

nn IIIIQ

trong cc n u l khong hu hn.Chng hn, U

+

=+=+=

1),(),(

n

naaaQ

V Qn v Q = U

=1kkP, cc Pk ri nhau nn

U II I U+

=

+

====

11)()(

kkn

kknnn PIPIQII

trong cc tp hp I kn hu hn v ri nhau theo chs

k = 1, 2, ...Theo phn va chng minh

+=

+

==

11 kk

kknn PPII I


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Cho n + , ta c

+

=+

1kkP

Do phi c

QPk

k =+=+=1

Vy, m l nh x - cng tnh trn i s N cc tp con ca .Theo nh l Hahn v thc trin nh x- cng tnh thnh o, ta

c mt - i s M cha N v mt o l thc trin cam t N ln M .

3. o Lebesgue trn nh ngha 3. o v- i s M nhn c khi thc trinnh x m trn i s N cc tp con ca c gi ln lt l oLebesgue v - i scc tp o ctheo ngha Lebesgue trn.

Cc tnh chto Lebesgue v - i s M cc tp o ctheo nghaLebesgue trnc cc tnh cht sau y.

1. l o .2. Tp khng qu m c trn c o khng.3. Tp m, tp ng trn l tp o c.4. Tp A l o c khi v chkhi vi mi > 0 tn ti

cc tp m G, tp ng F sao cho F A Q v (G \ F) < .

5. Nu A o c th cc tp hp t A , x0 + A ( t, x0) cngo c v ( t A ) = / t /( A ) , ( x0 + A ) =( A),trong

{ } { }aaxxatatA +=+= /,/ 00 Cc v da) Tp hp Q cc s hu t c o khng.b) Tp hp Cantor P0 trn [0, 1] xy dng theo cch di y c o khng.

Xt tp hp [0, 1].- Bc 1. Chia [0, 1] thnh ba khong bng nhau, bi khong

gia G1 = (1/3, 2/3).


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- Bc 2. Chia ba mi on cn li l [0, 1/3] v [2/3, 1] , bikhong gia ca chng, t G2 = (1/9, 2/9) (7/9, 8/9).

- v.v...Gi Gn l hp ca 2

n-1 cc khong bi bc th n ,

G =

U

=1k kGl hp ca tt c cc khong bi , P0 = [0,1] \ G.

Ta c cc tp Gn ri nhau v (Gn ) = 2n-1. 1/ 3n = 1/2 . (2/3)nKhi

+

=

+

====

1 132

21 1)()()(

n n

n

nGG

Vy (P0) = ([0, 1]) - (G) = 0.

rng tp hp P0 l tp khng m c v c o khng.$4. HM SO C1. Tp hp s thc mrng

Cho tp hp s thc = (- , + ).Ta b sung cho tp hp ny hai phn t l - , + , tp hp mi thu

c l [- , +] = (- , + ) {- , + } . Ta gi y l tp s thcmrng, k hiu l , vi cc quy c v php ton nh sau.

- < a < + vi mi a ;

a + (+) = (+) + a = + vi mi a (- , +];a + (-) = (-) + a = - vi mi a [- , +);

a . (+) = (+) . a = + vi mi a ],0( + a . (-) = (-) . a = - vi mi a ],0( + ;

a . (+) = (+) . a = - vi mi a (- , 0);a . (-) = (-) . a = + vi mi a (- , 0);0 . (+) = (+) . 0 = 0; 0 . (-) = (- ) . 0 = 0;

== + aaa ,0

+==+ Cc k hiu (+) + (-), (+) - (+), (-) - (-),

, 0a

vi mi a u khng c ngha.


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2. Hm s hu hn

nh ngha 1. Hm s f : A c gi l hu hn trn A nuf(A) .Cc v d1. Hm s f(x) = sinx l hu hn trn v f( ) = [-1, 1] .2. Hm s f(x) = x l hu hn trn v f( ) = .3. Hm s

=+

=

0

)1,0()(

1

xkhi

xkhixf x

l hm s khng hu hn trn [0, 1).3. Hm so c

Di y ta cho (X, M) l khng gian o c v A M .

nh ngha 2. Hm s f : A c gi l o c trn A nu{ } axfAxa )(/, M(4)

{ } axfAxa )(/, M


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Chng minh.t

{ }axfAxB = )(/ { }axfAxE = )(/ Khi ta c C = A \ B, E = A \D.Do B M C M v E M D M .

Suy ra (1) (2), (3) (4)nn ta ch cnchng minh (2) (3).- Trc ht ta chng minh

IU+

=

+

=

==

11

,n

n

n

n DCCD

trong

{ }nn

axfAxC 1)(/ +=

{ }nn

axfAxD 1)(/ >= Tht vy, ly x D th x A v f(x) > a. Theo tnh cht tr mt ca tp

s thc, tn ti 0n sao cho aaxf n >+0

1

)(

Suy ra 0nCx do U

+

=

1n

nCx

Ngc li, ly U+

=

1n

nCx th tn ti 0n sao cho 0nCx

Khi x A v 01)(n

axf + nn f(x) > a. Suy ra x D.

By gita ly x C th x A v axf )( nn vi mi n ta c

naxf 1)( > . Suy ra nDx vi mi n, do I

+

=

1n

nDx

Ngc li, ly I+

=

1n

nDx th nDx vi mi n, do x A v

naxf 1)( > vi mi n. Ly gii hn hai v ca bt ng thc cui


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cng ny khi n + , ta c )(lim)(lim1n

nnaxf =

++hay

axf )( . Do x C.Vy ta c cc ng thc v tp hp cn chng minh trn y.

- By gita chng minh (2) (3).Tht vy, gi s ta c (2), khi vi mi a vmi n ta c nC M

. M M l - i s nn D M . Vy (3) c tho mn.Ngc li, gi s ta c (3). Khi vi mi a vmi n ta c nD

M . M M l - i s nn C M . Vy (2) c tho mn.nh l chng minh xong.

H qu

1 Nu fo c trn A M v , BM th fo c trn .Chng minh.

Vi a ta c( ){ } ( ){ }/ /x f x a x f x a < = < M

2 f o c trn 1 , 2 , v f xc nh trn1

n

n

=

= U th fo c

trn .Chng minh.

Vi a ta c

( ){ } ( ) ( ){ }1 1

/ / /n nn n

x f x a x f x a x f x a

= =

< = < = <

U U M

do M l - i s.4. Cc tnh cht ca hm o c1 Nu f o c trn v c = const th cfo c trn .2 Nu f , go c v hu hn trn th f g+ , fgo c trn .

3 Nu f o c trn , 0 > th f o c trn .

4 Nu ( ) 0,f x x v fo c trn th1

fo c trn .

5 Nu f , go c trn th ( )max ,f g , ( )min ,f g o c trn .

6 Nu { }nf l dy hm o c trn th sup nn

f , inf nn

f , limsup nn

f

,

liminf nn

f

o c trn .

7 Nu { }nf hi t trn , nf o c trn th lim nn

f

o c trn .


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8 Nu f , go c trn th cc tp hp ( ) ( ){ }/x f x g x < ,

( ) ( ){ }/x f x g x , ( ) ( ){ }/x f x g x = u thuc M .9 Nu f o c trn th cc hm s

( )( ) ( )

( ) ( ){ }, 0

max ,00, 0

f x khi f xf x f xkhi f x

+

= =


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V d 8. Xt hm s

=

)7,4(4

]4,3[2

)3,1[1

)(

khi

khi

xkhi

xf

y l hm n gin trn [1, 7).Nhn xtCho hm n gin [ ]: 0;S + v 1 , 2 , , n l cc gi tr khcnhau i mt ca S.

t ( ){ }: , 1,k kx S x k n = = = .

Th th cc k ri nhau,1

k

k

=

= U v ( ) ( )1

,k

n

k

k

S x x x =

= .

V d 9. Xt hm n gin v d 8.

t 1 = [1, 3), 2A = [3, 4], 3A = (4, 7), 1 = 1, 2 = 2,

3 = 4. Khi cc tp hp ny ri nhau v)()()()(

321 321xxxxf AAA ++= vi mi x [1, 7).

Xt tnh cht ca hm n gin.

Cho ( ), M - khng gian o c, A M.nh l 3. Cho S l hm n gin trn

( ) ( )1

k

n

k

k

S x x =

= , AAn

k

k ==

U1

, k ri nhau, k khc nhau.

Khi So c trn khi v ch khi mi kA M.Chng minh.- Nu So c trn th

( ){ }: 1,k kx S x k n = = =M , - Nu 1 , , kA M th theo nh l 1 mi hm c trng k o ctrn . Khi hm ( )S x o c trn (v l tng, tch cc hm hu hno c).7. Cu trc ca hm o cnh l 4. Mi hm so c khng m trn u l gii hn ca mt dyn iu tng cc hm n gin o c trn .Chng minh.

Gi sf l hm o c khng m trn .t


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( )( )

( )

,

1 1, , 1,2,..., 2

2 2 2n n

n n n

n khi f x n

S x m m mkhi f x m n

= < =

th ( )nS x l dy n iu tng (theo n ) cc hm n gin o c trn .

Ta chng minh ( ) ( )lim ,nn

S x f x x

= .

Tht vy,- Nu ( )f x < + th vi n ln ta c ( )f x n< .

Do vi n ln tn ti s t nhin { }1,2,..., 2nm n sao cho

( )1

2 2n nm m

f x

< . V ( )1

2n nm

S x

= nn ( ) ( )1

2n nS x f x < vi n

ln.- Nu ( )f x = + th ( )nS x n= vi n .

Suy ra, ( ) ( )lim nn

S x f x

= + = .

Vy, ( ) ( )lim ,nn

S x f x x

= trong c hai trng hp.

$5. SHI T HU KHP NI

1.Khi nim hu khp ninh ngha 1. Cho khng gian o (X, M, ) v A M . Ta ni mttnh cht no xy ra hu khp ni trn tp hp A nu tn ti mt tphp B A , B M, (B) = 0 sao cho tnh cht xy ra ti mi x A\ B.

Ni mt cch khc, cc im x A m ti tnh cht khng xyra u thuc tp hp c o khng.

Hin nhin, mt tnh cht xy ra ( khp ni ) trn A th xy ra hu khpni trn A.

Sau y ta a ra mt vi khi nim c th thng s dng.nh ngha 2. Hai hm s f, g cng xc nh trn tp hpA M c gi l bng nhau hu khp ni trn A (hay tngng nhau trn A ) nu tn ti mt tp hp B A , B M, (B) = 0 saocho f(x) = g(x) vi mi x A \ B.

Khi ta k hiu f ~ g (trn A).V d 1. Hm s Dirichlet D(x) ~ 0 trn v D(x) = 0 vi mix \ Q , trong Q l tp o c v c o khng.V d 2. Hm s

=+=

0)1,0()(

1

xkhi

xkhixf x


19/58

tng ng vi hm s

=

=

01

)1,0()(

1

xkhi

xkhixg

x

trn [0, 1), v f(x) = g(x) vi mi x [0, 1) \ B, trong B = {0} l tp con ca [0, 1), o c v c o khng.V d 3. Hm s

=

Qxkhix

Qxkhixxf

\],0[cos

],0[sin)(

2

2

I

tng ng vi hm s g(x) = cosx trn [0, 2

] .nh ngha 3. Hm s fc gi l hu hn hu khp ni trn tp hp A M nu tn ti mt tp hp B A , B M,(B)= 0 sao cho f(x) vi mi x A \ B.V d 4. Hm s f(x) c cho v d 2 hu hn hu khp ni trn [0, 1).nh ngha 4. Hm s f c gi l xc nh hu khp ni trn tp hp AM nu tn ti mt tp hp B A , B M,(B) = 0 sao cho f xc nh trn A \ B.

V d 5. Hm s scp xxf1)( = xc nh hu khp ni trn .

nh ngha 5. Dy hm s { }nf c gi l hi t hu khp ni v hms f trn tp hp A M nu tn ti mt tp hp B A , B M, (B) =

0 sao cho )()(lim xfxfnn=

+

vi mi x A \ B.

V d 6. Dy hm s{ }nf xc nh bi

n

n

xx

xxxn xf 2

2

4sin3)( +

+=

hi t hu khp ni v hm s xxxxf

+= 432)( trn [-1, 1].

2.Shi thu khp ninh l 1. Cho khng gian o (X, M, ) v A M .Khi

(i) Nu f ~ g (trn A) v { }nf hi t h.k.n v f trn A th { }nf hi t h.k.n v g trn A.


20/58

(ii) Nu { }nf hi t h.k.n v f trn A v { }nf hi t h.k.n v gtrn A th f ~ g (trn A).

Chng minh.(i) V f ~ g (trn A) nn tn ti mt tp hp B A , B M, (B) =

0 sao cho f(x) = g(x) vi mi x A \ B.Mt khc, v { }nf hi t h.k.n v f trn A nn tn ti mt tp hp C A ,C M, (C) = 0 sao cho )()(lim xfxfnn =+ vi mi x A \ C.Khi (B U C) A, B U C M, (B U C) = 0 v vi mi

x (A \ B) I ( A \ C) = A \ (B U C) ta c

)()()(lim xgxfxfnn

==+

Vy { }nf hi t h.k.n v g trn A.(ii) Tng t, do { }nf hi t h.k.n v f trn A nn tn ti mt tp hp

B A , B M, (B) = 0 sao cho )()(lim xfxfnn =+ vimi x A \ B.

Li do { }nf hi t h.k.n v g trn A nn tn ti mt tp hp C A , CM, (C) = 0 sao cho )()(lim xgxfnn =+ vi mi x A \ C.

Khi , theo tnh cht duy nht ca gii hn ca dy s, vi mi x (A\ B) I ( A \ C) = A \ (B U C) ta phi c

)()()(lim xgxfxfnn

==+

.

M (B U C) A, B U C M, (B U C) = 0 nn f ~ g (trnA).nh l c chng minh.

Tnh l suy ra rng, nu ta ng nht cc hm s tng ng thgii hn ca dy hm hi t hu khp ni l duy nht.

nh l 2. (Egoroff) Gi s { }nf l mt dy hm o c, hu hnh.k.n, hi t h.k.n v hm s f o c, hu hn h.k.n trn mt tp hpA c o hu hn. Khi vi mi > 0, tn ti mt tp hp E o


21/58

c, E A sao cho (A \ E) < v dy hm { }nf hi tu vf trn E.

ngha: nh l Egoroff khng nh rng mi s hi t c th bin thnh hitu sau khi bi mt tp hp c o b tu .

Mi lin h gia hm o c v hm lin tc trn - Nu A l tp o c theo ngha Lebesgue trn v hm s f :A l hm lin tc trn A th fo c (L) trn A.

Tht vy, nu a l mt s thc bt kth v f lin tc trn A nntp hp

B = { x A : f(x) < a } = 1f (- , a)l mt tp mtrong A.

Mt khc, do A l khng gian con ca nn B = A I G , vi G l

mt tp mtrong .Suy raBo ctheo ngha Lebesgue trn .Vy foc (L) trn A.- Ngc li, mt hm so c (L) trn tp hp A cha chc l

hm lin tc trn A.Tuy nhin nh l di y s cho ta thy mt hm o c c th trthnhhm lin tc nu b qua mt tp hp c o b tu .nh l 3. (Lusin) Gi s

f l mt hm s hu hn xc nh trn tp hp A ;A l tp o c theo ngha Lebesgue v c o hu hn.

Khi f o c (L) trn A khi v ch khi vi mi s> 0,

tn ti mt tp hp ng F A sao cho (A \ F) < v f lin tc trn F.

$6. SHI T THEO O

3.Khi nimnh ngha 1. Gi s( ), M , l khng gian o, M v f , 1f ,

2f , l nhng hm o c hu hn hu khp ni trn A. Dy { }nf cgi l hi t theo o n f v k hiu nf f

trn A, nu vi0 > ta u c

( ) ( ){ }( )lim : 0nn

x f x f x

= .

Ni cch khc, vi 00 0 n > > sao cho

( ) ( ){ }( )0: : nn n n x f x f x >


22/58

Ch : iu kin f , 1f , 2f , hu hn hu khp ni m bo cho

nf f xc nh hu khp ni trn A.

V d 1. Xt dy hm { }nf xc nh bi

1

, [0,1 )( )

12, [1 ,1)

nx khi x n

f x

khi xn

=

v hm s ( ) , [0,1)f x x x= .Th th nf f

trn [0,1).

4.Tnh duy nht ca gii hn theo onh l 1.a) Nu f , go c v f g trn A, nf f

trn A th

nf g trn A.

b) Nu nf f trn A v nf g

trn A th f g trn A.Chng minh.

a) V f g trn A nn tp hp ( ) ( ){ }:x f x g x = c o

( ) 0 = (v f , go c nn M ).Vi 0 > ta c( ) ( ){ }

{ } { }

( ) ( ){ }

( ) ( ){ }

( ) ( ){ }

:

\ : ( ) ( ) : ( ) ( )

\ :

\ :

:

= =

=

=

=

n n

n n

n

n

n

x f x g x

x A B f x g x x B f x g x

x f x g x

x f x f x

x f x f x

U

Suy ra

( ) ( ) ( ){ }( )

( ) ( ){ }( )

: ( )

: 0

+ =

=

n n

n

x f x f x B

x f x f x

khi n v nf f trn A..

Do ( )lim 0nn = . Vy nf g trn A.b)t


23/58

{ }{ }

{ }

0 : ( ) ( ) 0

: ( ) ( ) ,

: ( ) ( ) , 0

1

: ( ) ( ) ,

: ( ) ( ) ,2

: ( ) ( ) ,2

k

n n

n n

A x A f x g x

x A f x g x

A x A f x g x

A x A f x g x kk

B x A f x f x n

C x A f x g x n

= > =

=

= >

=

=

=

Ta c cc tp hp ny u o c v , ,nf f g o c trn A..

Ta cn chng minh 0( ) 0A = .

- Trc ht ta chng minh 01

k

k

A A+

=

= U . (1)

Ly 0x A , ta c x A v ( ) ( ) 0f x g x > .

Theo tnh cht tr mt ca tp s thc s tn ti s t nhin 0k sao cho

01( ) ( ) 0f x g x k > , suy ra 0kx A nn

1

kk

x A

+

= U .

Ngc li, ly1

k

k

x A+

=

U th tn ti s t nhin 0k sao cho

0kx A . Suy ra A v

0

1( ) ( )f x g x

k nn

( ) ( ) 0f x g x > , do 0x A

.

Vy (1) c chng minh. Khi ta c

0

1

( ) ( )kk

A A +

=

(2)

- By gita chng minh

n nA B C U , n

, 0 > (3)

hay


24/58

\ \ ( )

( \ ) ( \ ).

n n

n n

A A A B C

A B A C

=

=

UI

Tht vy, ly ( \ ) ( \ )n nx A B A C I ta c x A v( ) ( )

2nf x f x

< v ( ) ( ) 2n

f x g x

<

Suy ra

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )2 2

n n

n n

f x g x f x f x f x g x

f x f x f x g x

= +

+ < + =

Do \A A . Vy (3) c chng minh.Khi

( ) ( ) ( )n nA B C + (4)M

lim ( ) 0, lim ( ) 0n nn n

B C+ +

= =

v nf f , nf g

trn A, nn ly lim hai v ca (4) ta c

( ) 0, 0A

= >

Suy ra( )

10, 0, = = > k khi k

k

T (2) ta c 0( ) 0A = . nh l c chng minh. nh l ny cho thy gii hn ca dy hm s theo o l duy nht,

nu ta ng nht cc hm tng ng (tc l b qua tp hp c o

0).5.Mi lin h gia hi t hu khp ni v hi t theo o.

nh l 2. Nu dy hm s{ }nf o c, hu hn hu khp ni, hi t hukhp ni n hm sf o c, hu hn hu khp ni trn tp hp A c

o hu hn th nf f trn A.

Chng minh.Gi s v l hai s dng ty . Theo nh l grp, tn ti mt tp hp

con o c B ca tp hp A sao cho( )\

< v dy hm s{ }nf

hi

tu n hm sf trn tp hp B. Do tn ti mt s t nhin 0n sao cho

vi mi s t nhin n , nu 0n n th


25/58

( ) ( )nf x f x < vi mi x Khi

( ) ( ){ }: \nx f x f x vi mi 0n n ;nn

( ) ( ){ }( ): ( \ )


26/58

nh l 3. Nu { }nf hi t theo o n f trn A th tn ti dy con { }knf hi t hu khp ni n f trn A.Chng minh.

V nf f trn A, nn vi 0 >

( ) ( ){ }( )lim : 0nn x f x f x = Do vi n , tn ti nk sao cho vi nm k

( ) ( )1 1

:2m n

x f x f xn

,

Nh vy ta c dy con { } { }nk n

f f . t

( ) ( )1

:nn k

x f x f xn

=

,

1n

m n m

= =

= I U hay1

Cmm

=

= I ,

Cm nn m

=

= U

Ta chng minh ( ) 0 = v ( ) ( )limnkn

f x f x

= vi \x .

Tht vy, ta c ( )1

2n n < , nn

( ) 11 1

2 2n n n mn m n mn m

= ==

< =

U , ( ) ( ) 1

1C

2m m

< vi

m .

Do ( ) 0B=

.Nu \x A B th tn ti s t nhin 0m 0mx C . Do

nx vi 0n m . Tnh ngha tp hp nB suy ra \x th

( ) ( )1

nkf x f x

n < vi 0n m .

Vy ( ) ( )limnkn

f x f x

= vi \x hay dy con { }kn

f hi t hu

khp ni n f trn A (pcm).


27/58

CHNG II. TCH PHN LEBESGUE

$1. NH NGHA1. Tch phn ca hm n gin

Cho khng gian o ( ), M , , M v S l hm n gin, o

c trn A. Gi 1 , 2 , , n l cc gi tr khc nhau i mt ca S.

t ( ){ }: , 1,= = =k kA x A S x k n .

Th th cc k ri nhau,1

n

k

k

A A

=

= U v ( ) ( )1

, =

= kn

k A

k

S x x x A .

Khi ngi ta nh ngha tch phn ca hm S nh sau.

a)nh ngha 1. S 1 ( )

n

k kk A = c gi l tch phn ca hm n

gin, o c S trn tp hp A i vi o v k hiuA

Sd

hay ( )

A

S x d .

Vy

1

( )

n

k k

kA

Sd A =

= (1)b)Nhn xt

1.A

Sd l mt s khng m hu hn hoc v hn.2. Ta chng minh nh ngha tch phn bi cng thc (1) l hp l, nghal chng minh gi tr ca tch phn khng ph thuc vo cch biudin hm s S(x).

Tht vy, gi s hm n gin S(x) c hai cch biu din:

( ) ( )1

, =

= kn

k A

k

S x x x A ;

( ) ( )1

, =

= im

i B

i

S x x x A,

trong ,k iA B M,1 1

n m

k i

k i

A B

= =

= =U U ,

' ', , ', 'k k i iA A B B k k i i = = .


28/58

Ta cn chng minh1 1

( ) ( )n m

k k i i

k i

A B

= =

= .

Ta c1 1

( ) ( )

m m

k k k i k i

i iA A A A B A B

= == = =I I U U I ,

trong

' '( ) ( ) ( ) , 'k i k i k i iA B A B A B B i i= = I I I I I

Do 1

( ) ( )m

k k i

i

A A B

=

= I

1 1 1

( ) ( )n n m

k k k k i

k k i

A A B

= = =

= I

Tng t1 1 1

( ) ( )m m n

i i i i k

i i k

B B A

= = =

= I Xt mt cp ( , )k i , c hai kh nng:

+ k iA B = I , khi ( ) 0 ( )k k i i k iA B B A B = =I I

+ k iA B I , ly 0 k ix A B I th0 0( ) , ( )k i k iS x S x = = =

( ) ( )k k i i k iA B A B =I I .

Vy1 1 1 1

( ) ( )m n m n

k k i i k i

i k i k

A B A B

= = = =

= I I .Ty ta c iu phi chng minh.

V d 1. Cho hm Direchle trn [ ]0,1 :

[ ]

[ ]

1, 0,1( )

0, 0,1 \

khix QD x

khix Q

=

I.

Ta c [ ]( 0,1 ) ( ) 0Q Q =I


29/58

[ ]( 0,1 ) 0Q =I .Do [ ]( 0,1 \ ) 1Q = .

Vy

[ ]0,1

( ) 1.0 0.1 0D x d = + = .

V d 2. Cho hm s

11, 0

2( )

12, 1

2

khi xf x

khi x


30/58

Mt khc, vi kA ta c ( ) k kf x t = > , m nf f nn

vi n ln ( )n kf x t> , do , ,1

k n k n

n

x A x A+

=

U .

T ta c bao hm ,1

k k n

n

A A+

=

U .Bao hm ngc li l hin nhin v

, ,

1

: k n k k n kn

n A A A A+

=

U .

Vy , ,1

( ) lim ( )k k n k k nnn

A A A A

+

+=

= =U .

t ,1

( ) ( )k n

m

n k A

k

x t x

=

= th

n n n n

A A A

f f d f d f d (3)

Cho n + , ta c

,

1 1

( ) ( )m m

n k k n k k

k kA A

d t A t A t fd

= =

= = .

Ly gii hn ca (3) khi n + ta clim n

nA A A

t fd f d fd +

Li cho 1t ta c (2).Tng t ta chng minh c lim n

nA

g d fd +

=

(v lim nn

g f+

= ). T ta c (1) khi f l hm n gin.

b) By gita chng minh cho trng hp 0f l hm o c tu.


31/58

Ly cnh m , t { }min ,n n mh f g= th nh cng l

hm n gin. Mt khc, do limn n mn

f f g g+

= nn n mh g

khi n + . Theo phn va chng minh, ta clim n m

nA A

h d g d+

= Nhng v

lim limn n n n n nn n

A A A A

h f h d f d h d f d + +

limm nn

A A

g d f d+

.

Cho m + ta c lim limm nm nA A

g d f d+ +

hay lim limn nn nA A

g d f d+ +

.Bng cch tng t ta chng minh c bt ng thc ngc li. Vy ta

c (1) vi 0f l hm o c bt k.2. Tch phn ca hm o c bt ka) Trng hp hm so c khng m

Cho : [0, ]f A + l hm o c. Khi tn ti dy n iu

tng cc hm n gin o c 0nf hi t v f trn A .

nh ngha 2. Tch phn ca hm f trn A i vi o l s(hu hn hoc v hn)

lim nn

A A

fd f d+= (4)

Theo tnh cht 2 ca tch phn ca hm n gin th tch phn (4) c xcnh mt cch duy nht, khng ph thuc vo cch chn dy hm n gin

{ }nf .b) Trng hp hm o c c du bt k

Gi s f l hm o c trn A . Khi ta c

, , 0f f f f f+ + = .

Cc hm s ,f f+

c tch phn tng ng trn A l


32/58

,

A A

f d f d + +

Xt hiuA A

f d f d + + .

nh ngha 3. Nu hiuA A

f d f d + + c ngha (tc l khng c dng

), th ta gi n l tch phn ca hm o c f trn A i vio :

A A A

fd f d f d + += (5)

nh ngha 4. Nu tch phn (5) hu hn th ta ni f l hm kh tch trn tphp A .nh ngha 5. Khi X = v l o Lebesgue th tch phn nh

ngha nh trn c gi l tch phn Lebesgue, k hiu li l( )

A

L fdx hoc

( ) ( )

A

L f x dx .

c) Cc tnh chtn gin

Tnh ngha, ta c cc tnh cht sau y.

1. ( ) ,A

c d c A c const = =

2. ( ) ( ) ( )B B AA A

x d x d B A = =

3. ( ) ( ) ( )1 1 1

= = =

= = B Bk k

n n n

k k A k k

k k kA A

x d x d B A

4. Nu ( ) 0A = , f o c th 0f d =

5. Nu ( )A< +

, f o c v b chn trn A th f kh tch trn A.

Chng minh


33/58

4. Cho 0f . Nu ( ) 0A = th vi mi dy hm { }nf n gin tng vf ta c

0 0nA A

f d f d = =

5. ( )A < + , ( ) ,f x K x A th vi mi dy hm n gin { }nf tng vf , ta c nf K nn

( )nA A

f d K d K A = < + .T suy ra

( )lim nn

A A

f d f d K A

= < +

Nhn xt. T tnh cht 5 suy ra mi hm s b chn, lin tc hu khp ni trnkhong hu hn I u kh tch Lebesgue. Nh vy lp cc hm kh tchLebesgue trong bao gm tt c cc hm kh tch Riemann v cn bao gmnhiu hm s khc (nh hm Direchle chng hn).

$2. CC TNH SCP CA TCH PHN

mc ny ta lun gi thit cc hm s v tp hp c ni n u o c.1. Cng tnh.

nh l 1. Nu = th f d f d f d

= +

(vi gi thit v tri hoc v phi c ngha.)Chng minh.a) Trng hp f n gin trn .

( ) ( )1 1

,

= =

= = k

nn

k k

k k

f x x U

Ta c ( ) ( ) ( ) = = k k k k . V A, Bri nhau nn k, k ri nhau. Do

( ) ( ) ( )1 1 1

= = =

= = + =

= +

n n n

k k k k k k

k k k

f d

f d f d


34/58

b) Trng hp 0f trn .

Cho { }nf l dy hm n gin, nf tng vf th theo a)

n n n

f d f d f d

= +

Cho n ta c ng thc cn chng minh.c) Trng hp f bt k: Theo b)

( )1f d f d f d + + +

= +

( )2f d f d f d

= +

Nu f d c ngha th v tri ca mt trong hai ng thc trn hu hn

(nu chng hn v tri ca (1) hu hn th hai tch phn v phi hu hn vcc hiu s

f d f d+

, f d f d+

c ngha.)

Tr (1) cho (2) ta c iu phi chng minh.

Nu f d f d

+ c ngha th suy lun tng t.

H qu 1. Nu v f d

th f d

. Nu f kh tch trn A th

f kh tch trn E.

H qu 2. Nu ( ) 0 = th f d f d

= .Chng minh.- Nu A, B ri nhau th y l h qu trc tip ca nh l 1 v tnh cht 4 trong$1.

- Nu A, B khng ri nhau th ta vit ( )\ = v v

( )\ 0 = nn ta trli trng hp trn.2. Bo ton tht.


35/58

nh l 2. Nu f g th f d g d

= . c bit, nu 0f = hu khp

ni trn A th 0f d

= .

Chng minh.

t ( ) ( ){ }: f x g x = = th M v ( )\ 0 = (do f g ).Theo h qu 2 nh l 1

( )\

f d f d f d

= =

tng t g d g d

=

T suy ra f d g d

= .

Nhn xt. Tnh cht ny cho thy: khi thay i gi tr ca hm s ly tch phntrn mt tp hp c o 0 th gi tr ca tch phn khng thay i.

Do nu f o c trn tp hp ' vi ( )\ ' 0 = thd f khng xc nh trn \ ' ngi ta vn nh ngha

'

f d f d

=

nh l 3. Nu f g trn A th f d g d

. c bit nu 0f trn

A th 0f d

.

Chng minh.- Nu f , gn gin trn A th iu l hin nhin.

- Nu f , 0g trn A th c dy hm n gin { }nf tng n f , { }ng

tng ngsao cho n nf g . Khi

n nf d g d .Chuyn qua gii hn sc iu phi chng minh.


36/58

- Nu f , g ty th f g+ + , f g

nn

f d g d+ +

, f d g d

. Tr tng v ta c iuphi chng minh.H qu 3. Nu f kh tch trn A th n phi hu hn hu khp ni trn A.Chng minh.

t ( ){ }:x f x = = + . Theo h qu 1, f kh tch trn B nhng vik ta c ( )f x k> trn B nn

( ) ,f d k k

Bt ng thc ny chng vi k nu ( ) 0 = .Tng t ta cng chng minh c trng hp tp hp

( ){ }C :x f x= = s c ( )C 0 = .Vy f hu hn hu khp ni trn A.

H qu 4. Nu 0f trn A v 0f d

= th 0f = hu khp ni trn A.

Chng minh.

t ( )1

: , =

n x f x nn

. Ta c

( )

\

0

1 1,

= = +

=

n n n

n

n

f d f d f d f d

d nn n

Do ( ) 0n = .

Mt khc ( ){ }1

: 0 nn

x f x

=

= > = U .

Vy ( ) 0 = , suy ra 0f = hu khp ni trn A.3. Tuyn tnhnh l 4.

,A A

cfd c fd c const = =


37/58

Ni ring,( )

A A

f d fd = Chng minh.

- Nu f n gin th hin nhin ta c iu phi chng minh.- Nu 0f th c dy hm n gin { }n tng n fnn suy ra c dy

hm n gin { }ncf v:

a) nu 0c th { }ncf tng vcfv do

n n

A A

cf d c f d = , chuyn qua gii hn sc

A A

cfd c fd = b) nu 0c < th 0cf nn ( ) 0,( )cf cf cf + = = .

Theo nh ngha0 ( )

A A

cfd cf d=

Theo a)( )

A A

cf d c fd =

VyA A

cfd c fd =

- Nu f bt k th f f f+ = v

( ) ,( ) 0

( ) ,( ) 0

cf cf cf cf khi c

cf cf cf cf khi c

+ +

+ +

= =

= =

nn ta c th chng minh hm s f cho v d 4 khng lin tc ti mi

im [1, ] .x e Khi tp hp cc im gin on ca f chnh l

[1, ]e c o bng 1 0.e >

V d 5. Tnh[0,1]

lim nn

f d+ ,

trong { },nf n , l dy hm s xc nh bi

sin 1, (0,1]

( )1, 0

n

n

xkhi x

f x x nkhi x

+

= =

Gii.

- Ta c nf b chn trn [0,1] v

1( ) 1 , , [0,1]

n

nf x e n xn +

- ([0,1]) 1= < +

- Ta chng minh dy hm cho hi t hu khp ni trn [0,1] .

Tht vy, vi (0,1]x , tsinx

tx

= th (0,1)t . Suy ra

1

1 1lim ( ) lim lim 1

1lim 1 0. 0, (0,1]

n nn

nn n n

tn

tn t

n

f x t tn tn

t e xtn

+ + +

+

= + = + =

= + = =

{ }( )0 0 0f h k [0 1]

Hoc Phan Do Do Va Tich Phan

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