chapter 3 l p -space. preliminaries on measure and integration
TRANSCRIPT
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Chapter 3
Lp-space
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Preliminaries on measure and integration
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σ-algebra
AAA c \)2(
,)1(
11)3(
nnnn AA
Ω ≠ψ is a set
Σis a family of subsets of Ω with
Σis called σ-algebra of subsets of Ω
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measure space
0)()1(
)(
int)2(
11
1
additivityAA
disjoisA
nn
nn
nn
Ω ≠ψ is a set
μ:Σ→[0, ∞] satisfies
Σis aσ-algebra of subsets of Ω
(Ω , Σ, μ) is called a measure space
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measurable function p.1
Rwfwf )(;
f:Ω→R is measurable if
(Ω , Σ, μ) is a measure space
The family of measurable functions is a
real vector space.
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measurable function p.2
1nnf if
The family is closed under limit, i.e.
is a sequence of measurable functions ,
which converges pointwise to a
finite-valued function f , then f is
measurable.(see Exercise 1.1 and 1.3)
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Exercise 1.1
R
If f, g are measurable , then
(Ω , Σ, μ) is a measure space
f+g is also measurable.
Hint: for all
gfgfQ
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gfx
xgf
xgxf
xgandxfthen
Qsomeforgfx
thengfxIf
gfgfClaim
Q
Q
))((
)()(
)()(
,""
:
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.
,
lg
,,
)()(
)()(
)()(
)()(
,""
measurableisgfhence
gfgf
ClaimBy
gf
countableisQand
ofsubsetsofebraaisSince
QRgfthen
functionsmeasurablearegandfthatAssume
gfx
xgandxf
Qsomeforxfxg
xfxg
xgxf
thengfxIf
Q
Q
Q
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Exercise 1.3
1 1
1
m k knn mff
ffnn
limLet f1,f2,… be measurable and
(Ω , Σ, μ) is a measure space
and f(x) is finite for each
Hint: for all
Show that f is measurable
x
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1 1
1 1
1
1
1)(..1
),()(lim
1)(..1
)(
""
:
1:
m k knn
knn
n
nn
m k knn
mfw
mfw
knm
wftsk
wfwfSincem
wftsm
wf
fwanyFor
pf
mffClaim
RanyFor
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.
1
,1
,1
)(
1)(
,
1)(
1,1
,1
""
1 1
1 1
measurableisfHence
mffthen
Nnmm
f
nmeasurableisfSince
fw
wfm
wf
havewenlettingBy
knm
wf
kmsomeform
fw
thenm
fwIf
m k knn
n
n
n
knn
m k knn
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indicator function
Afor
(Ω , Σ, μ) is a measure space
χA is the indicator function of A
χA is measurable
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<W>
<W> denotes the smallest vector
subspace containing W in a vector space.
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Simple function p.1
AA :
are called simple functions
Elements of
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Simple function p.2
iA
k
iif
1 k ,,1
if the right hand side has a meaning
, where
A simple function can be expressed as
are different values and Ai = {f =αi} ,
we define then )(1
i
k
ii Adf
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Simple function p.3
fd
In particular
is meaningful if f is simple and
nonnegative, although it is possible that
df
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Integration for f 0,measurable≧
For f 0, measurable ,define ≧
gddf
fgsimpleg
0:sup
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f+ ,f-
0)(0
0)()()(
wfif
wfifwfwf
fff
If f is measurable, then
f+ and f- are measurable
0)(0
0)()()(
wfif
wfifwfwf
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Integration for measurable function p.1
For any measurable function f ,define
dfdfdf
if R.H.S has a meaning
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Integration for measurable function p.2
df
df
and
is finite if and only if both
df are finite
f is called integrable
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Integration for measurable function p.3
dfdfdf
is integrablef
f is integrable if and only if
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limsupAn , liminf An
k kn
nnn
AA1
suplim
k kn
nnn
AA1
inflim
Ω is a set and {An} is a sequence of
subsets of Ω. Define
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limAn
nn
nn
AA
inflimsuplim
nn
A
lim
then we say that the limit of the sequence
{An} exists and has the common set as the
limit which is denoted by
If
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An: monotone increasing
121 nn AAAA
n
nnn
AA1
lim
then
If
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An: monotone decreasing
121 nn AAAA
n
nnn
AA1
limthen
If
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Lemma 2.1
1nnA
)(lim1
nnn
n AA
be monotone increasing, then
If
(Ω , Σ, μ) is a measure space
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nn
k
n
kn
kkk
kk
k
k
kk
kk
n
kkn
kkk
AB
BBA
disjoisBSince
BAandBAthen
kAABandALet
limlim
int,
,2,1\
1
111
111
10
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Lemma 2.2
1nnA
)(lim1
nnn
n AA
be monotone decreasing, then
If
(Ω , Σ, μ) is a measure space
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)(lim
)(lim)(
)\(lim)(lim\
1.2
sin
\
1
11
1
111
1
1
nnn
n
nnn
n
nn
nnk
kn
n
n
nn
AA
AAAA
AABBAA
LemmaBy
gincreamonotoneisBthen
AABLet
NnFor
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Egoroff Theorem
A BAB ,
Let {fn} be a sequence of measurable function
and fn→f with finite limit on
(Ω , Σ, μ) is a measure space
then for any ε>0 , there is
such that μ(A\B)<ε and fn→f uniformly on B
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N
mEx
N
nn
n
nmmn
nCx
ECTake
Nmxfxf
andEAtsNthen
AElemmaBy
AEthen
nxfxfAxElet
pf
xfxfandCAAC
tsCandNegerForClaim
N
)()(sup
)\(..
)()(lim1.2
,2,1)()(;
:
)()(sup,)\(,
.int,0,0:
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.
1)()(sup
2
)\()\()\(
,
1)()(sup
2)\(
..
,0
1
11
1
BonuniformlyffHence
Nnm
xfxfAnd
EAEABA
thenEBtakeThen
Nnm
xfxfandEA
tsEwithAEandZNClaimBy
ZmFor
n
mnBx
mm
mm
mm
mm
mnEx
mm
mmm
m
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Monotony Convergence Theorem
ffnn
lim
dffd n
nlim
Let {fn} be a nondecreasing sequence of
nonnegative measurable functions
Suppose
(Ω , Σ, μ) is a measure space
is a finite valued,then
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Theorem(Beppo-Levi)
,sup dfnn
nasdffn
n0lim
Let {fn} be a increasing sequence of integrable
functions such that
fn f . Then f is integrable and ↗
(Ω , Σ, μ) is a measure space
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00)()(
)(
int
int
sup
)(lim)(
)(lim0
11
1
1
11
111
nasdffdff
dff
andegrableisf
egrableisff
dfdf
dffdff
TheoremConvergentMonotonyBy
ffffandff
n
n
nn
nn
nn
n
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Fatous Lemma
dfdf n
nn
ninfliminflim
Let {fn} be a sequence of extended real-value
d
measurable functions which is bounded from
below by an integrable function. Then
(Ω , Σ, μ) is a measure space
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dfdg
dgdfdf
TheoremeConvergencMonotoneBy
functionegrableanbybelowfromboundedis
andgnondecreaisgthenfgLet
nn
nn
nnnk
kn
nn
nknk
n
inflimlim
liminfliminflim
.int
sin,inf
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Remark
dfdf nn
nn
suplimsuplim
Let {fn} be a sequence of extended real-
valued
measurable functions and fn 0. Then ≦
(Ω , Σ, μ) is a measure space
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dfdf
dfdf
dfdf
LemmaFatouBy
f
nn
nn
nn
nn
nn
nn
n
suplimsuplim
suplimsuplim
)(inflim)(inflim
0
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Lebesque Dominated Convergence Theorem
gfn
dffd n
nlim
If fn ,n=1,2,…, and f are measurable functions
and fn →f a.e. Suppose that
a.e. with g being an integrable function.Then
(Ω , Σ, μ) is a measure space
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dffd
dfdfdf
LemmaFatouByfunctionegrableby
abovefromandbelowfromboundedisfSince
nn
nn
nn
nn
n
lim
inflimlimsuplim
.int
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Corollary
gfn
0lim dffn
n
If fn ,n=1,2,…, and f are measurable functions
and fn →f a.e. Suppose that
a.e. with g being an integrable function.Then
(Ω , Σ, μ) is a measure space
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0limlim
,
..0
dffdff
LDCTBy
fgffff
eaff
nn
nn
nn
n
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5
The space Lp(Ω,Σ,μ)
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For measurable function f, let
pp
pdff
1
pif 1
..:0inf eaMfMf
f is called the essential sup-norm of f.
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Exercise 5.1
Show that
..eaff
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..)(
\)(
,
\1
)(
0)()(,
\1
)(
0)(\)(
1..0
11
eafxfHence
Axfxf
havewenlettingBy
Axn
fxf
andAAthenAALet
Axn
fxf
AandAwhereAxMxf
andMn
ftsM
NnFor
nnn
n
n
nnnn
nn
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Conjugate exponents
1, qpIf are such that
111
qp
then they are called conjugate exponents
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Theorem 5.1(Hölder’s Inequality) p.1
1, qpIf are conjugate exponents,
qp
gfdfg then
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qp
qp
thenLet
qandperchangesimplyotherwise
thatassumemayWe
oreitherNow
forqp
haveweforp
Since
qpthenIfClaim
qp
p
p
p
qp
11
111
1
1
11
1
1
11
1
,
.int
,1
.11
11
,01)1(
,,0:
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qp
q
q
q
p
p
p
qp
q
q
q
p
p
p
qp
q
q
p
p
qp
gfdfg
dg
g
qd
f
f
pd
gf
gf
g
g
qf
f
pgf
gfhavewe
Claiming
gand
f
fthatNow
eagfhence
gfthatassumemayWe
111
11
..,
,,0
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Theorem 5.1(Hölder’s Inequality) p.2
kiLf ipi 1)(
More generally, if f1,f2,…,fk are functions s.t
with
11111
21
kpppp
then
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Theorem 5.1(Hölder’s Inequality) p.3
Pk Lffff 21
and
kpkpppffff
2121
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21
21
21
21
221
21
1
21
1
221
2
1
21
1
1
2121
2
2
21
21
2121
21
21
21
21
2121
21
21
2121
)(
,111
,,2
ppp
p
p
p
p
pppp
ppppp
p
ppp
pppp
p
ppp
ppp
ppp
p
ppp
p
ppp
pp
ffff
ff
ff
ff
InequalityHolderby
ff
ffff
andpp
ppp
thenppp
andLfLfkIf
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Theorem 5.2(Minkowske Inequality)
p1If f, g be measurable ,
whenever f+g is meaningful a.e. on Ω
pppgfgf
then
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)11
)1
1((
)111
(
,1
.1
1
1)1(
11
1
pp
qp
q
pp
gfgf
gfgf
gfgf
qp
gfdgf
gfdgf
dggfdfgf
dgfgfdgfgf
thenpcasetheconsidernowWe
porpwhenobviousisIt
ppp
ppq
pp
p
ppq
p
p
pp
qp
pp
qqp
pp
ppp
p
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is the family of all measurable function f with
pf
From Minkowski Inequjality, it is readily seen that
Lp(Ω, Σ, μ)
Lp(Ω, Σ, μ)
Lp(Ω, Σ, μ) is a vector space.
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Theorem
pLp is a normed vector space
with
p1for
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andLgf
gfgfgf
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pSuppose
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p
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ppp
pppp
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then
0:),,( p
p fLfN
if and only if f=0 a.e. on Ω
Let
Nf
![Page 63: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/63.jpg)
then Lp(Ω, Σ, μ) is a vector space which
consists of equivalent classes of Lp(Ω, Σ, μ)
Lp(Ω, Σ, μ)
Lp(Ω, Σ, μ) =Lp(Ω, Σ, μ)/N
w.r.t the equivalent relation ~definded by
f~g if and only if f=g a.e. on Ω
![Page 64: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/64.jpg)
is a Banach space.
p
Theorem 5.3
(Fisher)
Lp(Ω, Σ, μ) with norm
![Page 65: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/65.jpg)
negligibleisEei
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thatsuchNistherekegerpositiveGiven
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nasff
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andLf
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n
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p
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),,(
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kasffknowweagainLDCTby
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.),,(
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p
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k
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Theorem
nasffandLfLfLetpn
ppn 0,,
kn
fthen there is a subsequence
such that
pn
n
Lhwithkeaxhxf
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.)()()2(
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![Page 72: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/72.jpg)
eaff
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NleaxhxfandLh
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![Page 74: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/74.jpg)
f is called an essential bounded function if
f
Exercise 5.4
L∞ (Ω, Σ, μ) is a Banach space.
![Page 75: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/75.jpg)
fthen
eaxfxfthen
RinsequenceCauchyabexfthen
nmnwhenevereaffthen
nmnwheneverff
tsNnthen
LinsequenceCauchyabefLet
nn
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)(
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![Page 76: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/76.jpg)
Outer Measure
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Outer Measure
0)()1(
R2:
BAifBA )()()2(
Ω ≠ψ: a set
μis called outer measure on Ω if
)()()()3(11
additivitysubAAn
nn
n
![Page 78: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/78.jpg)
μ-Measurable p.1
AACandAB c \
)()()( BABAB C B
A subset A of Ω is called μ-measurable if
for any
i.e. for any
)()()( CBCB
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Exercise 1.1
otherwise
setfiniteisAifAofycardinalitA)(
Let μ:2Ω→[0,+∞] be defined by
(μis called the counting measure of Ω and
every subset of Ω is μ-measurable.
Show that μis an outer measure
![Page 80: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/80.jpg)
)()(
,inf:)(
)()(
,:)(
)2(
0)()1(
BA
thensetiniteanisBIfiiCase
BA
BofycardinalittheAofycardinalitthe
andsetfiniteaisA
thensetfiniteaisBIfiCase
BALet
thatobviousisIt
![Page 81: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/81.jpg)
reoutermeasuanisBy
AA
NnsomeforsetiniteanisA
thensetiniteanisAIfiiCase
AAei
Aofycardinalitthe
Aofycardinalittheand
nsetfiniteaisA
thensetfiniteaisAIfiCase
thenofsequenceabeALet
nn
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n
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n
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)(
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![Page 82: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/82.jpg)
.
)()()(
inf
,inf)(
)()()(
,)(
measurableisAHence
BABAB
andsetsiniteareBAorBA
thensetiniteanisBIfiiCase
BABAB
andsetsfiniteareBAandBA
thensetfiniteaisBIfiCase
Banyfor
ALet
measurableare
ofsubseteverythatshowtoNow
c
c
c
c
![Page 83: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/83.jpg)
Exercise 1.2 p.1
SAthenSAIfiii
SBthenABandSAIfii
Si
nnnn
11 ,)(
,)(
)(
Let S be a subset of 2Ω having the following
properties:
![Page 84: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/84.jpg)
Exercise 1.2 p.2
Define μ : 2Ω→[ 0, +∞] by
What are theμ-measurable subsets of Ω
Show that μis an outer measure
otherwise
SAifA
0)(
![Page 85: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/85.jpg)
Exercise 1.2 p.2
Define υ : 2Ω→[ 0, +∞] by
What are theυ-measurable subsets of Ω
Show that υis an outer measure
otherwise
SAifA
1
0)(
![Page 86: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/86.jpg)
11
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andSAthenSBIfCase
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measureouteranisthatshowTo
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SAorSAeither
ofsubsetsmeasurableaisAClaim
measureouteranisiiiiBy
AAthen
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thenSAIfCase
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)()()(
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ofBsubsetanyforthen
ofsubsetsmeasurableaisAIfpf
c
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11
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SAandSAthator
SAandSAthateither
ofsubsetsmeasurableaisAClaim
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AAthen
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thenSAIfCase
c
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1)()()(
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BABABthen
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BsubsetanyFor
SAandSAthatassumeMay
generalityofloseWithout
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BAandBAthator
BAandBAthateitherthen
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BABAB
ofBsubsetanyforthen
ofsubsetsmeasurableaisAIfpf
c
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![Page 92: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/92.jpg)
Exercise 1.3
A
Suppose μis an outer measure onΩ and
then the restriction of μto A
denoted byμ A(B)=μ(A∩B) for ∣ B
Show that μ A is an outer measure on∣Ω and every μ-measurable set is also
μ A –measurable.∣
![Page 93: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/93.jpg)
.
)()(
)()(
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11
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measureouteranisAHence
BABA
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BanyForiii
DADACACA
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measureouterisAthatshowTo
nn
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nn
![Page 94: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/94.jpg)
.
)()(
)()()()(
)()()(
.
)2(
measureAalsoisBHence
CBACBA
CABCABCACA
CBCBC
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setmeasurablebeBLet
measureAalsois
setmeasurableeverythatshowTo
c
c
c
![Page 95: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/95.jpg)
Properties of Measurable sets p.1
Suppose μmeasures Ω.
(1) If A is μ-measurable, then so is Ω\A=Ac
(2) If A1, A2 are μ-measurable, then so is
A1 A∪ 2
![Page 96: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/96.jpg)
))(())((
)()()(
,)()()(
,,)2(
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2121
21211
11
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BAABAA
ABAABABA
andBABAB
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thenmeasurableareAAIf
measurableisAHence
BABA
BABAB
Banyfor
thenmeasurableisAIf
c
ccc
c
c
ccc
c
![Page 97: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/97.jpg)
Properties of Measurable sets p.2
Remark :
By induction the union of finitely
many μ-measurable sets is μ-measurable.
This fact together with (1) implies that
the intersection of finitely many
μ-measurable set is μ-measurable.
![Page 98: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/98.jpg)
Properties of Measurable sets p.3
B
1jjA is a disjointed sequence of(3) If
then
μ-measurable sets in Ω and
11 jj
jj ABAB
![Page 99: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/99.jpg)
11
111
1
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jj
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n
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ABAB
hencenallfor
ABABABthen
AB
ABAB
ABABAB
thenegerpositiveabenLet
![Page 100: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/100.jpg)
Properties of Measurable sets p.3
1nnA
1jjA is a disjointed sequence of(4) If
is μ-measurable .
μ-measurable sets in Ω, then
![Page 101: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/101.jpg)
1
111
11
)(
\
\
,
njj
njj
n
jj
n
jj
jj
jj
ABB
ABABAB
ABAB
thenBLet
![Page 102: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/102.jpg)
.
)(\
)(
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1
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measurableisAHence
BABABthen
ABABB
thenABIfCase
BABAB
haveweinequalityabovethein
nlettingbyABIfCase
jj
jj
jj
jj
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njj
jj
jj
njj
![Page 103: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/103.jpg)
Properties of Measurable sets p.4
11 jj
jj AandA
1jjA is a sequence of(5) If
μ-measurable sets in Ω, then so are
![Page 104: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/104.jpg)
.)1(
,
.)4(
,
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measurableisAby
AASince
measurableisAby
AAASince
jj
c
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j
cj
iij
jj
![Page 105: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/105.jpg)
Properties of Measurable sets p.5
sets in Ω, then (Ω, Σ, μ) is a measure
(6) Let Σ be the family of all μ-measurable
space.
![Page 106: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/106.jpg)
Exercise 1.4 p.1
nnn
n AA
lim1
121 nn AAAA(i) If
is an increasing μ-measurable sets in
Ω, then
![Page 107: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/107.jpg)
mm
m
nnn
mnn
m
nm
n
m
nmn
nnn
nn
nn
nn
nn
nnn
A
AAAA
BBBA
thenBA
andmeasurableofsequencedisjoaisB
thenAABandALet
lim
limlim
lim
,
int
,\
111
1
1111
11
1
10
![Page 108: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/108.jpg)
Exercise 1.4 p.2
nnn
n AA
lim1
121 nn AAAA(ii) If
is a decreasing μ-measurable sets in
Ω with μ(A1)<+∞, then
![Page 109: Chapter 3 L p -space. Preliminaries on measure and integration](https://reader035.vdocuments.site/reader035/viewer/2022081501/56649ec65503460f94bd2345/html5/thumbnails/109.jpg)
nnn
nn
nnn
nn
nnn
n
nn
nnn
n
nnn
n
nn
nn
AA
AAAA
AAAA
AAAAAA
BBand
measurableofsequencegincreaanisB
thennforAABLet
limlim
limlim
lim\
lim\lim\
lim
sin
,,2,1\
1
11
1
11
1
111
1
1
1
1
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Regular
BA
B
A measure μ on Ωis called regular if for each
there is a μ-measurable set
such that μ(A)=μ(B)
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Exercise 1.5
nnn
n AA
lim1
121 nn AAAAIf
is a sequence of sets in Ω and μis a
regular measure on Ω, then
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nnj
n
nnj
n
n
n
jj
jn
nn
nn
nnnn
nnj
nj
nj
j
nnj
n
nn
nnnn
nn
nn
AAhence
AAthen
nallforAAABut
A
BBBB
CCBA
CC
iExercisebytheninsetsmeasurable
ofsequencegincreaaisC
thenBBBCLet
BAthatsuch
ABsetmeasurableaisthere
NnanyforregularisSince
lim
lim
lim
lim)\(lim
lim
lim
)(4.1,
sin
),\(
)()(
,
1
1
11
11
111
1
1
11
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Theorem
The family Σμ of μ-measurable subsets
of Ω is σ- algebra and μ=μ ︳ Σμ is
σ- additive . i.e.
(Ω, Σμ, μ) is a measure space.
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Premeasure
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premeasure
If Ω is a nonempty set,
G a class of subsets of Ω containing ψ, and
τ: G→[0,+∞] satisfy τ(ψ)=0.
τis called a premeasure.
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Outer Measure constructed from τ
1
1
1
inf)(i
i
AC
GCCA
ii
ii
For a premeasure τ,
Define μ:2Ω→[0,+∞] by
Then μ measures Ω and is called the
outer measure constructed fromτby
Method I.
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Example 2.1 The Lebesgue measure on Rn
Let G be the class of all oriented rectangles
in Rn with ψ adjoined and let
τ(I)=the volumn of I if I is an oriented
rectangle
τ(ψ)=0
the measure on Rn constructed fromτ by
Method I is called the Lebesgue measure
on Rn.
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Exercise 2.1
GI
For ε>0, Let Gε be the class of all open oriented rectangles in Rn with diameter <εand
τε (I)=volume of I for
Show that the measure on Rn constructed
from τε by Method I is the Lebesgue
measure.
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Exercise 2.2 p.1
Let μ be the Lebesgue measure on Rn
(i) Show that μ(I)=volume I if I is an open
oriented rectangle.
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)(
)(,
)(
)()(
.tan
1
11
IIHence
IIIIBut
IIthen
II
IIwithIsequenceanyforthen
glerecorientedanbeILet
nn
nnnn
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Exercise 2.2 p.2
(ii)Show that every open oriented rectangle
is μ-measurable and hence so is every
open set in Rn
( μ-measurable set is called Lebesgue
measurable set in this case.)
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)(inf)(inf
)(inf)(inf
)()(inf)(inf)(
.tan
cn
IBIGI
n
IBIGI
cn
BIGI
n
BIGI
cnn
BIGI
n
BIGI
n
n
III
IIII
IIIIIB
RofBsubsetanyFor
RinglerecorientedanbeILet
c
n
n
n
n
n
n
n
n
n
n
n
n
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Exercise 2.2 p.3
(iii) If A and B are subsets of Rn and
dist(A,B)>0, then
μ(A B)=μ(A)+μ(B)∪
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Metric spaces
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Metric Space
Mzyxzyyxzxii
yxyx
andMyxxyyxi
,,),(),(),()(
0),(
,0),(),()(
Let M be a nonempty set and
ρ:MXM→[0, ∞) satisfies
ρis called a metric on M
(M,ρ)is called a metric space.
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Example 1 for Metric Space
nn
n
ii
n
Raaaif
aawhere
Ryxyxyx
,,
,),(
1
21
1
2
Let M=Rn and let
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Example 2 for Metric Space
nii
niRyxyxyx
,max),(
1
Let M=Rn and let
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Example 3 for Metric Space
],[,)()(max),( baCgftgtfgfbta
Let M=C[a,b] (-∞<a<b<∞) and let
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Example 4 for Metric Space
)(,)()(max),( KCgftgtfgfKx
Let M=C(K), where K is a compace set in Rn
and let
Unless statement otherwise, C(K) will denote
the metric space with the metric so defined.
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Example 5 for Metric Space
pifgf
pifdgfgf
LgfFor
pp
p
1),(
),,(,1
Let M= Lp (Ω, Σ, μ)and let
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Converge
Mx 0
nxNn 0
Let (M, ρ) be a metric space.
A sequence
0nn
is said to converge to
if for any ε>0, there is
such that ρ(xn ,x0)<ε whenever
Since x0 is uniquely determined, x0 is denoted by limn→∞ xn
If limn→∞ xn exists, then we say that{ xn } converges in M.
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Example 3.1 for converge
)(xfn
1,0Cfn converges if and only if
converges uniformly for 1,0x
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Cauchy sequence
Nn 0
Mxn A sequence
is called a Cauchy sequence if for any
ε>0 there is
0,),( nnmwheneverxx nm
such that
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Exercise 3.1
nx
Mxn Show that if converges, then
is a Cauchy sequence.
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22
),(),(),(
,2
),(
0,lim
00
0
00
0
0
mnmn
n
nn
xxxxxx
nnmanyforthen
nnwheneverxx
thatsuchNnisthere
anyforthenxxIf
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Complete Metric space
A metric space M is called complete
if every Cauchy sequence in M converges
in M
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Examples for Complete
)(KC
nRK (1) Let be compact, then
is complete.
(2) Lp(Ω, Σ, μ) is complete.
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Normed vector space p.1
Ex
x
Let K=R or C and let E be a vector space
over K. Suppose that for each
there is a nonnegative number
associated with it so that
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Normed vector space p.2
Eyxyxyxiii
ExKxxii
xxi
,)(
;,)(
00)(
Then E is called a normed vector space
(n.v.s) with norm
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Normed vector space p.3
Eyxyxyx ,),(
Then ρ is a metric on E and is called
the metric associated with norm
Let E be a n.v.s and
Unless stated otherwise, for a n.v.s., we
always consider this metric.
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Banach space
Both C(K) with K a compact subset of Rn
and Lp(Ω, Σ, μ) are Banach spaces.
A normed vector space is called a
Banach space if it is a complete metric space
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Continuous mapping
10 Mx
),())(),(( 0102 xxwheneverxTxT
Let M1 and M2 be metric spaces with
metrics
ρ1and ρ2 respectively.
A mapping T: M1→M2 is continuous at if for any ε >0 , there is δ>0 such that
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Open and Closed set in a metric space
Gx
),( yxwheneverGy
A set G in a metric space is called an open
set if there is δ>0 such that
The complete of an open set is called
a closed set.
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Exercise 3.2 p.1
10 Mx
2121
2022
)(:)(
,)(
GxTMxGT
settheGxTwithMG
Let M1 and M2 be metric spaces with metrics
ρ1and ρ2 respectively and let T: M1→M2
(1) Show that T is continuous at
if and only if for any open set
contains an open subset which contains x0 .
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.
),())(),((
),()),(()(
),())),(((
))),(((),(
..0
))),(((
,)),(()(
)),((,0""
)())),(((),(
),()),(()(
..0,
)),((..0
,)(
""
0
0102
010
0101
01
10
10
01
1
00
20
21
01
0
00
0
20
2022
xatcontinuousisTHence
xxwheneverxTxT
xxwheneverxTBxT
xxwheneverxTBTx
xTBTGxxB
tsthenGxwith
xTBTGsetopenanisthere
thenxTBxTwith
MinsetopenisxTBFor
GTxTBTxB
xBxwheneverxTBxT
tsxatcontinuousisTSince
GxTBts
thenGxTwithMinsetopenanbeGLet
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Exercise 3.2 p.2
22 MG
)( 21 GT
(2) Show that T is continuous on M
if and only if for any open set
is an open subset of M1 .
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.
),())(),((
),(),()(
)),((),(..0
)),((
),(
0,""
)(
)(),(
),()()),((
),()),(()(
..0,
)),((..0
)(),(
""
1
0
0102
00
01
0
101
0
20
10
121
21
0
021
01
00
0
20
2021
0
22
MoncontinuousisTTherefore
xatcontinuousisTHence
xxwheneverxTxT
xBxxTBxT
xTBTxBts
MinopenisxTBTx
MofsubsetopenanisxTB
anyforthenMxLet
MofsubsetopenanisGTHence
GTxB
xBxGTxTBTx
xBxxTBxT
tsxatcontinuousisTSince
GxTBts
GxTthenGTxLet
MinsetopenanbeGLet
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Exercise 3.3
.
,)(
;,)(
setindexanyisIwhere
OAthenOAIfii
OBAOBAi
IiiIii
Let O be the family of all open
subsets of a metric
space. Show that
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OBAHence
openisBA
BAxB
BxBandAxB
thenLet
BxBandAxB
ts
BxandAx
BAxanyFori
),(
),(),(
,,min
),(),(
..0,
)(
21
21
21
BABA
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OAHence
openisA
AxB
AxBtsthen
IisomeforAx
thenAxLetii
Iii
Iii
Iii
i
i
Iii
),(
),(..0
,)(
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4.Carathéodory measure
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Carathédory Measure
)()()( BABA
0),(inf),(
yxBAdistByAx
If Ω is a metric space, then a measureμ is c
alled Carathédory measure if
whenever
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Example 4.1
The Lebesgue measure on Rn is a Carathéd
ory measure.
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Lemma 4.1
121 nn AAAALet
nnn
n
cnn
AA
ThenAA
sup
.0),(
1
1
be an increasing sequence of subsets of Ω
and for each n
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thenDDdisthavewe
nmandnanyforassumptionBy
AADAADADLet
Athatassumemaywe
AAthatshowTo
AAObviously
mn
nnn
nn
nnn
n
nnn
n
,0),(
2,
,\,,\,
sup
,sup
sup,
112211
1
1
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nnj
j
jnj
n
njjn
njjn
njjn
jj
ii
ii
nn
k
k
iik
AA
havewenlettingby
DA
DA
DAAAA
NowDSimilarly
DThenkeachfor
AA
DDDD
sup
,
.
.
sup
1
1
1
111
12
112
12
1121231
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Theorem 4.1
If μ is Carathédory measure onΩ, then
every closed subset of Ωis μ- measurable.
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BABABA
havewenlettingbythusneachfor
BABABAand
BBB
BtoappliedLemmabyhence
nnBBBdistNow
BBBB
ObviouslyBB
haveweclosedisFceThen
nFxdistBxB
letNneachFor
FBandFA
letandsetclosedabeFLet
nn
nn
nnn
n
nn
nn
nn
n
c
)()(sup)(
,,;
)()(
sup
),,(1.4
0)1(
1)\,(
,.
,sin,
1),(;
,
,
1
1
121
1
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Borel sets
B(Ω) is the smallest σ-algebra of subsets
Elements of B(Ω) are called Borel sets of
Ω.
of Ω that contains all closed subsets of Ω
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Corollary 4.1
If μ is Carathédory measure onΩ, then
all Borel subsets of Ωareμ- measurable.
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Lebesgue Measure p.1
RA
n
n
AII
I
n
n
inf
Ω =R , I: open finite interval of R
Define L(A)
then L is a Carathédory measure.
L is called the Lebesgue measure.
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Lebesgue Measure p.2
RA
nIII 21
(R, ΣL ,L)
Similar construction on Rn with
I replaced by n-dimensional intervals
Ln is a Carathédory measure.
Ln is called the Lebesgue measure on Rn .
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Regularity of Measure
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Regular measure
BA
B
A measure μ on Ωis called regular if for each
there is a μ-measurable set
such that μ(A)=μ(B)
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Borel regular measure
BAB
A measure μ on Ωis called Borel
if every Borel set is μ-measurable.
It is called Borel regular if it is Borel an
d for every
such that μ(A)=μ(B)
there is a Borel set
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Radon measure
A measure μ on Ωis called Radon
measure if it is Borel regular and
μ(K)<∞ for each compact set K.
We already known that Carathéodory m
easure is Borel .
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Theorem 6.1
ALet μ be a Borel regular on a metric
space Ω and suppose
is μ-measurable andμ(A)<∞
Then μ ︱ A is a Radon measure.
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CDthen
CCAEAEADD
AEADSince
DACAC
andsetBorelaisDthenAEDLet
CAEthatsuch
CAEsetBorelaisthereCnowLet
BorelisAthatassumemayWe
CBCHence
CACABAC
ACBACBCBCB
haveweCFor
BAAB
BAtsABsetBorelaisThere
regularBorelisthatshowtoremainsIt
measureBorelais
setmeasurableissetmeasurableeverySince
setcompacteveryforKClearly
ALet
c
c
)(
)(
,
,
.
,
.
\
\
,
0)()(\
)()(..
.
.
,
.
.
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Measure Theoretical Approximation of Sets in Rn
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Lemma 7.1 p.1
CB \
BC
Let μ be a Borel measure on Rn and
B is a Borel set
(i)If μ(B)<∞ , then for each ε >0 there is a
closed set such that
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Lemma 7.1 p.2
BU \
BU (ii) If μ is Radon measure , then for each
ε >0 there is an open set
such that
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1
111
1
11
\
\\\
,
2\
,0
,:1
\
0
.,
iii
iii
ii
ii
ii
iii
ii
iiii
n
CA
CACACA
andclosedisCthenCCLet
CAthatsuch
ACclosedaisthereieachforFix
AAthenAIfClaim
CAthatsuchACset
closedaisthereeachforthatsuchRA
measurablethosealloffamilythebeLet
setBorelfiniteaisvALet
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CAandAC
thenCCletandCA
thatsoellysufficientmChoose
CA
CACACA
ThenClaimof
prooftheinasiCChooseFor
AAthenAIfClaim
m
ii
m
ii
iii
iii
ii
ii
m
ii
m
i
iiii
\
,\
arg
\
\\\lim
.1
,2,1,0
,:2
00
11
0
1
1111
11
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GAhence
AAClaim
ASince
AClaim
GAAthenGAIfClaim
setsopenallcontainsGandGAGAThen
AAGnowLetsets
openallcontainsthatClaimbyfollowsitsets
closedofunioncountableaswrittenbecanset
openeveryandsetsclosedallcontainsSince
i
ci
c
ici
iiii
c
c
1
1
11
1
,
.2
,:3
:.
2
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.0
),0(int
)(
)(
.\\
0,,
.
mradiusandcenterwithballopenthe
BUletmegerpositiveeachfor
iiproveTo
iprovesThis
CBCB
thatsuchBCsetclosedaisthere
forhenceGBparticularIn
setsBorelallcontainsGThen
mm
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1
1
11
1
\\
\\\,
\
,\
2\\\\
..\)(
0\
\
mmm
mmm
mmm
mm
mmm
mmmmm
mm
mm
m
BCU
BCUBUNow
UCUBUB
andopenisUthenCUULet
CBUBCU
tsBUCsetclosedaisthereiby
forsoandUBUwith
setBorelaisBUThen
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Theorem 7.1Approximation by open and compact sets
nRA
openisGAGGA ,);(inf)(
Let μbe a Radon measure on Rn, then
(1) For
(2) If A is μ-measurable on Rn, then
compactisKKAKA ,);(sup)(
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.
),(
,,:inf
,:inf)(
)(
.
.)(
,\
,\..
01.7
.
.)(
obviousisinequalityreversethebecause
iestablishwhich
openisUAUU
openisUBUUBAThen
BAwithABsetBorelaisThere
arbitaryisAnowLet
holdsithatshowwhich
AAUAUhence
AUtsAUsetopenanisthere
forLemmaBy
setBorelaisAthatfirstSuppose
AthatassumemayWei
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closedisCACCAhenceand
CAwhichfrom
UCCRCA
andACclosedisCUCLet
UwithAUsetopenanisthere
givenforiBy
measureRadonaisTheoremBy
APut
AwithmeasurablebeALetii
cn
c
c
,:)(sup)(
)(0
,\\
,,,
,0)(
.1.6
,
)(
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kkkkk
k
kk
k
k
ACwithACsetclosedaisthere
aboveprovediswhatBy
AmeasureRadonaisSince
AA
andmeasurableisAeachThen
kkxkAxAletAIf
2
1
,
)(,
,2,1,1:,
1
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,2,1:sup
,:sup
.
,,
2
1lim
1
1
1111
1
nC
compactisKAKKThus
neveryforCisso
compactisCeachboundedisCeachSince
ACCC
andACNow
n
kk
n
kk
kk
kkk
kk
kk
n
kk
n
kk
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Exercise 7.1 p.1
nRA
AH
(i)Show that the Lebesgue measure on Rn
is a Radon measure.
(ii) Let show that there is a Gδset
such that Ln(H)=Ln(A),where
Ln denotes the Lebesgue measure on Rn.
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.1.4
,
.)(
BorelisLCorollaryby
measureCaraeodoryaisLSince
ROnmeasureLebesguethebeLLeti
n
n
nn
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Exercise 7.1 p.2
nRA
AM
(iii) Let be Lebesgue measure
show that there is a Fσset
with Ln(M)=Ln(A).
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Exercise 7.1 p.3
RRf n :(iv) Let be Lebesgue measurable
show that f is equivalent to a Borel
measurable function.
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Theorem
nRA
openisGAGGA ,);(inf)(
Let μ=Ln be the Lebesgue measure on Rn, then(1) For
(2) If A is Lebesgue measurable on Rn, then
compactisKKAKA ,);(sup)(
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(A, Σ︳ A, μ)
A
dffd AA
(Ω, Σ, μ): measure space
(A, Σ︳ A, μ ) is a measure space
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Lp(Ω)nR
Lp(Ω, Σ,μ)= Lp(Ω)
Σ: the family of Lebesgue measurable
subsets of Ω
μ: the Lebesgue measure
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Cc(Ω)
)( cCf then f is continuous and
Cc(Ω) is the space of all continuous
functions with compact surport in Ω i.e. if
closurexfx 0)(;
is a compact set in Ω
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Lemma
)( cCg
nR
B
such that
Let B be a measurable subset of Ωwith
=Lp(B)<∞, then for any ε>0, there is
pBg
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ppp
c
BG BFG B
FG BFG B
F BB
c
p
p
FBFG
BGFG
dxgdxg
dxgdxg
dxgdxgthen
Gxxg
Fxxg
g
satisfyingfunctioncontinuousabegLet
FGthatsuchGGF
withGsubsetopenanisthere
thenandFB
tsofFsubsetcompactaisthere
anyFor
c
c
22\\
\
,0)(
,1)(
;10
2
2
..
,0
\
\\
\
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Corollary
)( cCg
ii
k
iBi Bandf
i0,
1
such that
Let
be a simple function on Ω, then for any ε>0
p
fg
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i
k
ii
k
i pBii
k
i pBiip
k
iBiip
c
k
iii
ipBici
kg
ggfg
andCgthenggLet
kgtsCg
kiFor
i
ii
i
11
11
1
)(
)()(
)(,
..)(
,,2,1
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Theorem
Cc(Ω) is dence in Lp(Ω), 1 p<∞≦
nR
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0)()(lim)()(lim
min
)(2)()(
)()()()(lim
..
:Pr
2
..)(2
,0:
)(
1
pk
k
pk
k
pppk
kkk
kk
ppp
p
c
p
p
xuxfxuxf
TheorematedDoLebesgue
xforxuxuxfthen
xforxuxfandxuxf
ts
functionssimpleoffsequenceaisThere
Claimofoof
gffugu
gf
tsCgisthere
futhatsuchonffunction
simpleaisthereanyForClaim
LuLet
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Chapter IV
Lp space
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IV 1
Some result for integration
which one must know
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Theorem(Beppo-Levi)
,sup dfnn
nasdffn
n0lim
Let {fn} be a increasing sequence of integrable
functions such that
fn f . Then f is integrable and ↗
(Ω , Σ, μ) is a measure space
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Lebesque Dominated Convergence Theorem
gfn
dffd n
nlim
If fn ,n=1,2,…, and f are measurable functions
and fn →f a.e. Suppose that
a.e. with g being an integrable function.Then
(Ω , Σ, μ) is a measure space
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Fatous Lemma
dfdf n
nn
ninfliminflim
Let {fn} be a sequence of extended real-value
d
measurable functions which is bounded from
below by an integrable function. Then
(Ω , Σ, μ) is a measure space
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Theorem IV.3 (Desity Theorem)
Cc(Ω) is dense in Lp(Ω), 1 p<∞≦
nR
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Theorem IV.4(Tonelli)
openRN :1
1 openRN :2
2
measurableRF :: 21 1..),(
2
oneadyyxF
dxyxFdydyyxFdx
1221
),(),(
)( 211 LF
Suppose that
and that
Then
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Theorem IV.5(Fubini) p.1
)(),( 21 yLyxF
1x
)(),( 11
2
xLdyyxF
)( 211 LFSuppose that
and that
then for a.e.
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Theorem IV.5(Fubini) p.2
)(),( 11 xLyxF
2y
)(),( 21
1
yLdxyxFand that
Similarly, for a.e.
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Theorem IV.5(Fubini) p.3
21
1221
),(
),(),(
dyxF
dxyxFdydyyxFdx
Furthermore, we have
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IV 2
Definition and elementaryproperties of the space Lp
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Exercise 5.1
Show that
..eaff
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..)(
\)(
,
\1
)(
0)()(,
\1
)(
0)(\)(
1..0
11
eafxfHence
Axfxf
havewenlettingBy
Axn
fxf
andAAthenAALet
Axn
fxf
AandAwhereAxMxf
andMn
ftsM
NnFor
nnn
n
n
nnnn
nn
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