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1 1
LEBESGUE OUTER MEASURE
Unit Structure :
1.0 Objective
1.1 Introduction
1.2 Algebra
1.3 Extension Measure
1.4 Lebesgue outer measure
1.5 Properties of outer measure
1.6 Summary
1.7 Unit End Exercise
1.0 OBJECTIVE
After going thr ough this chapter you can able to know that
Concept of Algebra, Measurable set.
Extension measure in n
Lebesgue measureable set
Lebesgue outer measure & its properties.
1.1 INTRODUCTION
In this chapter w e shall fist study such a verified theory function d -
dimensional value based on the notation of a measure, and then we
shall use this theory to build a stronger and more flexible theory.
Now if we want to partition the range of a function, we need same
way of measuring how much of the domain is sent to a particular
region of the partition, To set a feeling function what we are aiming
function let us assume that we want to measure the volume of
subsets 3,ACand that are denote the volu me of A by A.
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2 Then function we have
i) Ashould be non -negative number as .
ii) 0 it will be convenient to assign a volume to the empty
set.
iii) If 12, ,.....,nAA A are non overlapping disjoint sets then
1 1nnn nAA
This means that the volume the whole is equal to the sum of the
volume of the parts. This problems leads us to the theory of
measures where we try to give a notation of measure to subsets of an
Euclidean space.
Defenition :
The Euclidean norm on n is 12221....d xx x .
The distance between ,nxy is xy
1.2 ALGEBRA
Definition :
Let X be a set. A collection A of subsets of X is called a algebra
of the following hold.
i) A
ii) AA X AA
iii) 12
1, ,....,i
iAA A A A
Note :
The pair ,XA is called measurable space and elements of A are
called measurable sets.
Example 1 :
Let 1,2,3X and 11,1 , 2 , 3, , ,bX 21,2,3, 3 , ,bX.
Check whether 1b and 2b are both algebras or not.
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3 Solution :
I) Let 1,2,3Xand 1bis not Algebra.
Since it does not contain 1C.
II) 2b is Algebra since it satisfies all condition of Algebra
i.e. 1Xb
2b 21,2b & 21,2Cb 2b is Algebra.
Example 2 :
A measure on a topological space X whose domain is the Borel
algebra is called a Borel measure.
Example : For every xX, the Dirac measure is given by
1
0xif x AAif x A
Definition :
Let be a set function whose domain in a class A of subsets of a set
X and whose values are non -negative extended reals, we say that is contably additive if
1 1kkk kAA
whenever, kA is a
sequence of painoise disjoint set in A whose union is also in A.
Theorem :
Let be a finitely additive set function, defined on the Algebra
A. Then is countably additive iff it has the following property : if
nAA and 1nnAA Anti for each positive integer n, and if
1n
nAA then
1limnnnnAA
.
Proof :
Suppose is countable additive Let nAbe a sequence of elements
in A s.t. 12
1,......,i
iAA A A A munotes.in
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4 s.t.
1liminniAA
Define 11BA
1KK KBA Afor 2K
Examples 3:
Let ;iAi I be collection of Algebra. Show tha iiIA is a Algebra, but iiIA is not in general.
Solution :
Let iiIAi A
To show that A is a Algebra
a) If A
iA is Algebra, iI
iAi I
i
iIAA
b) Let AA iiIAA iA is Algebra iI For C
iiAA A AiI
ciiI
cAAAA
c) Let , 1,2....kAA k
then kA Ai i i munotes.in
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5 1
1
1ki
kkiki I
k
kAA iAAAA
iiIAA is a Algebra
Now, we have to show that iA is not a Algebra.
Let 1,2,3X
Let 1,, 1 , 2 , 3AX 2,, 3 , 1 , 2AX
then 1A & 2A are Algebra but 12AA is not Algebra. 121AAbut 121,3AA.
Clearly Bi A i and 'Bi s are pairwise disjoint we first show that
1k
k
iA Bi
By induction on ‘k’
The result is trivial when 1k
Assume the result is true for 1k
i.e. 1
1
1k
k
iA Bi
Now 1
11kkkiiBi Bi B
11kk k
kAA A
A The result is true for k. by introduction is true for all k
11k
k
iA Bi k munotes.in
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6 Note that
11 1k
k
kk iA A Bi
1kkB is countably additive, we have
1 1kkk KAB B
1
1lim
lim
limn
knk
n
knk
nnBBA
Conversely,
Suppose whenever if 123 ...., ,r A A A Ai B Ai AA
Then
1limnniAi A
T.S.T. is countably additive
Let nA be a pairwise disjoint sets in A.
Define
1k
k
iB Aithen kBA and 12.......BB By hypothesis, we have
1limnniBi B
But
11 1ikii KBi A
1iAi munotes.in
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7
11
1
1
1lim
lim
limii
nn
n
ni
n
ni
iAi Bi
BAiAiAi
Theorem :
Let A be a Algebra, If ,uv are measures on A, ,0tt and
AA hen the following are measures on A.
a) defined by EE E E A
b) t, defined by ,tt E E A
Proof :
a) defined by ,EE E E A is a measure
on A. & are measure on A. They are countably additive non -negative set function. E is also countably additive non -negative set functi on
whose domain is A. is a measure on A.
b) tEt E
is a measure on A
is countable additive non negative set function whose
domain in A. for EA tEt E and t is also countably additive non -negative set
of function whose domain is A t is measure on A. munotes.in
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8 1.3 EXTENSION MEASURE
Defin ition :
Let X be a set, nA Exterior measure or outer measure on X is a non -
negative, extended real valued function *whose domain consist of
all subsets of X and which satisfies :
a) *0
b) (Monotonicity) if AB then **AB
c) (Countable sub -additivity)
For any sequence nA of subsets of X, we have
**
1 1nnn nAA
Theorem :
Let C be a collection of close d rectangle of n, For RC, let R
denote the volume of R. If *is defined by
*
1 1inf ; ,kk
k kAC C C k A
For ,nAAthen *is exterior measure on n.
Proof :
T.S.T. * defined by *
1inf ;kkkAC C is closed rectangle
where nA is on exterior Measure on n.
We first shows that
;kk kV C C is closed set A C
Where nA
Let kR rectangle with side length ‘k’ and centre origin.
Then
1nk
kR munotes.in
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9 for any
1nkkAR kRcovers A
1;kkkCC closed rectangle
1k
kAC
We now show *0
Let 0
Let 0, ..... 0,nnyyR be a rectangle in n with &RR R covers By definition of *, *
This is true for any 0
*0…………………………………………. (1)
Let nAB
T.S.T.**AB
If kCCovers B, then kC covers A
11 1
11::
inf : inf :kk kk
kk kkk kkkkCB C CA CCB C CA C
**AB …………………………………….. (2)
Let nA be a sequence of subsets of n we show that
**
1 1nnn nAA
Let 0 by the definition of * a cover 1iniR of nA such that munotes.in
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Analysis II
10 *
12innn
iRA
Then
11innjR
covers
1nnA
*
1 11
*
1
*
11
*
122nn
n ni
n n
nnn
nn
n
nAR
A
A
A
*
1nnA ………………… …………………. (3)
From (1) (2) & (3) * is an exterior measure on n
Note :
By above lemma, the exterior measure lemma attempts to describe
the volume of a set nE by approximating it fr om outside. The
set E covered by rectangle and if the covering gets finer, with fewer
rectangles overlapping the volume of E should be close to the sum of
the volumes of the rectangles.
1.4 LEBESGUE OUTER MEASURE
Definition : *is called the Lebesgue exterior (or outer) measure on n and is
denoted by *m.
Now the consequences of the definition of exterior measure on n.
1) If kR are countably many rectangles and kER then *kmE V R
2) For a given 0 there exist countable many rectangle kR with kERsuch that **
kkmE R mE E .
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11 Example 4:
Show that exterior (or outer) measure of a closed rectangle is its
volume i.e. *mR V R where R is a rectangle or a 0nb in.
Solution :
Let R be a closed rectangle in n
*tst m R V R
Note that R covers R
by definition of *mR, we get
*mR V R……………………………………. (1)
Let 0
By defini tion *,mR a countable cover iR of closed rectangles
of R.
*
12ii
iRm R
For each i choose an open rectangle iS such that iiRS and
12ii iVS VR
Then
11iiiiRR S 1iiS is an open cover of R R is compact this open cover has a finite sub cover say
1miiRS (after renaming)
We have
11miiiiVR VS v S munotes.in
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12
1
1
1
*
*2222i i
i
i
iVR
VR
mR
mR
This is true for any 0
*VR m R
From (1) & (2)
*VR m R
Example 5:
Show that exterior (or outer) measure of an open rectangle in n is
volume.
Solution :
Let iS be an open rectangle them iiRS where iS is closed
rectangle iS is a cover of R.
by definition *
imR V S V R……………. (1)
Let 0 be iR be a countable cover of closed rectangle of R such
that *
12i
iVR m R for each i choose an open rectangle iS
such that iiRS & 12iiVR
Then
11iiiiRR S 1iiS is an open cover of R R is compact. This open cover has a sub cover say
1miiRS (after renaming)
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13 We have
11miiiiVR VS VS
11
1
*
*2222i i
i
i
iVR
VR
mR
mR
This is true for any 0
*VR m R ……………………………….. (2)
From (1) & (2)
*VR m R
Example 6:
Show that exterior measure of a point in n is zero.
Solution :
Let 12, ,......,nn aa a a
To show that *00m
Let 0 then the closed rectangle.
11
11
11
22,22, ......................22nn
nnRa a
aa
Covers a By definition of *ma, we have *ma V R
This is true for any 0
*00m
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14 1.5 PROPERTIES OF OUTER MEASURE
Exterior measure has the following properties.
i) (Empty set) The empty set has exterior measure *0m.
ii) (Positivity) we have *0mA for every subset A of n.
iii) (Monotonicity) If nAB, then **mA mB.
iv) (Finite sub -additivity) If jjJA are a finite collection of subset
of n then **jjjJ jJmA m A
v) (Countable sub -additivity) if jjJA are a countable collection of
subsets of nthen **jjjJ jJmA m A
vi) (Translation invariance) If E is a subset of n and nx then **mx m .
Let ,nnxE
**tst m x m
Let 0, by definition of *m
a countable cover iR of closed rectangles in n for s.t.
*
1i
iVR m E ………………………………… (1)
We now show that
1iixE xR
Let axE a xy
1iiax yE R iaxR for some i iax R for some i munotes.in
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15
1iiax R
1iixE xR By definition of *m, we have
*
1iimx E V x R ………………………………… (2)
We now show that ii iVx R VR
Let , ..... ,i iu iu iu iuRa b a b then 11 11, ...... ,i i i in n in nxR x a x b a x b x
1ni ij i ij ijVx R b x a x
1nij ij ijba V R ……………………………… (3) By 1,2,3 we get
**
11ii
iimx E V x R V R mE
**mx E mE
This is true for any 0 **mx E mE ………………………………….. (4)
Let &Ex E y x
Then by (4)
**my E mE **mx x Em x E **mE mx E ……………………… ……………. (5)
By (4) & (5) **mx E mE munotes.in
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16 Theorem :
Show that there are uncountable subset of whose exterior
measure is zero.
Proof :
Define canter set as follows
Let 00,1C
trisect 0C and remove the middle open interval to get 1C.
i.e. 110, 2 3,13C
0,1 \ 1 3,2 3
repeat this procedure for each interval in 1C we get 2C 20,1 \ 1 3,2 3 \ 1 9,2 9 \ 7 9,8 9112 20, , ,7 999 33C
repeating this procedure at each stage we get a sequence of subsets iC of 0,1for 0,1,2i
Note that each kC is a compact subse t of and 02,CC C
The Cantor set ‘C’ is defined as
0iiCC
C because all end points of each rC is inc and also C is
uncountable
We now com pute
**
01
**
21
2
2
**
32 3
2
23211, 1332
9
1213321133122133 3mC mC
mC mC
mC mC
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17 in general,
22*23 13
22122 21 .....33 3 322 32 2......33 33 3k
k k
k
kmC
2
1
**
*
11213322
2 3313
221133
2
3
2
3k
k
k
k
kCC kmC mC kmC k
letting k, we get
*
*000mC
mC
Theorem :
Show that exterior measure of n is infinite.
Proof :
Let 0M and R be a rectangle s.t. VR M
note that n By monotonicity of *m **nmR m
But *mR V R M *nmM
This is true for any 0M *nm
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18 Theorem :
If E and nF such that ,0dE F then show that ** *mE F mE mF.
Proof :
Let ,nEF be s.t. ,0dE F tst ** *mE F mE mF. By
countable subodditivity property ** *mE F mE mF.. (1)
Let 0
By the definition of *,m countable Ri of closed rectangles in n
for EF such that *
iV Ri m E F . ………………. (2)
We categorize the collection Ri into 3 types :
1) Those intersecting only E
2) Those interescting only F
3) Those intersecting both E & F
Note that if a rectangle intersect both E & F, then ,0dR dE F subdivide such the rectangles into rectangles
whose diameter is less than ,dE F.
This subrectanlges intersect either E or F not both. We can have a contable collection 2R of rectangles which
intersects either E or F but not both.
Let 1;i Ii R E
2;i Ii R F
12II ,iiIR covers E, we have 1*iiImE V R
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19 Similarly, 2*iiImF V R 12**
1iiiI iI
i
imE mF V R V RVR
*mE F (by (2))
This is true for any 0 ***mE mF mE F ……….. (3)
From (1) & (3) ** *mE mF mE F
Theorem :
If a subset nE is a countable unit of almost disjoint closed
rectangle .
i.e.
1iiER then show that *
1iimE R.
Proof :
Let
1iiER where iR’s are almost disjoint closed rectangles.
tpt *
1iimE R
By countably subadditivi ty proposition of
** *
1 11ii ii iimE m R mR V R
(R is rectangle *mR V R)
Let0, by definition of *,m a countable cover iR of closed
rectangle n for E s.t.
*
1i
iVR m E
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20 For each i, choose open rectangle iS s.t. iiSR &
2iiiVR VS
Note that ,0ijdSS for ij ** *ij i jmS S mS mS for ij …………………….. (1)
Using (1) finite no. of times, we get **
1 1k kiiimS m S
1ii
k
i
iSRE iSE By monotonicity
**
11kk
ii
iimE mS V S k
Let k
*
11
12ii
ii
i
imE V S V R iVR
This is true for any 0
*
1iimV R ………………………………………….. (2)
From (1) & (2)
*
1iimV R
Theorem :
Show that
1) If *0mA then **mAB mB
2) If *0mA B then show that **mA mB
3) ** *\mA B mA mB munotes.in
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21 Proof :
1) As BA B
By monotonicity
**mB mAB …………………………………. (1)
Also by countable subadditive of *m ** *mAB mA mB
*mB ……………………………………… (2)
From (1) & (2) **mAB mB
2) If *0mA B tst **mA mB ** *\\\\wk A B A B B AmA B mA B mB A
given that *0mA B **0mA B mB A
but **00mA B mA B
*0mA B **0mA B mB A
WKT **mA mAB
**mA mAB
similarly we show that
**mB mAB **mA mB
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22 3) ** *\mA B mB mA
Proof :
Since A and B are measurable sets CA is also measurable and we have /BAB A A B
/CBA B A is measurable.
&CBAis measurable \BAB A union of disjoint measurable sets ** * *\\mAB A mA mB A mB ** *\mB A mB mA
Theorem :
Let nE show that **inf ; &m E m E open
Proof :
Let nE
tst **inf ;nm E m E and open in
Let be open in n s.t. E
Then by monotonicity of *m, **mE m *mE is lower bound of *;,m open **inf ; ,m E m E open ……………………. (1)
Let 0, then by definition of *m an countable cover iR of closed rectangle of E s.t. *2i
iVR m E
For each im let iS be open rectangles containing iR s.t. 12iiVR VS i munotes.in
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23
Let
1iS then is open &
11ii ER S
** *
1 1
1iii
i
imm S m SVS
11
1
*
*2222i i
i
iVR
VR
mE
mE
This is true for any 0. **
*inf ; ,mmm is open **mm E
Theorem :
For every subset E of ,n a zG
Subset G of ns.t. GE & **mG mE
Proof :
Let nE
we first show that **infmE m i E and is open subset of n
Let 0,
Then for each ,kk open in n & kE s.t. **2kk mm E munotes.in
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24 let
1kkG G is G-set and GE By monotonicity **mE mG ……………………………………… (1)
Note that kG k ** *2kk mG m mE
This is true for any 0 **mG mE …………………………………… (2)
By (1) & (2) **mG mE
Thoerem :
There exist a countable collection jjJA of disjoint subset of
such that **jjjJ jJmA m A
Solution :
Consider rational and realy ;xx
We known that any two cosets are either identified or disjoint.
We now show that if A then 0,1A
Let Ax
Let q be rational number in ,1xx
then 0,1xq
Also, xqx A 0,1 0,1 xqA A
For each \A choose munotes.in
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25 0,1AxA
Let ;A Ex A
By construction 0,1E
Let
1,1qXq E
We now show that 0,1 1,2X
Let 1,1q Note that 0,1E for any ,1 , 2xE qx
This is true for any 1,1q
Theorem :
There exist a finite collection jjJA of disjoint subset of such
that **jjjJ jJmA m A
Proof :
Consider & /xx
We known that any two cosets are either identical or disjoint.
We now show that if A then 0,1A
Let Ax
Let q be a rational number in ,1xx then 0,1xq
Also, xqx A 0,1 0,1 xqA A
For each \A choose 0,1AxA
Let /A Ex A
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26 By construction 0,1E
Let
1,1qXq E
We now show that 0,1 1,2X
Let 1,1q
Note that 0,1E for any ,1 , 2xE qx
This is true for any 1,1q
There exist a finite collection jjJA of disjoint subset of ¡ such
that **jjjJ jJmA m A
U
Consider Q¡ ||QxQ x ¡¡
We know that any two cosets one either identical or disjoint.
We know show that if |QA¡ then 0,1AQI
Let Ax Q
Let q be a rational number in ,1xx then 0,1xq .
Also xqxQ A 0,1 0,1xq A A Q II
For each |QA¡ choose 0,1AxAI.
Let ||AQEx A¡
By construction 0,1E
Let
1,1qQXq
IU
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27 We show that 0,1 1,2
Let 1,1qQI
Note that 0,1E ,1 , 2xE qx 1,2qE
This is true for any 1,1qQI
Let 0,1y
Then 0yy y A (say) but AxA
,0 , 1 1 , 11,10,1 0,1 1,2A
AA
Ayx y
yx y x
q
yqx qE
yx
XX
By monotonicity of *m ** *0,1 1,2mM X m *13mx ………………………………………. (1)
1,1qxq E by countable subadditive and tran slation
invariance of *m, we get. ** *
1,1 1,1qqmX mq E mE
By *10mX
*0mE By Aritimedian property
n s.t. *1mEn
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28 Let I be a finite subset of 1,1 with cardinality 3n.
Then * 133qImE nn by (1) **
qImx mq E
Theorem :
Let &0nE show that ** nmE m E
Proof :
To show that **,0nmE m E
Let 0, by definition of *,mE a countable cover of iR of closed
rectangle in n, for E s.t. *
iVR m E
11iiiiERE R
Let 11, .... ,i i i in inRa b a b
1
1
1
11,....., ; ,
,....., ; ,
,....., ; ,
, ..... ,i n j ij ij
n j ij ij
n j ij ij
i i in inRx x x a b
xx x a bxx x a bab ab
iR is a closed rectangle niiVR V R
1iiER by monotoricity & countable additive property we get
**
11 1
*nii inn
imE mR V R V RVR m E
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29 This is true for any 0 ** nmE m E …………………………………….. (1)
let 11EE by (1) *1 * * 1
**11
nmE m EmE m E
**nmE m E ………………………………… (2)
From (1) & (2) ** nmE m E
1.6 SUMMARY
In this chapter we have learned about.
definition of -Algebra, bored algebra
measure on a set.
The extension Measure
Lebesgue outer Measure * on n
Properties of lebesgue outer measgure.
1.7 UNIT END EXERCISE
1) Let ,,,Xa b c d and 1,,AXd and 2,, ,AXd ,,abc check whether 1A & 2A are both algebra or not. Also
check wheter 12 AA is an algebra or not.
2) Show that exterior measure at any countable subset of n is
zero. Justify the converse?
3) Show that the outer mesuration interval is its length.
4) Show that if FI is a collection of -Algebra on X then F is also a -Algebra on X. munotes.in
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30 5) If a subset nE is a countable union of almost disjoint
closed rectangle then show that *
1iimE R.
6) If 1A and 2A are measurable subsets of the closed interval ,ab
then 1A-2A is measurable and if 12 AAthen 12 1 2m A A mA mA.
7) Show that for any se t A, **mA m A x where ;Ax yx yA
8) Show that for any set A and any 0, there exist an open set
O such that 0A and **0mm A .
9) Compute the Lebesgue out er measure of 12 3B
10) Prove that if the boundary of k has outer measure zero
than is measureable.
11) Let be an arbitary collection of subsets of a se t. Show that for
a given AC there exists a countable sub -collection AC of
C depdending on A such that AAC.
12) Check that *is an outer measure on R. Not
i) Let X be any seet and *:0 ,PXbe given by
i) *0A if A is countable
=1 otherwise
ii) *0then be on infinite set1A if A finiteX
if otherwise
iii) *0A if A
= 1 otherwise
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31 2
LEBESGUE MEASURE
Unit Structure :
2.1 Objective
2.2 Introduction
2.3 Lebesgue Measure
2.3.1 Properties of measurable sets
2.4 Outer Approximation by open sets
2.5 Inner approximation by closed sets
2.6 Continuity from above
2.7 Borel Cantelli Lemma
2.8 Summary
2.9 Unit End Exerises
2.1 OBJECTIVE
After going through this chapter you can able to know that
Construction of Lebesgue measure in n.
Lebesgue Measurable set in n.
Properties of measurabl e sets.
Existance of non -measurable sets.
2.2 INTRODUCTION
In the previous chapter we have studied about Lebesgue outer
measure *m is not countability additive and it cannot be measure. So
that we have to cover with subset of nfor which *mis countably
additive this subclass a collection at Measurable sets. Now we shall
define lebesgue measure of a set using the lebsgue outer measure
and discuss properties of lebesgue measure se t.
2.3 LEBESGUE MEASURE
Definition - (Lebesgue measurability)
Let E be a subset of n we say that E is Lebesgue measurable, or
measurable if we have the identity munotes.in
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32 ** *mA mAE mA E
2.3.1 Properties of measurable set s :
Following are the properties of measurable sets :
a) If E is measurable, then CnEE is also measurable.
b) Any set E of exterior (or outer) measure zero is measurable. In
particular, any countable set is measurable.
c) If 12&EE are measurable, then 12EE and 12EE are
measurable.
d) (Boolean algebra property) If 12, ,...nEE E are measurable then
1njE &
1njE are measurable.
e) (Translation in variance) If E is measurable & nx then xE
is also measurable, and mx E mE.
Lemma : (Finite additivity)
If 1k
j ijJEi E are a finit e collection of disjoint measurable sets
and any set A, we have
**jjjJ jJmA E mAE
Further more we have
jjjJ jJmE m E
Proof :
We prove by induction on K
The result is trivial when K=1
Assume result is true for k -1
We prov e result for K
Let
1kiiEE munotes.in
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Lebesgue Measure
33 tpt **
1kiimAE mAE
Now kE is measurable we have for nAE. ** *CkkmAE m AE E m AE E
But kAE E=kAE
kEE 1
1CCkkk
i
iAE E A EEAE
** *
**
1
*
1kik
ki
i
k
i
imAE mAE mA EmAE mAE
mAE
The result is true for K
By introduction, it is true for ‘n’.
ii) Put nA
Theorem :
If AB are two measurable sets then BA is also measurable & mBA mB mA
Proof :
tst BA is measurable.
Suppose A & B are measurable intersection of two measurab le set is measurable & complement
of a measurable set is measurable. CBA B Ais measurable
Note that BAB A
which is a disjoint union. munotes.in
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Analysis II
34 mis finitely additive mB mA mB AmBA mB mA
Example 1 :
Let A be a measurable set of finite outer measure that is contained in
B show that ** *mB A mB mA A is measurable
By definition for this B ** *
** *mB mBA mB AmB mA mB A *mA we get ** *mB A mB mA
Example 2 :
Suppose AEB where A & B are measurable sets of finite
measure show that if mA mB then E is measurable. A & B are measurable CBA B A is measurable.
Note that BAB A AB.
which is a disjoint union. m is finitely additive, we get
**
*0
|00mB mA mBA
mBA mB mA
AEB E AB A
mE A mB A m B A
mE A
EA is measurable EAE A is measurable
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35 Example 3 :
Show that if 1E & 2E are measurable then
12 12 1 2mE E mE E mE mE
Solution :
Suppose 1E & 2E are measurable not that 12 1 2 1EE E E E which is a disjoint union.
By finite additie property of ‘m’ 12 1 2 1mE E mE mE E ……………………….. (1)
also 21 2 2 1EE EE E
which i s a disjoint union.
By finite additivity of ‘m’ 21 22 1mE mE E mE E ……………………….. (1) 21 2 1 2mE E mE mE E
subs in 1 12 1 2 1212 12 1 2mE E mE mE mE EmE E mE E mE mE
Theorem :
Let 1kkE be a countable disjoint collection of measurable sets
prove that for any set A, **
1 1kkkmA E mAE .
Proof :
Let 1kkE be countable collection of disjoint measurable sets.
Let nA
tpt **
1 1kkkmA E mAE .
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Analysis II
36 By countable subadditivity property of *m we get,
**
11kkmA E m AE
*
1kkmAE ………………………….. (1)
Also by finite additive property of m, we get
**
11
*
1
*
1kkkk
m
k
k
m
k
kmA E mA EmA E
mAE
This is true for all ‘m’
**
1 1kkk kmA E mAE
……………………….. (2)
from (1) & (2)
**
1 1kkkmA E mAE
Theorem :
Show that the union of a countable collection of measurable set is
measurable .
Proof :
Let 1kkA be a countable collection of measurable sets and
1kkEA.
tst E is measurable.
Define 1,BA & for 2k
1
1kkk iBAA
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Lebesgue Measure
37 Since finite union of complement m -set are measurable kBis measurable.
Clearly kB’s are pairwise disjoint
1
11 1
1
11
1
11
12 1 3 12
1....k
kk i
kk
Ck
ki
ki
Ck
ki
k
CC C
k
kBA A
AA
AA
AA A A AA
AE
Example 4 :
Show that the intersections of a countable collection of measurable
set is measurable. Let A be a subset of n and for n.
Define
1nk
kFB E
kBS are measurable nF is measurable By definition ** *
**C
nnCC C Cnn n
CC
nmA mAF mAFFEF E A F A EmAE mAF
** *CnmA mAF mAE …………………….. (1)
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Analysis II
38 Now
**
1n
nkmAF mA B
*
1
*
1
*
1
*
1n
k
K
n
k
K
n
k
K
n
k
KmA BmA BmAB
mAB
By (1)
** *
1nCk
kmA mAB mAE LHS is independent of n, we have
** *
1nCkmA mAB mAE
But
**
1
*
1
*
1
** *k
k
kCmAE mA B
mA B
mABmA mAE mAE
As CAA EA E by countable subadditivity proposition of *m. ** *CmA mAE mAE ………………………. (3)
By (2) & (3) ** *CmA mAE mAE By definition E is measurable. munotes.in
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Lebesgue Measure
39
Example 5 : Countable additive
If jjJE are a countable collection of disjoint measurable sets then jjJE is measurable and jjjJ jJmE m E Without loss of generality we ma y assume J suppose 1kkE be a countable collection of disjoint measurable set we first
show that kEE measurable let nkFE.
then by previous exercise we get E is measurable.
We now show that
1kmE mE
By subadditivity proposition of *m
**
1
*
1kkmE m E m EmE
1kkmE ………………………… (*)
By finite additivity property and monotonicity of *m
we have as nFE
1
1nnkk
nkkmE mF m EmE LHS is independent of n, we get
1kkmE mE …………………………. (**) By countable additivity
1kkmE mE
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Analysis II
40 Example 6 :
Show that every closed and open rectangles in n are measurable. Let R be a closed rectangle
tst R is measurable
Let 0, Let nA
by definition of *mA a countable collection of closed rectangle 1iiR such that
1iiAR and *
1i
iVR m A ……………………………. (1)
we dec ompose each iR into finite union of almost disjoint rectangle 1, ,....,iiikRS S such that 1
1k
ii
jijRR S.
1
iiRR R R and CijSR By finite add itive property of M.
1
1
1
11 1 1k
ii
j
k
ii
j
ii
iii jij
ijijmR mR mS
VR VR VS
VR VR VS
Note That 1iiR lover AR
111 1ii iii iAR R R R R R
,ijSi j covers CAR
*1*
1 1CiiVR m R m A R and **
,Cij
ijmSm A R munotes.in
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Lebesgue Measure
41
**
,
*
,,ij
ijij ijij ijmAR m SmS VS
By (1)
*
1
11 1
**i
i
ki ijii jCmA V RVR VSmAR mAR
This is true for any 0 ** *CmA mAR mAR By definition R is measurable.
Example 7 :
Show that every open and closed subsets of n are measurable. Let maxiKK
Let G be an open subset of n consider the grid of rectangle in n
of side length one and whose vertices have integer co -ordinates.
TST G is measurable. Number of rectangle in grid is countable and one almost disjoint
we ignore all these rectangle contained in CG.
Now we have two types of rectangle (1) Those rectangle contained
in G (2) Those rectangle intersect with G & CG.
Let C = set of all rectangle contained in G.
We bisect type (2) rectangle into two rectangle each of its side
length is ½.
Repeat the process iterating this process for arbitrarily many times
we get a constable collections c of almost disjoint rectangle
contained in G.
By construction
RCRG
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Analysis II
42 Let xG G is open
We can choose sufficiently small rectangle in the bisection
procedure that contains x is entirely contained in G.
RC
RC
RCxRGRGR
G is countable union of closed rectangle and hence G is
measurable.
2.4 OUTER APPROXIMATION BY OPEN SETS
Let nE such that E is measurable iff for 0, there is an open
set containing E for which *mE . Suppose E is measurable
Let 0
Suppose *mE By the definition of *mE a countable collection of open re ctangles iR such that
1iiER
and *
1i
iVR mE .
Let
1iiR which is countable union of opensets. is open in n and E is open, it is measurable E is measurable EE which is a countably disjoint union ***
** *mm E m EmE m m E munotes.in
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Lebesgue Measure
43 But
**
1 11
**
1ii ii ii
i
iRm m R V RmE V R m E
Suppose *mE
For each k kKEE Rwhere
kRrectangle with centre origin and side length K
For each k
Then **
ki imE mR VR K by first case for each K, k open in n such that
*2kk k kkEEmE .
Let
1kk which is countable union of open set. is open and E
**
*
1
*
1
*
1
*
1
12CCk
k
k
k
k
k
kk
k
k
kmE m EmEmEmE
mE
Conversely suppose for a given 0 open set E such that *mE .
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Analysis II
44 Tst E is measurable
Let nA is open is measurable &* *mA mA mA
Note that AE A A E which is a disjoint union.
** *
** ** *
** *
*mA E mA m A EmAE mA E mAE mA m A EmAE mA mA
mA
This is true for any 0 ** *mAE mA E mA E is measurable.
Exercise 8 :
Let nES.T., E is measurable iff for each 0 there is sG set G
conlaining E for which *0mG E.
Proof : suppose E is measurable By outer approximation by an open set.
For each n, an open set kE s.t.
*1kmEk
Let
1kkG, then G is a G set ant EG munotes.in
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Lebesgue Measure
45
**
1
*
1
*
1
*
*
1k
KCK
K
C
k
K
C
K
KGmm K EEmEmE
mE
mEk
This is true for all k *0mG E
Conversely, suppose G set GE
s.t. *0mG E
tst E is measurable
Let nA G is countable int of measurable
Set G is measurable. By definition ** *CmA mAG mAG
Note that AE AG A G E
Which is a disjoint union ** *mA E mA G m AGE
** * *
** *
*
*0mAE mA E mAE A G mAG EmAG mA G mG E
mA
mA
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Analysis II
46 2.5 INNER APPROXIMATION BY CLOSED SETS
Theorem :
Let nE S.T. E is measurable iff for each 0, there is a closed
set FE for which *mE F E.
Proof :
Suppose E is measurable CEis measurable
Let 0 By outer approximative by open seet an open set CE s.t. * CmE
Let CEF is closed & FE.
Now ** * CmE F mEF mE
**
*CC
CmE m E
mE
Conversely suppose for 0, closed set FE such that *mE F E
Tst E is measurable
Let nA F is measu rable
By definition ** *mA mAF mA F
Note that AE AFF AF which is disjoint union. munotes.in
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Lebesgue Measure
47
***
**
** *
** *
*mAE mAF m AEFmAE mA EmAF m AEF mA EmAF mE F mA F
mA
Example 9 :
Let E be a set of finite outer measure show that there is an F set F
& a G set G s.t. FEG & ***mF mE mG.
[Ans] E is measurable for given each k open set kG and closed
set kF such that kkFE G and *1kkmGFk.
Let
1kkGG&
1kkFF.
Then G is SG set and F is F set and FEG.
We now show that ***mG mE mF GEG E which is
disjoint union. ** *mG mE mG E
Now CGE G E
1
1C
k
k
C
k
k
kk
kkGE
GEGE GEGF
**1kk mG E mGFk
This is tr ue for all k *0mG E **mG mE ……………………………………. (1) munotes.in
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Analysis II
48
***
1
1
1
** 1C
C
k
kCCkk
k
k
k
k
kk
kkEFE F
mE mF mE F
EF E F E FEF E FEF
EF
GF
mE F mGFk
This is true for all k *0mE F
Example 10 :
Let E be a set of finite outer measure show that if E is not measure,
then there is an open set containing E that has finite outer
measure and for which ** *mE m m E .
Solution : Since E is not measurable
00 for any open set containing E. *0mE …………………………………… (1) E has finite outer measure.
By definition a countable collection of open rectangles 1iiR
such th at
1iiER and *01i
iVR mE .
Let 0
1iiR 0E & 0 open. By (1) *0mE …………………………. (2)
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Lebesgue Measure
49 By countable sub additivity of *m
** *0011
** *
00 0
** *
00ii
iimm R V R m E
mm Em E
mE m m E
2.6 CONTINUITY FROM ABOVE
Theorem :
If 1kkB is a descending collection of measurable set and 1mB then
1limkkkkmB m B
Proof :
12BB…. Be collection of measurable sets and 1mB
tst
1limkkkkmB m B
Let 11kkAB B k then 12AA…… and kA’s are measurable
(kB’s are measurable)
11
11 1Ckk kkk kAB B B B
1
1CkkBB
1
1Ck
kBB
Let
1kkBB
11
1C
k
kAB BB B By continuity from bel ow 1 limkkmB B mA ………………………… (*)
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Analysis II
50 B and 1B are measurable 11mB B mB mB and 1kkmA mBB
=1kmB mB By (*)
11
1
1lim
lim
lim . . limkk
kkkk kkkkmB mB mB mB
mB mBmB mB i e B mB
Example 11 :
Show by an example that for continuity from aboe the assumption 1mE is necessary. Let ,kBk then 12...BB and kkmB we now show
that
1k
kB.
Let
1,kk
kxB x B k k
,xkk is bounded by x, which is not possible.
1
0 limk
kkkkBmm B m B
Example 12 :
Show that the continuity of measure together with finite additivity of
measure implies countable additivity of measure. Let kE be a countable collection of disjoint meas ure sets.
Let
1kkiiAE munotes.in
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Lebesgue Measure
51 Then kA’s are measurable and 12AA……
Also
11 1 1kki kkk i kAE E
By continvity from below,
1limkkkkmA m A
.
But by the fi nite additive property
1 1
1 1
1
1 1lim limk k
ki i
i i
kkk ikkk i
k
i
i
kk
k kmA m E mEmE m A m EmE
mE m E
Definition :
For a measurable set E, we say that a property holds atmost
everywhere on E, or it holds for almost all xE, provided there is a
subset 0E of E for which 00mE and the property holds for all 0xE E.
2.7 BOREL CANTELLI LEMMA
Let 1kkE be a countable collection of measurable sets for which
1k
kmE . Then almost all nx belong to Atmost finitely
many of the 'kEs.
Proof :
Let 0E be the subset of n such that 0 :nkEx x E for
infinitely many}
0
1kki kEE
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Analysis II
52 We sow that 00mE
Let kkkiFE
Then 12FF……. and 01k
kFE
Note that
1i
imE
Let
1iiLm E
1
lim
limikik ik
ikik
ikikmF m E m EmE
mE
1
1
1
1lim
limkiikik i
k
ikimE mELm E
1
0iiLm ELL
00mE
Example 13 :
Show that there is a non -measurable subset in .
Solution :||Qx Q x
WKT any two c osets are either identical or disjoint.
We now show that
If |AQ then 0,1A
Let AxQ munotes.in
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Lebesgue Measure
53 Let q be a rational number in ,1xx then0,1xq
Also xqxQA 0,10,1xqA
A
For each AQ choose 0,1AxA
Let AEx A Q
By construction 0,1E
Let
1,1 qXq E For any ,1 , 2xE qx 1, 2qE
This is true for any 1,1qQ
Let 1,1y then 0yy yQA (say)
but AxA AA Ayx qQ x A x yQ for some qQ
,0 , 1
1,1
1,10,1 0,1 1, 2A
A
Ayx
yx
qQ
yq x q E
yX X X
By monotonicity of *m ** *0,1 1, 2mm x m
*13mx
If E is measurable then qE is measurable and mE mq E
1,1 1,1 qQ qQmE m q E
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Analysis II
54 1,1 qQmX mE
1,11313qQmX
mE
If 0mE then 1,10qQmE
103 and if 0mE then 1,1 qQmE
Which is contradictin to (1) E is not measurable.
2.8 SUMMARY
In this chapter we have learned about.
Lebesgue measureable sets.
Construction of Lebesgue measurable sets in n
Properties of Lebesgue measurable sets
Non-measurable sets
2.9 UNIT END EXERISES
1. Show that the intersection of a countable collection of
measurable sets is measurable.
2. Show tht every open and closed subset of n are measurable.
3. Show that a set E is measurable if and only if for each 0,
there is a closed set F and open set for which FE and *mF
4. Let E be a measur able set in n and mEshow that for any 0 there exist a compact set kE such that *mE K E.
5. If 1kkA is an ascendin g collection of measurable sets then
1limkkkkMA m A
6. The outer measure of , the set of all rational number is ‘0’.
7. Prove that the outer measure of countable set is zero.
8. Show that the outer Measure of an interval is its length.
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55 3
MEASURABLE FUNCTION
Unit Structure :
3.0 Objective
3.1 Introduction
3.2 Measurable Function
3.3 Properties of Measurable Function
3.4 Egoroff’s Theorem
3.5 Lusin’s Theorem
3.6 Summary
3.7 Unit End Exercise
3.0 OBJECTIVE
After going through this ch aper youcan able to know that
Measurable function
Properties of measurable function.
Concept of simple function
3.1 INTRODUCTION
In the previous chapter we have studied about Lebesgue measure of
sets of finite and infinite measures. Now we can discuss Lebesgue
Measurability of functions. The definition of measurability of
function applies to both bounded and unbounded functions. We also
discuss simple function and its Approximation.
3.2 MEASURABLE FUNCTIONS
Definition : We say a function ‘f’ on n is extended real valued if it
take value on .
Definition : A property is said to hold almost everywhere on a
measurable set E provided it holds on 0EE, where 0E is a subset of
E for which 00mE
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Analysis II
56 Example 1 : Let f be a function defined on a measurable subset E of n. Then the following are equivalent.
1. For each real number C, the set :xE f x C is measurable.
2. For each real number C, the set ;xE f x C is measurable.
3. For each real number C, the set ;xE f x C is measurable.
4. For each real number C, the set ;xE f x C is measurable.
Solution : 12
Suppose for any C ,xf x Cis measurable ………. (*)
Let C
tst ;xf x C is measurable
Note that 1:;1xE f x C xE f x Cnn which is a
measurab le as countable intersection of measurable set is measurable
(by (*)) :xE f x C is measurable 23
Suppose :xE f x C is measurable ;;CxE f n C nE f x C which is measurable as
complement of m easurable set is measurable. ;xE f x C is measurable. 34
Suppose ;xE f x Cis measurable.
Let C
tst;xE f x Cis measurable. munotes.in
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Measurable Function
57 Note that
1;;1xE f x C xE f x Cnn which is measurable as
countable intersection of measurable set is measurable set. ;xE f x C is measurable. 45
Suppose ;xE f x Cis measurable.
tst;xE f x Cis measurable.
Note that ;;CxE f x C xE f x C which is measurable as
complement of measurable set is measurable. ;xE f x C is measurable.
Definition : An extended real -valued function ‘f’ defined nE is
said to be Lebesgue measurable or measurable, if its domain E is
measurable and it satisfies one of the above four statement i.e. For
each real number C, the set ;xE f x C is measurable.
Example 2 : Show that a real vaued function that is co ntinuous on
its measurable domain is measurable.
Solution :
Let ‘f’ be a continuous function
tst ‘f’ is measurable
Let C
Note that, 1;,xE f x C f C but ,C is open subset
of and :fE is continuous. 1,fC is open in E 1,fC G E for some G is open subset of n but any open -
subset of n is measurable and E is given as measurable. 1,fC G E is measurable munotes.in
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Analysis II
58 1;,xE f x C f C is measurable By definition f is measurable.
Example 3 : Let f be an extended real valued function on E. Sho that
1) F is measurable on E and fg a.e. on E then g is measurable
on E.
2) For a measurable subset D of E, f is measurable on E iff the
restriction of F to D and ED are measurable.
Solution : Suppose f is measurable and fg a.e.
Let :Ax E f xg x
Then as fga.e. we have 0mA
tstgis measurable.
Let ,;Cx E g x C ;; ExA g x C x g x CA ;; ExA g x C x f x C f gA ;;fgxA g x C xE f x C E A
But ;xA g x C A and 0mA any subset of measure zero set is measurable ;xA g x C is measurable f is measurable ;xE f x C is measurable E & A are measurable 0mA EA is measurable ;;xA g x C xE f x C E A is measurable ;xE g x C is measurable g is measurable. munotes.in
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Measurable Function
59 2) ;;fxE x C xD f x CD
;xE f x C D
For |
|;|
|fDf ExE x CD
E
|;;|E
D
EDxf x C
xE f x C
Converse ;; ; Exf x Cx D f x Cx f x CD ;xD f x C is measurable and ;Exf x CD is
measurable.
As union of measurable set is measurable fis measurable.
3.3 PROPERTIES OF MEASURABLE FUNCTION
Let f and g be measurable func tion on E that are finite a.e. on E
show that
1) (Linearity) for any '' and '', Fg is measurable on F.
2) (Product) fg is measurable on E.
Solution :
Let 0 : Ex E f x and gx then as f and g are
finite a.e. on E we have 00mE the restriction 0|fgE is measurable. any exte nsion of ''fg as an extended real valued function to all
of E is also measurable.
Without loss by generality , we may assume that ‘f’ and ‘g’ are finite
all over E.
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Analysis II
60 Now we first show that ''f is measurable fo r some .
If 0 then fis a zero function then for any C. ::
0
0xE Fx C xE f x Cif C
E if C and E are measurable ;xE Fx C is measurable F is measurable.
Suppose 0 ::xE Fx C xE f x C
;0;0C xE f x
C xE f x
…………………………(*) f is measurable and C & are red numbers. * is measurable ixE fx C is measurable fx is measurable f is measurable ………………………. (1)
We now show that fg is measurabl e.
Let C
If fg xC fx gx Cfx C gx Q is dense in , then is an rQ such that fx r C gx ;; :
rQxEf gx C xE f x r xE g x Cr Q is countable and :xE f x ris measurable & :xE g x Cr is measurable munotes.in
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Measurable Function
61 countable union of measurable set is measurable :xEf gx C is measur able fg is measurable …………………………… (2)
From (1) & (2) fg is measurable.
2) tpt fg is measurable
Note that 22212fg f g f g ,fg are measurable ,fgf is measurable it is enough tst
square of measurable function is measurable.
Let 0C
Then 2;;;xE fx C xE f x C xE f x C C
Which is union of two measurable set. by definition, 2f is measurable,
If 0C 2;xE fx C E which is measurable. In both the case 2f is measurable fgis measurable.
* Compos ition function fog
Example 3:
Let g be measurable real valued function defined on E and f a
continuous real valued function defined on all of show that the
composition fog is a measurable function on E.
Solution :
Given; Let ‘g’ be measurable function and ‘f’ be continuous function
on .
Let ;gE be measurable and :fR be a continuous munotes.in
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Analysis II
62 Let C
tst :fog is meas urable
Note that ;x E fog x C 1 11,, fog C g f C ,C is open subset and f is continuous 1,fC is open in . 1,0fC for some open subset O of . O is open in , we can write
1,iiiOa b
11 1
1
1
1
1,,
,
::ii
i
ii
i
ii
igfC g a b
ga bxE g x a xE g x b
:ixE g x a is measurable and :ixE g x b is
measura ble. countable union of measurable set is measurable set. :x fog x Cis measurable fog is measurable function on E.
Check your Progr ess :
If f is measurable, then show that
1) kf is measurable for all integer 1K
2) f is measurable for a given constant
3) f is measurable for a given constant
4) f is measurable
5) sup ,inf , lim sup lim infnn A n nnnfn fn fn fn are measurable. munotes.in
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Measurable Function
63 Definition :
For a sequence nf of functions with common domain E,a function
f on E and a subset A of E, we say that
1) The sequence nfconverges to ‘f’ point wise E, on A provided limnnfn f x for all xA
2) The sequence nf converges to ‘f’ point wise a.e. on A provided
it converges to F pointwise on ABwhere 0mB
3) The sequence nf converges to ‘f’ uniformly on A provided for
each 0,N such that nff on a for all nN.
Theorem :
Let nf be a sequence of measurable function on E that converges
point-wise a.e. on E to the function f, show that f is measurable.
Proof :
Let 0E be a subset of E with 00mE and nff on 0EE. 00mE & we have ‘f’ is measurable on E iff 0|fEE is
measurable. By replacing E by 0EE we may assume that the nf converges
to f on E
tst f is measurable
Let C
tst ;xE f x C is measurable ; ; lim
nxE f x C xE f x C but lim
nfx C iff ther e are natural nos. n and k for which
1
jfx C j kn
1;;j xE f x C xE fx Cn
1,kn munotes.in
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64 note that 1;j
jkxE fx Cn is measurable.
Countable union of measurable set is measurable ;xE f x C is measurable.
Simple Functions :
Definitions :
A real-valued functions defined on a measurable set E is said to be
simple if it is measurable and takes only a finite number of values.
If is simple, has domain E and takes the distinct values 1....,nCC
then
1knkEkC on E, where ;kkEx E x C .
This particular expression of is a linear combination of
characteristic functions is called the canonical representation of the
simple function .
Theorem : The simple Approximation Lemma
Let ‘f’ be a measurable real valued function on E. Assume ‘f’ is
bound on E. Then for each 0, there are simple function and E defined on E which h ave the following approximation
properties : EEf and 0EEE on E.
Proof :
Suppose :fE R is bounded measurable nf fis bounded, 0Msuch that fx M x E
Let ,cd be an open interval s.t. ,fE c d (fis bounded)
Let 0
Consider the partition
0, ....ndCy y y of ,cdwith 1,1kkyy k n
Define 1
11,kknnEk E E Ekkyy where 1
1,kk kEfy y
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Measurable Function
65 Note that 1
1,kk kEfy y
1
1
1;,
;
;:kk
kkkkxE f x y y
xE y f x yxE f x y xE f x y
which is measurable. (fis measurable) kEare measurable, 1kn & Eare measurable and takes only finite number of values & E are simple functions.
Let ,xE f x c d k s.t. 1kkyf x y 1 Ek k Exy f xy x …………………. (1) EExf x x
Also by (1) 1 0EE k kxx y y
Theorem : The Simple Approximation Theorem
An extended real valued function ‘f’ on a measurable set E is
measurable if and only if there is a sequ ence n of simple
functions on E which converges point -wise on E to f and has the
property that nf on E for all ‘n’.
If ‘f’ is non negative, we way choose n to be increasing.
Proof :
Suppose f is measurable
Case (1) Assume 0f
Let ,n Define ;nEx E f x n
Then |nfEis a bounded function. By simple Approximation Lemm a for 1,n simple functions & E such that |
nfEn and 10nnn . munotes.in
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66 We extend n on E defining nxn if fx n construct the
sequences n.
We now show that nf pointwise on E
(1) If ‘f’ is finite
N such that fx
11N
NN
NN N
NxE
xf x x
fx x x xNfx x nn
nxf x as n
(2) If f
fx N for any N limn
nnxnxf
Case (2) ‘f’ is any measurable function
Define 1max ,0xff x 1
1min ,0 fx f xfx f x f x f and f are non -negative measurable function. By Case (1), a sequence of simple functions &nn s.t.
nf pointwise and nf pointwise.
nnf pointwise n and n are simple function n nn a’s also a simple function n.
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67 3.4 EGOROFF’S THEOREM
Theorem Statement (Assume E has finite measure )
Let nf be a sequence of measurable functions one that converges
pointwise on E t o the real valued function f. Then for each 0
there is a closed set F contained in E for which nffuniformly
on F and mEF.
Proof :
Since nff pointwise on E, for 0, and ,xEK such that jfx f x j K ………………………… (1)
Since we want to get a region of uniform convergence, we
accumulate all xEfor which the same N holds for a fixed E.
For any pair k & n define
1:,n
kjEx E f x f x j Kn
Not all nkE are empty otherwise it will contradict pointwise
converges of nfxE. jf and f are measurable nkE is measurable.
Note that from fixed n 1nnkkEE and
1n
k
kEE By the confinuity of measure. lim nkKmE mE mE is finite, i.e. mE, for the above, 0, such that
12n
k nmE ME 12nn
k nmEE by countable additivity).
By construction for each nnkxE
1jj nfx f n kn …………………… (2) munotes.in
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68 Let nnkAE
We show that nff uniformly on A
Let 0 choose 0
01nn
By (2)
001jj nfx f n kn on 00nnkE 00nKnAE
001jj nfx f n kn on A
nff uniformly on A.
Now CmEA mE A
nCn
k mE E
1
1
1
122n
nCn
k
n
n
k
n
n
nmE E
mEE
nnkE are measurable and countable intersection of measurable set is
measurable. A is measurable. a closed subset F of A s.t. 2mAF
22mEF m EA AFmEA mAF
nff uniformly on A & FA
nff uniformly on F.
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Measurable Function
69 Examples 4 : Let f be a simple function defined on E. Then for each 0, there is a continuous function g on and a closed set F
contained in E for which fg on F & mEF.
Solution :
Let f be a simple function defined on E
Let f takes the values 1,.....,naa be the distance values taken by ‘f’.
1iniEifa
Where :iiEx E F x a
Note that
1niiEE 'kas are distinct 'kEs are disjoint f is mea surable 'ksF are measurable
Let 0
For each ,1 ,kk nkE is measurable closed subset kF of kE
such that kkmE Fn
Let
1njjFF Fis closed
1
1
11
1C
n
C
k
k
n
C
k
k
nn
C
kj
kj
n
C
kj
jmEF mE F
mE F
mE F
mE F
mE F
1
1
1
11n
k
n
C
kj
k
n
kk
k
nn
kk
kkmE F
mE F
mE Fn
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70 nn
Define :gF by igx a if ixF 'iEs are disjoint 'iFsare disjoint gis well defined and fg
on F we now show that ‘g’ is continuous on f then
11,ik
ikFF F F and kxF. an open interval KIF containing ''x 1IF 0k
kk ygy a y Igy gx a a I
g is continuous at x.
This is true for any xF g is continuous on F.
We can extend this continuous function ‘g’ on the closed set F to a
continuous function on .
Let the new function be ‘g’ then ‘g’ is continuous on and gF
on f and mEF.
3.5 LUSIN’S THEOREM
Statement :
Let f be a real valued m easurable function defined on E then for each 0, there is a continuous function g on and a closed set F
contained in E for which fg on f and |mE f.
Proof :
Let f be a real valued measurable function defi ned on E.
1) mE is finite by simple Approximation theorem a sequence n of simple
function on E such that nfand nf on E n. munotes.in
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Measurable Function
71 for each n there is a continous function ''ng on and a
closed set fnf conained in E for which nng on nf &
12n nmEF .
nf pointwise on E
By Egoroff’s theorem a closed set 0f contained in E such that nF uniformly on 0F and 02mEF .
Let
0nhFF
F is closed as countable intersection of closed sets.
Each n is uniformly on 0FF F n is continuous f is continuous on F
i.e. fF is continous.
We can extend fF to a continuous function ‘g ’ on .
Then fgon F
and CnmEF mE F
0
00
1
1
1
1|
2 2
22n
n
n
nn
nn
n
n
nmE F
EmE FFmE F m E F
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72 3.6 SUMMARY
In this chapter we have learned about
Concept of measurable functions.
Properties of me asurable functions
Simple functions & ith Approximation Theorem
Egoroffs Theorem and LUSIN Theorem of Measurable function.
3.7 UNIT END EXERCISE
1. Pure that “every continuous function is measurable”.
2. Show that the sum and Product of two simple function a re simple
function
3. Show that if ,0 ,f is differentiable, than 1f is
measurable.
4. Prove that if f is a measurable function X, than the set 1fx X f x is measurable.
5. Prove that if :0 , 1f is continous atmost everywhere than f
is measurable.
6. State and prove Egoroff’s Theorem of measurable function.
7. State and Prove Lusin’s Theorem of real valued measurable
function.
8. If ‘f’ is measurable then show that 1fCis measurable, C.
9. If f is measurable then show that ff is measurable.
10. Show that A is Measurable if and only if the set A is
measurable.
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73 4
LEBESGUE INTEGRAL
Unit Structure :
4.0 Objectives
4.1 Introduction
4.2 Lebesgue Integral of Simple function
4.3 Definition
4.4 The General Lebesgue Integral
4.5 Summary
4.6 Unit End Exercise
4.0 OBJECTIVES
After going throu gh this chapter you can able to know that
Lebesgue integral
Lebesgue integral of a simple function
Lebesgue integral of a bounded measurable function
The general Lebesgue integral
4.1 INTRODUCTION
We have already learned simple functions, measurable functions.
Now here we are going to discuss. Lebesgue integral on this
function. Lebesgue integral over come on the class of all
Riemannintegrable functions & the limitation of operations. So now
we defined the general notation of the Lebesgue integral on n step
by step.
4.2 LEBESGUE INTEGRAL OF SIMPLE FUNCTION
Definition :
For a simple function with canonical representation
1iniEixa Xdefined on a set of finite measure E, we define the
integral of over E by
1niiiEam E
.
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Analysis II
74 Example 1 : Let 1niiE be a finite disjoint collection of
measurable subset of a set of finite measure E. For 1,inLet
ia.
If
1ni Eiia on E, than
1niiiEam E
.
Solution :
Let
1iniEias.t. 'iEs are pairwise disjoint which may not be in
canonical form.
Let 1kjjbbe distinct elements of ,.....inaa.
Define
jjiiIFE where :ji jIi a a.
Note that 'jFs are disjoint. jjiiImF mE
1jkjFjb is a canonical re presentation of . By definition
1kjjjEbm F
1jkjiji Ibm E
1niiiEam E
4.2.1 Theorem (Properties of integral simple function)
Let and be simple functions defined on a set of finite measure.
Then
1) Linearity : For any ''and ''
EE E
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75 Proof :
Let
1iniAia and
1jnjBjb be canonival representation of
and respectively.
,1 ,1ij i jCA B i n j m
then
11nmi ijijaC and
11nmi ijijbC …………. (1) By definition
11nmi ijijEam C
and
11nmj ijijbm C
By (1)
11nmi j ijijab C By definition
11nmi j ijijEab m C
11 11
11 11nm nm
i ij j ij
ij ij
nm nmi ij i ijij ij
EEamC bmCam C bm C
2) Monotonicity
If on E then
EE
Proof :
Suppose on E
tst
EE
Let 0f By linearity property munotes.in
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Analysis II
76 0EE E Ef
EE
3) Additivity :
For any two disjoint subset ,AB E with finite measure,
AB A B
Solution : ABAB E
AB
EABEE
AB
4) Triangle inequality : If is a simple and
EE.
Solution : Let be a simple function and
1ni Aiia be canonical
representation of .
Then
1ni Aiia which is a simple function.
By Definition
1
1n
i
iE
n
i
iEa m Aia m Ai
1n
i
ia m Ai (by triangle inequality) munotes.in
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Lebesgue Integral
77
1
1n
i
i
n
i
i
i
Ea m Aia m Aia
5) If a.e. on E, then
EE
Solution : Suppose a.e. on F
Let 0;Ex E a x
Then 00mE and on 0;EE
Let
1ni Aiia and
1nj Bjjb be canonical representation of
and representation. By definition
1n
i
iEa m Ai
00
10011
0
1|n
ii i
i
nn
ii ii
ii
n
ii
iam A E A E Eam A E am A E EOa m A E E
0| EE E
Similarly
0 EE E on 0EE
EE
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78 * Lebesgue integral of a bounded measurable function on a set of
finite measure.
We now extend the notion of integral of simple function to a
bounded measurable function on a set of finite measure.
Let ‘f’ be a bounded real -valued function defined on a set of finite
measure E. We define the lower and upper Lebesgue integral
respectively, of ‘f’ over E to be sup :
Esimple and f on E
and inf :
Esimple and f on E .
Since ‘f’ is bounded by the monotonicity property of the integral for
simple functions, the lower and upper integral are finite and the
lower integral the upper integral.
4.3 DEFINITION
A bounded function ‘f’ on a domain E of finite measure is said to be
Lebesgue integrable over E if its upper and lower Lebesgue integrals
over E are equal. The common value of the upper and lower
integrals is called the Lebesgue integrals or simply the integral, of ‘f’
over E and is denoted by
Ef.
Example 2 : Show that a non negative bounded measurable function
on a set E of finite measure is integrable E of finite measure is
integrable over E.
Solution : Let ‘f’ be a bounded measurable function defined on E.
where mE. By simple Approximation Lemma
For ,nsimple function n and n such that nfn and 10nnn .
11
nn n n
EE E EmEnn
But, sup ; ,nEsimple f and inf ; ,nsimple f munotes.in
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Lebesgue Integral
79 0 inf ; ,
Esimple f sup ; ,
Esimple f
1
nn
EEmEn
This is true for any n and mE
inf ; ,
Esimple f
sup ; ,
Esimple f fis Lebesgue integrable over E.
Example :
Let ‘f’ be a bounded measurable function on a set E of finite
measure. Show that if 0Ef then 0f a.e.
Solution : Suppose 0Ef and 0f
tst 0fa.e.
Let 1;nEx E f xn then 1
nExf xn.
By monotonicity,
1010
0nE
E
n
nxfn
mEn
mE
But 0
1;0nnEx E f x E 0mE 0f a.e. over E.
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80 4.3.1 Properties of integral of bounded func tion :
Theorem : Let ‘f’ and ‘g’ be bounded measurable functions defined
on a set of finite measure E then
1) Linearity : for any '' and
EE Efg f g
Proof : Let ,fg be bounded functions, ,
tst
EE Efg f g
It is enough tst
EEff and
EE Efg f g
If 0 then 0f
0
EEff
Suppose 0 f is bounded fis bounded f is lebesgue integrable.
Let 0
Ef upper lebesgue integrable of ''f
inf : &
inf : &
inf : ,
inf : ,E
E
Eis simple fsimple fsimple fsimple f
Ef
Let 0
Similarly for lower Lebesgue integral of f munotes.in
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Lebesgue Integral
81 EEff
We now show that
EE Efg f g
Let 1 and 2 be simple functions on E such that, 1f and 2g then 1 + 2 is a simple function and 12fg f and g are bounded fg is bounded. fg is Lebesgue integrable By definition 12 1 2inf ; ,
E
EE Ef g f g is simple
This is true for any 1, 2simple with f, and 2g fgis lower bound of
12 1 1 21 2
12 121 2
11 1 22 2;,, ,
inf ; , , ,
inf ; , inf ; ,EE
EE E
EE
EE Ef g simplef g f g simplef simple g simplefg
fg f g
For the reverse inequality
Let 1 and 2be simple function for which 12&fg on E then
12fg and 12 is simple munotes.in
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Analysis II
82 12
12sup ; ,
EE
E
EEf g f g simple
This is true for any 1, 2 simple with 12 &fg
Efg is upper bound of
12 121 2
12 121 2
11 1 22 2;,, ,
sup ; , , ,
sup ; , sup ; ,EE
EE E
E
EE
EE E
EE Ef g simplef g f g simplef simple g simplefg
fg f g
fg f g
2) Monotonicity : If fg on E, then
EEfg
Proof
Suppose f and g a re bounded mesurable function on a set E of finite
measurable function and fg
tst
EEfg
Let 0hfg h is non-negative bounded function. By linearity
EE E Egf g f h munotes.in
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Lebesgue Integral
83 h is bounded & 0h
h where 0 simple function
But sup ; ,
EEh simple h 0* 00EE
EE E
EEhm E
gf h
gf
3) Additivity : For any two disjoint subsets, ,AB E with finite
measure.
AB A Bfff
Proof :
Let ‘f’ be bounded measurable function on a set E of finite measure
and A,B disjoint subsets of E .
tst
AB A Bfff f is bounded measure. ,,AB A Bff are bounded measurable functions. AB A BAB E EABE
AB
EE
AB A Bffffff
fff
4) Triangle inequality : Let f be a bounded measurable function on a
set of finite measure E, Then
EEff.
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84 Proof :
Let f be bounded measurable function on a set E of finite measurable f is measurable and bounded on E.
Note that fff By monotonicity and linearity
EE E
EEfffff
Example :
Let nf be a sequence of bounded measurable functions on a set of
finite measure E. Show that if nffuniformly on E, then
limnn
EEff
Solution : Let nf be a sequence of bounded measurable function
on a set E of finite and nffuniformly on E
tst limnn
EEff
i.e. n
EEff
nff uniformly on E for a given 0, 0n 0,nxE fx f x nnmE
i.e. 0nff n nmE on E
For 0nn
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Lebesgue Integral
85 Now nn
Eff f f
n
Eff
mE
mEmE
By definition
limnn
Eff.
Example 5 :
Show by an example that the pointwise convergence alone is not
sufficient to the passage of the limit under the integral sign.
Solution : Example
Let 0f, function on 0,1E
Let 10, 0kKk as k
kf pointwise 1.0 ,1.1k
EKmkKk
0Ef
k
EEf
Example 6 :
Let f be a bounded measurable function on a set of finite measure E.
Assume g is bounded and fg a.e. on E,
Show that
EEfg munotes.in
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Analysis II
86 4.4 THE GENERAL LEBESGUE INTEGRAL
For an extended real -valued function ‘f’ on E, the positiv e part f
and the negative part f of f defined by ,0f x ma f xand ,0f x ma f x x E
Then f and f are non-negative functions on f ff f on E and ff f on E
Thus f is measurable iff f and f are measurable.
Example 7 :
Let f be a measurable function on E, show that f and f are
integrable over E iff f is integrable over E.
Ans. Suppose f and f are integrable
Ef &
Ef
But ff f
EE E Eff f f f f is integrable
Conversely, suppose f is integrable
Ef
But ff & ff
EEff f is integrable
Similarly f is integrable.
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Lebesgue Integral
87 Definition :
A measurable function f on E is said to be integrable over E if f is
integrable over E i.e.
Ef. If ‘f’ is integrable over E, then we
define the integral of ‘f’ over E by
EE Eff f
Example :
Let ‘f’ be integrable over E. Show that f is finite a.e. on E and
0 EE Eff where 0EE and 00mE
Solution :
‘f’ is integrable on E f is integrable
Ef
Note that f is non negative integrable function.
We now show that f is finite a.e. on E.
Note that ;xE f x
;xE f x x ;;xE f x xE f x nn
But by chebychev’s Lemma ………………………. (*)
1;nmxE f x n fn f is integrable,
Ef is finite
i.e. f ;0;0mxE f n n
mxE f n munotes.in
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Analysis II
88 fx is finite a.e. on E ff, we get f is finite a.e. on E
Let 0EE s.t. 00mE By definition
EE Eff f
00||EE EEff (&ffare non-negative integrable
functions)
00EE EEff f
Example 9:
Define 23101fx xx
00x
Show that f is Lebesgue integrable on 0,1 and 1
23
013dxx. Find
also ,2fx
Solution :
231x as 0x
So f is unbounded in 0,1 its Lebesgue integrability define
231,fx nx if 3211xx
= n if 321Ox n
0 if 0x
Now 32
321|11
00 1 |,,,n
nf x n dx f x n dx f x n dx munotes.in
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Lebesgue Integral
89 32
321|1
2|3
01 |
13
321
11 231 3n
nndx dxxnnn n
by definition of the Lebesgue integral of on bounded functions
11
00lim ,
2lim 33n
nf x dx f x n dxn
Lebesgue integrable define f or 2n
231,2fxx if 2311xz
2 if 2310xz
0 if 0x
4.5 SUMMAR Y
In this chapter we have learned about
Introduction concept of Lebesgue integral.
Lebesgue integral of complex valued Measurable functions
Lebesgue integral at a simple function.
Lebesgue integral on bounded Measurable function general
Lebesgue integ ral
4.6 UNIT END EXERCISE
1. Show that for a finite family 1knf of measurable functions
with common domain E, the functions 1....nMax f f and 1....nMin f f also are measurable.
2. Show that the sum and produ ct of two simple functions are
simple. munotes.in
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Analysis II
90 3. For every non -negative and measurable function f on 0,1 then
show that
0,1 0,1inff dm dm.
4. Prove that a measurable function 10,1fx L if and only if
120 , 1 ;2nn
nmx fx
5. If 10,1fLfind 21
2
0lim log 1
kfxK dxK
6. Let f be a Lebesgue integrable function on X use the positive and
negative part of f to prove that
xxf dx f dx.
7. Let f be a non -negative measurable function on X and suppose
that fMfor some constant M prove that
Exf dx f dx for
8. Calculate Lebesgue integral for the function
1
2where x is rationalfxwhere x is irrational
9. Evaluate 5
0f x dx if
00 111 2 3 422 3 4 5x
fx x x
xx
by using Riemann and Lebesgue definition of the integral.
10. Show that if f is a non -negative measurable function then 0fa.e. on a set A iff 0Af dx
11. If 1fx x if 01x 9then f is not Lebesgue integrable in 0,1
12. Let F be a non -negative measurable function on and suppose
that fMfor some constant M. Prove tha t Efd m E for
any measurable E.
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91 5
CONVERGENCE THEOREMS
Unit Structure :
5.1 Introduction
5.2 Measurable Functions
5.3 Lebesgue Theorem on Bounded Convergence
5.4 Limits of Measurable Functions
6.5 Fatou’s Lemma
5.6 Lebesgue integral of non -negative measurable function
5.7 The Monotone Convergence Theorem
5.8 Dominated Convergence Theorem
5.9 Lebesgue integral of complex valued functions
5.10 Review
5.11 Unit End Exercise
5.1 INTRODUCTION
In this section we analyze the dynamics of integrability in the case
when sequences of measu rable functions are considered. Roughly
speaking a “convergence theorem” states that integrability is
preserved under taking limits. In other words, if one has a sequence 1nnf of integrable functions, and if ‘f’ is some kind of a lim it of
the 'nfs then we would like to conclude that ‘f’ itself is in tegrable,
as well as the equality limnnffsuch results are employed in
two instances.
i) When we want to prove that some function ‘f’ is integrabl e. In
this case we would look for a sequence 1nnf, of integrable
approximation for f.
ii) When we want to construct and integrable function in this case,
we will produce first the approximates and then we will examine
the existence of the limit.
The first convergence result, which is some how prim ote, but very
useful in the following.
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Analysis II
92 5.2 MEASURABLE FUNCTIONS
Theorem :
Let ,,XA be a finite measure space, let , GC o and let :0 , 9 , 1nfX n be a sequence of measurable functions satisfying.
1) 12.... 0ff
2) lim ,nnfx x X Then one has the equality lim 0nn
Af dx
.
Proof :
Let for each 0 and each intege r 1n, the set ;C
Ke nAx X f x obviously, we have ,0 , 1nAA n we
are going to use the following case .
Claim I :
For every 0, one has the equality lim 0nnA.
Fix 0, Let us first observe that (a) we ha ve the inclusion
12CCAA………………………………….. (II)
Second using (b) we clearly have the equality
1k
kA . Since is
finite using continuity property we have
1lim 0n
nnnAA
Claim II :
For every 0, and every integer 1,n one has the inequality 0nn
Xf du a A x .
Fix and n and let us consider the elementary functions. nAnAnh ax x where EnBX A obviously, since x the
function nh is elementary integrable. By cons truction we clearly
have 0nnfh, so using the properties of integration, we get munotes.in
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Convergence Theorem
93 0nn n
XXf dx h dx a A BaA X
Using claim I & III it follows immediately that 0 lim inf lim supnnnn
XXfd fd X
Since the last inequality hold for arbitary 0, we get lim 0nn
Xf du
5.3 LEBESGUE THEOREM ON BOUNDED
CONVERGENCE
Statement :
Let nfbe a sequence of functions measurable on a measurable
subset ,Aa bsuch that limnnfx f xthen if there exists a
constant M such that nfx Mfor all ‘n’ and for all ‘x’, we have limnn
AAf x dx f x dx.
Proof : limnnfx f x and nfx M fx M
The function ‘f’ is bounded and measurable
Hence Lebesgue integrable.
Now we shall show that lim 0nn
Af x f x dx
For a given 0, we define a partition A into disjoint measurable
sets 'kAs as follow s : 1: , , 1,2,3,.....kk n nAx f f f f k K
munotes.in
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Analysis II
94 In particular, 1
21: ; 1, 2,3,.....: ; ; 2,3, 4,.....n
nAx f f n
Ax f f f f nn
Clearly, 11 1kk kKK K n
nnAn n n nAA A APQm m P Q mP mQ
Now
nnnn n
APQf f dx f f dx f f dx …………. (1)
For each ‘n’, we have
nff on nP and 2nnff f f m on nQ
Thus, 2nn nAf f dx mP M mQ
As n, limnnmP mA and lim 0nnmQ
Thus, n
Af f dx mA being an arbitrary val ue limnn
AAf x dx f x dx
Example 1 :
Verify Bounded Convergence.
Theorem for the sequence of functions 1;1 ,
1n nfO x n
xn .
111n nfx nxn and x
Each nf being bounded and me asurable, the limit function. munotes.in
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Convergence Theorem
95 211lim lim
1nxnnfxexn
It is also bounded and measurable. Now
11
1
0
01
11
11
11 11n
nnxdx nnn xn
nnnn
1
0lim
1
11
lim 11 11nnnndx
xn
nnnn
1
lim 11 1nnn
nnn n
n 111eee
Similarly,
11 1
00 0
1
011lim
1
111x
n xn
xdx dx e dxe x
n
eeee
Hence Bounded convergence theorem is verified.
5.4 LIMITS OF MEASURABLE FUNCTIONS
If : , ;1,2,....nfn is an finite sequence of functions then
we say that :,f is the pointwise limit of the sequence nnf if we have limnnfx f x for each x. munotes.in
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Analysis II
96 For any sequence :,nf we can define lim supnnf as the
function with value at ‘x’ given by
lim sup lim supnknnk nfx fx
Something th at always makes sense because supk
knfx decreases n
increases or atleast does not get any bigger as n increase. Suppose
that nf is a sequence of real number. Let A be the set of numbers
such that nff for some subsequence knf of nf. f is called a limit point of nf, so A is the set of all limit points of nf. Then suprem um and infimum of A are denoted by the
following lim inf inf , lim sup supnnnnfA f A .
5.5 FATOU’S LEMMA
Statement :
If nf is a sequence of non -negative measurable functions, then for
any measurable set E. lim inf lim infnnnn
EEf dx f dx
Proof : We write lim inf
nfx fx
We recall that for any x, liminf inf infnnfx f where Ex is the set
of all limit points of nfx.
nff pointwise convergence on E
nff pointwise on 1,0EmEE
nff pointwise on 1E
1EE and 10mE
We may assume nff pointwise on E 'nfs are non -negative measurable and nff fis non -negative and measurable. munotes.in
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Convergence Theorem
97 Now to show that lim infnn
EEff
Let h be a bounded measurable function of finite support such that
0hf 0mE where 0;0Ex E h x h is bounded choose M such that hx M on E for n Define minnh,nhf.
Clearly 0nh is measurable bounded function and nhM. We can
now show that nha pointwise on 0E.
For 0xE h x f x
Case I :
hx f x 0fx h x
nffpoint wise on E for 0fx h x
0n such that 0nnfx f x n 00min ,n
nn
nnfx f x fxhx f x f x nhx h f h x n n
nhh point wise on 0E
Case II :
hx f x
Then nnhx fx on fx n
nffpoint wise on 0E
nhf h pointwise 0E
By bounded convergence Theorem
For the bounded sequence nhrestricted to 0E munotes.in
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Analysis II
98 We have
00limnn
EEhh
00
00lim lim
0, ; 0
lim lim inf lim infnnnn
EE E E
nnnnn
EE E Ehh h h
h on E E h on E Ehh h f
This is true for any bounded measurable function with finite support
such that 0hf By definition of
Ef
lim infnh
EEff
5.6 LEBESGUE INTEGRAL OF NON -NEGATIVE
MEASURABLE FUNCTION
Definition :
Let f be a measurable function defined on E. The support of ‘f’ is
defined as sup ; 0fx E f x .
Definition :
A measurable function f on E is said to vanish outside a set of finite
measure if a subset 0E of E for which 0mE & 0f on 0EE. It is convenient to say that a function that vanishes outside a
set of finite measure has finite support. We have defined the integral of a bounded measurable function ‘f’
over a set of finite measure E. But mEand f is bounded and
measurable on E with finite. Support we can define its integral over
E by
0 EEff where 0mEand 0fon 0EE.
Definition :
For a non -negative measurable function f on E we define integral of
‘f’ over E by sup :EEhh bounded; measurable of finite suppor t
and 0h f on E. munotes.in
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Convergence Theorem
99 Chebychev’s Inequality :
Statement :
Let f be a non -negative measurable function on E then for any 0.
1;
Emx Efx f
Proof :
Let: Ex E f x
Case I : nm for each n define ,nEE n n. Then nnE.
Then n is bounded measurable function nmnEE and nf
Note that 1nnEE and
1n
nEE By continuity of measure. lim
limnmmn
nn
EEE
n is bounded on E and nf by definition
Ef, we get
n
EE
m
EfEf
Both side
1
EmE f
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Analysis II
100 Case II : mE
Define Eh then h is bounded measurable function hf by definition of
Ef, we get EmE h f
1
1;E
EmE fmx Efx f
5.7 THE MONOTONE CONVERGENCE THEOREM
Statement : Let nf be an increasing sequence on non -negative
measurable functions on A. If limnnfx f x then limnn
AAff.
Proof :
Letnf be an increasing sequence of non -negative measurable
functions and limnnff x i.e. it is convergent at pointwise to
f on A.
Now to show that limnn
AAff.
nff pointwise on A and 1 nnff n nnff on A
n
AAff on A
supn
AAff
lim supnn
AAff……………………………. (I)
By the Fatou’s lemma
lim infnn
AAff ……………………………. (II)
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Convergence Theorem
101 From I & II we get
lim inf lim sup
limnnnn
AA A
nn
AAff fff
5.8 DOMINATED CONVERGENCE THEOREM
(Generalisation of Bounded Convergence Theorem)
Statement : Let nf be a sequence of measurable function on E.
Suppose there is a function ‘g’ that is integrable over E and
dominates nf on E in the sense that nfg on E for all n. If
nff pointwise almost everywhere on E,then f is integrable over
E and limnn
EEff.
Proof : nnfg on E and nff pointwise on E.
fg g
fg f is measurable
nfg and 0n fgg f and ngf gf pointwise By Fatou’s lemma
liminf
liminf
limsupnnEEnEEgf gfff
limsupn
EEff …… ………………………… (I)
Similarly 0n gf & ngf gf pointwise on E. By Fatou’s lemma,
lim infnEEgf gf munotes.in
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Analysis II
102 liminfnEE Egfg f
lim infnEEff ……………………………. (III)
From I & II we get
liminf limsup
limnn
EE E
nn
EEff fff
Example 2 :
Check the convergence of 1;
0;nfx x nnxn
Solution : Let 1;nfx x nn
0xn
Then 0nfxuniform ly on but 2;nf dx 1, 2,3,.....n
1lim lim 0nnnfxn where xn
0 when xn lim 0nnfx uniformly on the whole real time.
Now, 211 1
22mmfxf xmm m
Whenever 12M
Now 100 2nn
n
nnf x dx dx dx dxn .
This emplies that uniform converges of nfx is not enough for lim limnnnnff munotes.in
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Convergence Theorem
103 This equality is Lebesgue integration.
In general, is only due to dominated convergence of the sequence nfx. However on the set of finite measure uniformly convergent
sequence of bounded func tion are bounded convergent.
5.9 LEBESGUE INTEGRAL OF COMPLEX VALUED
FUNCTIONS
If f is a complex valued function on nEwe may write as fx u x i x where &uvare real functions called the r eal and
imaginary part of f.
A complex valued measurable function, :f u iv on E is said to be
integrable if 22
EEfx u x v x and the integral of ‘f’
is given by
EEEfu i v
Theorem :
Show that a complex v alued function is integrable if and only if both
of its real and imaginary parts are integrable.
Proof :
Suppose :f u iv is integrable
22
22 2
22f
uvuu u u vuu v
uis integrable
Similarly vis integrable
Conversely
Suppose &uv are integrable
u and V
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104 By Minkowski’s inequality
22 2 2fu vuv u vfu v f is integrable.
Definiti on :
A measurable function :,nfE C E is said to be an 1L function
if
Ef.
Note : 1 nL {set of all complex valued function on n}
Definitio n : A family G of integrable function is dense in 1nL if
for any f and 0gG so that
Efg
Example 3:
Show that the continuous function of compact support is de nse in 1nL.
Solution :
To show that : The continuous function of compact support is dense
in 1nL.
i.e. tst for any 1fL and 0. a continuous function ‘g’ on n with compact support such that ,fg i.e. fg .
Let 1nfL
We may assume ‘f’ is real valued becaue we may approximate its
real and im aginary part independently.
In this can we write ff f.
Where 0f and 0f It is enough to show the result 0f. 0f can be approximated by integrable simple functions. munotes.in
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Convergence Theorem
105 It is enough to show that the result for an integrable simple
functions. nA integrable simple functions is a Linear combination of
characteristic function.
It is eno ugh to show for Ef where E is a measurable set of finite
measurable.
Let 0 E is measurable a compact set k and an open set of n such
that KE and | mk
By Urysohn’s Lemma a continuous function :gk such that 0gon |k& 1g
on K g is continuous function with compact support 1E gf g |Ekand 0E g on outside |Ek 1| |
nEkgf m E k m k continuous function of compact support such that gf . Continuous function of compact support is dense in 1nL.
Example 4 :
Let 1nfLshow that ff
Solution : Let 1fL to show that ff
Let zf
If 0z then clearly 0fz z f ff
If 0z
Define zz munotes.in
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Analysis II
106 1 and zz fz z f f
Let fu i v
By definition 0fu i vfu i vfv
fu ……………………………………….. (I) uu f f f
By Monotonicity property uf ………………………………………….. (II)
By (I) and (II) ff proved
Example 5 :
Show that 1nL is complete in its metric.
Solution :
Let nf be a Cauchy sequence in 1nL for 0 0,n such
that 01,mnff n n n for each k
We can choose kn such that for ,kmn n112mnkff and 1kknn
then the sequence knf has the property that
112kknnkff
.
Construct the series 12 132
11
1....kknnnnn
nn n
Kfn f x f x f x f x f x
fx f x fx
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Convergence Theorem
107 and
11
1kk nn n
Kgx f x f x f x
Let kSg denote the thk partial sum of the series g then.
111
1iik
kn n n
iSg f x f x f x
Then kSg is a seq uence of non -negative function converges
pointwise to g. 1 kkSg S g n By Monotone Convergence Theorem g is integrable and limknSg g
Note that fg fg ( g is integrable) f is integrable f is 1nL
Let kSf denote the thk partial sum of the series of f , then
111
1ii
kK
kn n n
i
nSf fx f x fxfx
kSf f pointwise
knff pointwise
Now we show that
knff in 1nL
Note that
knff g k
By Dominated convergence The orem
1lim 0lim 0k
knn
nnff
ff
knff in 1nL munotes.in
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Analysis II
108 nf is Cauchy and has convergent subsequence knfconverges of f.
We get nff Every Cauchy sequence in 1L is convergent. 1L is complete in its metric. Proved
5.10 REVIEW
In this chapter we have learnt following points.
Limits of Measurabl e function
Bounded convergence theorem of measurable function
Monotone convergence theorem of measurable function.
Fatou’s lemma of measurable function
Dominated convergence Theorem
Complex valued measurable function
Compactness of 1nL
5.11 UNIT END EXERCISE
1. show by an example that the inequality in Fatou’s lemma may be
a strict inequality.
Example : Consider a sequence of function nnf defined on 0,1
by 220,11nnxfx xnx.
i) Show that nf is uniformly bounded on 0,1 and evaluate
22
0,1lim1nnxdxnx
ii) Show that nf doesnot converge uniformly on 0,1
Solution :
1) For all n for all 0,1x we have 2212 0n x nx and
2210nx
Hence 221012nnxfxnx munotes.in
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Convergence Theorem
109 Thus fx is uniformly bounded on 0,1
Since each nf is continuous on 0,1
f is Riemann integrable on 0,1
In this case Lebesgue integral and Riemann integral on 0,1.
Consider
1
22 220
0,111nx nxdx dxnx nx
Put 221nx t
21
0112nt dtx
22
22
0,1log 1 1log 112 2nnxdx nnx n n
Using 1L Hospitalrule we get
2log 1log 02nnn
Hence
22
0, 1lim 01nnxdxnx
ii) For each
22
0,10,1 lim 01nnxxnx
Hence nff pointwise on 0,1
Now to show that nf does not converges to 0f uniformly on 0,1.
We find a sequence nx in 0,1.
Such that 0nx and xnfx00f as n, taking 1nxn
then 12nfx .
Thus 1lim 0 02nnnfx f
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Analysis II
110 Example 2 :
Evaluate 2
01limn n
x
nxe dxn
Solution : We know that
lim 1nxnxen and 1111nnxx
nn .
Also we have 1nxxen
21nxxxeen by Dominated convergence then to the function 1nxxen with
the dominating function xe
2
0
2
0,1
0
2
0,1
0
0lim 1
lim 1 1
lim1 11n n
x
n
n
x
n
n
x
n
xxe dxn
xx e dxn
xx e dxn
e dx
2) Show by an example that monotone convergence theorem does
not hold for a decreasing sequence of functions.
3) Let 2:; 0nxfx x nn
= 0 ; otherwise
Evaluate
0limn
nnf x dx and
0limn
nnf x dx are these equal?
4) 1002gx x
= 1 112x munotes.in
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Convergence Theorem
111 2
21,0 11, 0 1k
kfx g x x
fx gx x
To show that 00lim inf lim infnnnnf x dx f x dx
5) If ;0 ,nfX is measurable for 1, 2,....n and
1n
nfx f xx X then show that
1n
nXXf dr f dr .
6) Use the dominated convergence theorem to find
1limnnf x dx where 31nxfxnx.
7) If nnab for all n, then show that lim inf lim infnnnnab .
8) State and prove bounded convergence theorem of measur able
function.
9) Use convergence theorem to show that 0,cosxft e t d ux
is continuous.
10) Use the dominated, convergence theorem to prove that
22 1
0lim 0nx
nn xe dx
11) Use the dominated convergence theorem to show that
21
2 2
2
2lim 1n
x
n
RRxdx e dxn
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112 6
SPACE OF INTEGRABLE FUNCTIONS
Unit Structure
6.0 Objective
6.1 Introduction
6.2 Signed Measures
6.3 Hahn decomposition theorem
6.4 Complex valued Lebesgue measurable functions
6.5 The space L1 (µ) of integrable functions
6.6 Let’s sum u p
6.7 Unit end exercise
6.8 List of References
6.0 OBJECTIVE
After going through this chapter you will able to know:
Signed measures is to generalize the concept of a traditional measure
in measure theory to allow for negative values.
The Hahn decomposition theorem is to provide a fundamental result in
the theory of signed measures.
Studying complex -valued Lebesgue measurable functions is to extend
the notion of measurability and integration to functions whose range is
the set of complex numbers.
Studying th e space
(µ) of integrable functions plays a central role in
measure theory, functional analysis, and various fields of mathematics
and applied sciences.
6.1 INTRODUCTION
Till now our measures have always assumed values that were greater than
or equal to 0. In this chapter we will extend our de finition to allow for both
positive and negative values. signed measures extend the concept of
measures by allowing them to take both positive and negative values.
This decomposition provides a clear separa tion of the positive and
negative components of the signed measure, enabling a deeper
understanding of the measure's behaviour on different subsets of the space.
The sets A and B are unique up to null sets, meaning that any measurable munotes.in
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Space of Integrable Functions
113 subsets of A and B with measure zero can be added to or removed from A
and B without affecting the positivity/negativity of the measure.
The space L1 (µ) of integrable functions is a fundamental concept in
measure theory and functional analysis. It provides a rich f ramework to
study functions that are Lebesgue integrable with respect to a given
measure µ on a measurable space. The integration theory based on the
Lebesgue integral allows for a broader and more flexible class of functions
compared to the traditional Ri emann integral.
6.2 SIGNED MEASURES
A signed measure is a mathematical concept used in measure theory,
which is a branch of mathematics that deals with the study of measures.
Measures are used to assign a notion of size or volume to subsets of a
given set. In traditional measure theory, measures are non -negative,
meaning they take values in the real numbers and are non -negative for all
sets. However, in certain applications and contexts, it becomes useful to
work with more general measures that can take pos itive or negative
values, and these are referred to as signed measures.
Definition: Le(,)beameasurablespace.Asignedmeasureon(,)is a
function : such that
i) µ takes on atmostone of the values − ∞ or ∞.
ii) µ(∅)=0.
iii) If 1nnE is the sequence of pairwise disjoint set, then 11()nnnnEE
Additionally, a signed measure can be decomposed into its positive and
negative variations:
i) {() , 0 }EMax E is positive variation.
ii) { ( ),0}EMax E is negative variation.
With these variations, the signed measure can be written as the difference
of two non -negative measures:
()EE
It's important to note that a signed measure can take the value of positive
infinity, negative infinity, or be finite.
6.3 HAHN DECOMPOSITION THEOREM
The Hahn decomposition theorem plays a significant role in understanding
signed measures and provides a powerfu l tool for analysing their
properties and behaviour. It highlights the duality between positive and
negative parts of a signed measure, enabling deeper insights into the munotes.in
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Analysis II
114 structure of measures and their applications in various mathematical
contexts.
The Hah n decomposition theorem is valuable in several ways:
1. It allows us to analyze the positive and negative parts of a signed
measure separately, which can be beneficial in various applications.
2. It is a key step in proving other important results in measure th eory,
like the Jordan decomposition theorem.
3. It is used to establish the Radon -Nikodym theorem, which is a
fundamental result connecting measures and integrals in a more
general setting.
To prove the Jordan decomposition of a signed measure, we first show
that a measure space can be decomposed into disjoint subsets on which a
signed measure is positive or negative, respectively. This is called the
Hahn decomposition.
Definition: Let µ be a signed measure on ( X, ). A pair { P, N } of
elements in for which P is positive N is negative, P ∪ N = X and P ∩ N
= ∅ is called a Hahn decomposition of X with respect to µ.
Radon -Nikodym theorem for signed measures:
Let (, ,) be a
- finite measure space and v a finite signed measure
on measurable space (, ) that is absolute continuous with respect to.
Then there is a function fthat is integrable over X with respect to and
()Ev E fd for all .
Function fis uniqu e upto a set ofmeasure zero.
Definition: Two measures and v on measurable space (, )are
Mutually Singular if there are disjoint A and B in M. For which XA B and () () 0Av B
Definition : The decomposition of signed measure ν on measure space (, )into the difference of two (nonnegative) measures given in the
Jordan Decomposition Theorem is called the Jordan decomposition of v
Lemma: Suppose that ν is a signed measure on a measurable
space(,)XA. If A ∈ and 0 < ν(A)< ∞, then there exists a positive subset
P ⊂ A such that ν(P)> 0.
Proof: First, we show that if A ∈ is a measurable set with |ν(A)| < ∞,
then |ν(B)| < ∞ for every measurable subset B ⊂ A. This is because ν takes
at most one infinite value, so there is no possibility of cancelling an
infinite signed measure to give a finite measure. In more detail, we may
suppose without loss of generality that ν : A → [−∞, ∞) does not take the munotes.in
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Space of Integrable Functions
115 value ∞. (Otherwise, consider −ν.) Then ν(B) ∞; and if B⊂A, then the
additivity of ν implies that
ν(B) = ν(A) − ν(A \ B)−∞
since ν(A) is finite and ν(A \ B∞.
Now suppose that 0 < ν( A) < ∞.
Let1= inf { ν(E) : E ∈ A and E ⊂ A}.
Then −∞ ≤ 1 ≤ 0, since ∅⊂ A. Choose A1⊂ A such that 1 ≤ ν(A1) ≤ 1/2
if 1 is finite, or µ( A1) ≤ −1 if 1 = −∞. Define a disjoint sequence of
subsets { Ai⊂A :i ∈N} inductively by setting 1= inf { ν(E) : E ∈ and E ⊂ A 111\i
jA
}
and choosing
⊂ A111\i
jA
such that 1 ≤ ν(Ai) ≤ 1/21
if −∞ <1≤ 0, or ν(Ai) ≤ −1 if 1= −∞.
Let 1,iiBA
P = A \ B.
Then, since the
are disjoint, we have
1() ( )iivB vA
As proved above, ()vB is finite, so this negative sum must converge. It
follows that ()ivA ≤ −1 for only finitely many i, and therefore 1 is
infinite for at most finitely many i. For the remaining i, we have ()1 / 2 0iivA .
So iconverges and therefore 1→ 0 as i → ∞.
If E⊂P, then by construction ν(E) ≥ 1 for every sufficiently large i ∈ N.
Hence, taking the limit as i → ∞ , we see that ν(E) ≥ 0, which implies that
P is positive. The proof also shows that,
since ()vB≤ 0, we have ν(P) = ν(A) − ν(B) ≥ ν(A) > 0 ,
which proves that P has strictly positive signed measure.
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Analysis II
116 Hahn decomposition theorem:
Statement: If ν is a signed measure on a measurable space (,)XA, then
there is a positive set P and a negative set N for ν such that P ∪ N = X and
P ∩ N = ∅. These sets are unique up to ν-null sets.
Proof: Suppose, without loss of generality, that ν(A) < ∞ for every A ∈A.
(Otherwise, consider −ν.)
Let m = sup{ ν(A): A∈A such that A is positive for ν},
and choose a sequence {iA : i∈N} of positive sets such that ν(iA) → m as
i → ∞. Then, since the union of positive sets is positive,
1iiPA
is a positive set.
Moreover, by the monotonicity of ν, we have ν(P) = m. Since ν(P) ∞, it
follows that
m ≥ 0 is finite.
Let N = X \P. Then we claim that N is negative for ν.
If not, there i s a subset A′⊂N such that ν(A′ ) > 0, so by above Lemma,
there is a positive set P ′⊂A′ with ν(P ′ ) > 0. But then P ∪P ′ is a positive
set with ν(P∪P ′ ) > m, which contradicts the definition of m.
Finally, if P ′, N′ is another such pair of positive and negative sets, then P \
P ′⊂P ∩ N ′ ,
soP \ P ′ is both positive and negative for ν and therefore null, and
similarly for P ′ \ P.
Thus, the decomposition is unique up to ν-null sets.
Remark: It is generally the case that the Hahn decomposition is not
unique.
In fact, let X = [0, 1] and let = P(X) . If 12 is the point mass at 1/2, then
if P = {1/2 } and N = [0, 1] \ {1/2}, then { P, N} is a Hahn decomposition
of [0, 1] with respect to µ. However, 1P = [0, 1/2 ] and1N = ( 1/2 , 1] is
also a Hahn decomposition.
In fact, if { P, N } is a Hahn decomposition of X with respect to µ and if
M∈ is null, then { P∪M, N \M} is a Hahn decomposition of X with
respect to µ.
Furthermore, if {11,PN} and {22,PN} are Hahn decompositions of X with
respect to µ, then
M = P1∆P2 = (P1∩ N2) ∪ (N1∩ P2) = N1∆N2 is a null set. munotes.in
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Space of Integrable Functions
117 Furthermore, since E ∩ P1 \ P2 ⊆ P1 ∆ P2 and E ∩ P2\ P1 ⊆ P1 ∆ P2, it
follows that
µ(E ∩P1) = µ(E ∩ P1 ∩ P2) = µ( E ∩ P2) for each E∈ .
Similarly, µ(E ∩ N1) = µ(E ∩ N1 ∩ N2) = µ(E ∩ N2) for each E∈ .
What is true however, as we shall see, is that every Hahn decomposition
induced a decompositi on of µ into the difference of two positive measures
and that any two Hahn decompositions induce the same decomposition
of µ.
Theorem (Jordan decomposition): If ν is a signed measure on a
measurable space (X,A) , then there exist unique measures ,:vv A
[0, ∞], one of which is finite, such that vv vand vv.
Proof: Let X = P ∪ N where P, N are positive, negative sets for ν.
Then v(A) = ν(A∩P), v(A) = −ν(A ∩ N) is the required decomposition.
The values of v are independent of the choice of P, N up to a ν-null set,
so thedecomposition is unique.
We call v and v the positive and negative parts of ν, respectively.
The total variation | ν| of ν is the measure |ν| = vv. We say that the
signed measure ν is σ-finite if | ν| is σ-finite.
6.4 COMPLEX VALUED LEBESGUE MEASURABLE
FUNCTIONS ON d
Complex valued Lebesgue measurable functions on d are an essential
concept in measure theory and functional analysis.
Complex valued measurable functions: A fun ction :df is
complex -valued measurable if both its real and imaginary parts, Re( )fand ()lm f are measurable functions.
We then say that f is Lebesgue integrable if the function 22() () ()fx u x v x
(which is non -negative) is Lebesgue integrable in the sense defined
previously.
It is clear that () ()ux f x and () ()vx f x
Also, if ,0ab it has 11 122 2()ab a b so that () () ()fx u x v x . munotes.in
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118 We can deduce from these straightforward inequalities that a complex -
valued function is only integrable if its real and imaginary parts are
integrable . Then, the Lebesgue integral of f is defined by () () ()f x dx u x dx i v x dx
Finally, if E is a measurable subset of d,and f is a complex -valued
measurable function on E, we say that f is Lebesgue integrable on E if XEf
is integrable on d, and we define EX Eff .
The collection of all complex -valued integrable functions on a measurable
subset E⊂ dforms a vector space over . Indeed, if f and g are
integrable, then so is f + g , since the triangle inequality gives |( f + g)(x) | ≤
|f(x)| + |g(x)|, and monotonicity of the integral Then, at that poin t, the
Lebesgue necessary of fis characterized by
EE Efg d f d g d
Also, it is clear that if a ∈ and if f is integrable, then so is af. Finally, the
integral continues to be linear over.
Appro ximation of Lebesgue integrable functions by continuous
functions:
The main concept of the approximation is to demonstrate that a sequence
of continuous functions may roughly approach any Lebesgue integrable
function, both in terms of pointwise and integra l convergence. Because
continuous functions are easier to deal with and have a lot of attractive
qualities that make them conducive to analysis, this finding is very
relevant.
Formally, let f be a Lebesgue integrable function on a measurable set
E⊆n. Then, the approximation theorem states that for any ε > 0, there
exists a continuous function g on n such that the Lebesgue measure of
the set where f and g differ (i.e., {x ∈ E: |f(x) - g(x)| > ε}) is arbitrarily
small. In mathematical terms, we can find a sequence ( gn) of continuous
functions converging to f almost everywhere on E: lim ( ) ( ) 0nnfx g xd
where μ is the Lebesgue measure.
6.5 THE SPACE
(µ) OF INTEGRABLE FUNCTIONS
An important observation about the algebraic properties of integrable
functions is their formation of a vector space. The fact that this vector
space is complete within the appropriate norm is an essential analytic fact.
Definition: If ( X, , µ) is a measure space, then the space 1()LX consists
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119 1Lff d
where we identify functions that are equal a.e. A sequence of functions
1{( ) }nfL X
converges in 1L , or in mean, to 1()fL Xif
1 0nnLf f f f d as n
We also denote the space of integrable complex -valued functions f:X→
by L1(X). For definit eness, we consider real -valued functions unless stated
otherwise; in most cases, the results generalize in an obvious way to
complex -valued functions.
Let us consider the particular case of L1(d). As an application of the
Borel re gularity of Lebesgue measure, we prove that integrable functions
on d may be approximated by continuous functions with compact
support. This result means that L1(d) is a concrete realization of the
comple tion of Cc(d)with respect to the L1(d)-norm, where
Cc(d)denotes the space of continuous functions f :d →with
compact support. The support of f is defined by {: ( ) 0 }dsupp f x f x
Thus, f has compact support if and only if it vanishes outside a bounded
set.
Properties of L1 integrable functions:
i) Measurability: L1 integrable functions are required to be
measurable. This mean s that the set { x: |f(x)| >M} must be
measurable for all M> 0.
ii) Linearity : If f(x) and g(x) are both L1 integrable functions, then any
linear combination of these functions, such as af(x) + bg(x) , where a
and b are constants, is also L1 integrable.
iii) Triangle Inequality: The integral of the absolute value of the sum
of two L1 integrable functions is less than or equal to the sum of their
individual integrals:
∫|f(x) + g(x)| dx ≤ ∫|f(x)| dx + ∫|g(x)| dx
iv) Dominating Function: If |f(x)| ≤ g(x) almost everywhere on the
measurable set E, and g(x) is L1 integrable, then f(x) is also L1
integrable.
v) Convergence in Measure: If a sequence of measurable functions
{nf(x)} converges to f(x) in measure, and each function nf(x) is L1
integrable, then the limit function f(x) is also L1 integrable:
lim ( ) ( ) 0 ( )nnf x f X dx f x dx
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120 vi) Product of
Functions: If f(x) is an L1 integrable function and
g(x) is a bounded measurable function, then the product f(x)g(x) is
also L1 integrable.
vii) Change of Measure : If f(x) is an L1 integrable function over a
measurable set E, and there exists a measurable function h(x) such
that h(x) ≥ 0 and ∫h(x) dx < ∞ , then the integral of f(x) with respect to
the measure defined by h(x) is also fin ite.
viii) Approximation: L1 integrable functions can be approximated by
simple functions (finite linear combinations of indicator functions).
ix) Density in L1: Under appropriate conditions, continuous functions
with compact support are dense in L1 space.
Proposit ion: The space L1 () is linear (over) and if f ∈ L1 () the
real and imaginary parts, Re f, Im f are Lebesgue integrable as are there
positive parts and as is also the absolute value, f.
Proof: We first consider the real part of a function f∈ L1 (). Suppose nf∈cc () is an approximating sequence. Then consider ng = Renf.
This is absolutely summable,
since nngf and ( ) Re ( )nn n
nnfx f g f x
Since the left identity holds a.e., so does the right and hence Re f∈L1 ().
The same argument with the imaginary parts shows that Im f∈ L1 ().
This also shows that a real element has a real approximating sequence and
taking positive parts that a positive function has a positive a pproximating
sequence.
Indeed, if f, g ∈ L1 () have approximating series nf and ng then nh = nf + ng is absolutely summable, nnnnnnhf g
And ( ) ( ), ( ) ( ) ( ) ( )nn n
nn nfx f x g g x hx f x g x .
The first two conditions hold outside (probably different) sets of measure
zero, E and F, so the conclusion holds outside E ∪ F which is of measure
zero. Thus f + g ∈L1 (). The case of cf for c∈ is more obvious.
The proof that | f| ∈L1 () if f∈L1 () is similar but perhaps a little
trickier. Again, let {nf} be a sequence as in the definition showing that
f∈L1 (). To make a series for | f| we can try the ‘obvious’ thing. Namely
we know that munotes.in
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121 1() () ()n
jj
jjf x f x if f x
so certainly, it f ollows that
1() () ()jj
jjf x f x if f x
So set
1
11 1 1() () ; () () ()kk
kj j j j gx fx g x fx fx x
.
Then, for sure,
11() () () ()NN
kj j
kj jg x f x f x if f x
So equality holds off a set of measure zero and we only need to check that
{gj} is an absolutely summable series.
The triang le inequality in the ‘reverse’ form || v| − |w|| ≤ | v −w| shows that,
for k > 1,
1
11() () () ()kk
kj j k
jjgx fx fx fx
Thus ,
kk
kkgf
so the gk’s do indeed form an absolutely summable series and holds almost
everywhere, so | f| ∈L1 ().
Riesz -Fischer theorem:
The Riesz -Fischer theorem, also known as the Riesz representation
theorem, is a fundamental result in functional analysis that establishes the
connection between certain types of continuous linear functional s and
elements of a Hilbert space. This theorem is named after the Hungarian
mathematician Frigyes Riesz and the German mathematician Ernst
Fischer, who both made significant contributions to its development.
Theorem: The vector space L1 is complete in i ts metric.
Proof: Suppose {nf} is a Cauchy sequence in the norm, so that nmff
→ 0 as n, m → ∞.
The plan of the proof is to extract a subsequence of {nf} that converges to
f, both pointwise almost everywhere and in the norm.
Under ideal circumstances we would have that the sequence {nf}
converges almost everywhere to a limit f, and we would then prove that
the sequence converges to f also in the norm. munotes.in
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122 Unfortunately, almost everywhere convergence does not hold for general
Cauchy sequences. The main point, however, is that if the convergence in
the norm is rapid enough, then almost everywhere convergence is a
consequence, and this can be achieved by dealing with an appropriate
subsequence of the original sequence.
Indeed, consider a subsequence 1knKf of {nf} with the following
property:
12, 1kk
n nkf f for all k
The existence of such a subsequence is guaranteed by the fact that
1kn nkff
whenever n, m ≥ N(
), so that it suffices to take nk = N(2-k). We now consider the series whose convergence will be seen
below,
1 1
1() () ()k n n nk
kfx f f x f x
And
1 1
1() () ()k n n nk
kgx f f x f x
And note that
11 1
11() () 2
kk
n n nk n
kkff x f x f
.
So the monotone convergence theorem implies that g is integrable, and
since
≤ g, hence so is f. In particular, the series defining f converg es
almost everywhere, and since the partial sums of this series are precisely
the
(by construction of the telescopic series), we find that () () . .nkfx f x a e x
To prove that nkf → f in L1as well, we simply observe that | f −nkf | ≤ g for
all k, and apply the dominated convergence theorem to get
10nk Lff as .0k
Finally, the last step of the proof consists in recalling that {nf} is Cauchy.
Given , there exists N such that for all n, m>N we have /2nmff .
If kn is chosen so that kn>N, and /2nkff , then the triangle
inequality implies
n n nk nkff ff f f
whenever n>N. Thus {nf} has the limit f inL1.
Hence proved.
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123 6.6 LETS SUM UP
In this chapter we have learn:
Signed measures are particularly useful when dealing with functions
and distributions that have both positive and negative components or
values. They find applications in various areas, such as the study of
integration with respect to signed measures, Lebesgue -Stieltjes
integration, complex analysis, and the study of distributions in
functional analysis.
The Hahn Decomposition Theorem addresses the decomposition of a
measurable space into two disjoint sets, where the positive and
negative components of a signed measure ar e concentrated separately.
The space L1 (µ) mathematicians and scientists gain access to a
versatile class of functions that are essential for understanding the
behaviour and properties of integrable functions with respect to a
given measure µ.
6.7 UNIT E ND EXERCISES
1) A function f : −→ is an element of L1 () if and only if it is
measurable and there exists F∈L1 () such that | f| ≤ F almost
everywhere .
2) Show that there are f∈L1 (d) and a sequence {nf} with fn
∈1()dL such that 1nLff→ 0, but nf(x) → f(x) for no x.
3) Show that f ∗ g is integrable whenever f and g are integrable, and that
11 1() () ()'dd dLL Lfg f g
with equality if f and g are non -negative .
4) State and prove Riesz -Fischer theorem .
5) Prove that if f and g are integrable functions on X, and | g(x)| ≤ |f(x)| for
all x in X, then f and g are both in L1(µ) if and only if f is in L1(µ).
6) Let ν be a finite signed measure and let µ be a measure on ( X, ).
Then ν µ if and only if for any ε > 0 there exists δ > 0 such that
|ν(E)| < ε when µ(E) < δ.
7) Show that if D is a Lebesgue measurable subset of d, then {: }d DLE L E D .
8) Prove that “ The space Cc(n) is dense in 1()nL. Explicitly, if
f∈1()dL, then for any > 0 there exists a function g∈Cc(n) such
that 1"L fg .
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124 9) If (X, , µ) is a measure space and ,vv: → [0, ∞] are measures,
one of which is finite, then show that ν = vv: is a signed measure.
10) Suppose that ν is a signed measure on a measurable space (X, ). If A
∈ and 0 < ν(A)< ∞, then there exists a positive subset P ⊂ A such that
ν(P)> 0.
6.8 LIST OF REFERENCE
E. M. Stein and R. Shakarchi, Real Analysis, Princeton University
Press, 2005.
S. Lang, Real and Functional Analysis, Springer -Verlag, 1993
G. B. Folland, Real Analysis, 2nd ed., Wi ley, New York, 1999.
F. Jones, Lebesgue Integration on Euclidean Space, Revised Ed.,
Jones and Bartlett, Sudberry, 200 6.
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