hoc phan do do va tich phan
TRANSCRIPT
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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
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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.
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(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
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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 >
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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
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{ }{ }
{ }
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
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\ \ ( )
( \ ) ( \ ).
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
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( ) ( )nf x f x < vi mi x Khi
( ) ( ){ }: \nx f x f x vi mi 0n n ;nn
( ) ( ){ }( ): ( \ )
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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).
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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 = = .
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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
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[ ]( 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
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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.
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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
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,
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
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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
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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.
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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.
-
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- 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 = =
-
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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]