Ma hwa e&So Compu ing10(2003)5-16
Ex ensionso Se Func ions¤
S.O chinniko 1and J.C.Falmagne2
1Ma hema icsDep .SanF anciscoS a eUni .
SanF ancisco,CA94132
2Dep .o Cogni i eSciences.Uni .o Cali o nia
I ine,CA92697
se gei@s su.edu, jc @uci.edu
Abs ac
We es ablishanecessa yand su±cien condi ion o a unc-
ion de¯nedonasubse o analgeb a o se s obe ex endable o
aposi i eaddi i e unc ionon healgeb a.I isalsoshown ha
hiscondi ionisnecessa yand su±cien o a egula unc ion
de¯nedona egula subse o heBo elalgeb a o subse so a
gi encompac Hausdo ®space obe ex endable oameasu e.
1991 Ma hema icsSubjec Classi¯ca ion:28A60
1In oduc ion
As anda dme hodo cons uc ingameasu eina gi ense Xis o
de¯ne¯ s anaddi i e unc iononanalgeb aAo subse so Xand
henex end his unc ion oameasu eon he¾{algeb a gene a ed
byA.This`ex ension p oblem' isanimpo an pa o he classical
measu e heo y.S anda dexamplesincludeHahn'sex ension heo em
and heBo elmeasu ein[0;1](c . [3,III.5]).
In hepape ,wea e conce nedwi h he ollowingp oblem:Le
Xbease and Abeanalgeb a o subse so X.Gi enasubse S
¤Thiswo kis suppo ed byNSFg an SES{9986269 oJ.{Cl. Falmagne.
5
6S.O chinniko &J.C.Falmagne
o Aand a eal alued unc ion®onS,¯nd necessa yand su±cien
condi ions o ® obe ex endable o a posi i eaddi i e unc ion¹on
A.
The ollowingcondi ionisins umen al inou ea men o he
ex ension p oblem:
X
A2F
n(A)ÂA(s)¸0;8s2X)X
A2F
n(A)®(A)¸0;[R]
o any¯ni e amilyFµS,whe e coe±cien sn(A)'sa ea bi a y
in ege sand ÂAs ands o he cha ac e is ic unc iono ase AµX.
Weshow ha condi ion[R] isnecessa y(Sec ion2)and su±cien
(Sec ion3) o ® obe ex endable o a posi i eaddi i ese unc ion.
In he casewhenXisa¯ni ese ,as onge esul isalsoes ablished
in Sec ion3.To ob ain hese esul s,weonlyassume ha Xisa¯ni e
uniono elemen so S( hisassump ionisd oppedin he caseo a
¯ni ese X).
Wemakeaddi ionalassump ionsabou hequad uple(X;A;S;®)
when ea ing he ex ension p oblem o measu esinsec ions4 and
5.In bo hsec ions,Xisacompac Hausdo ®space.In Sec ion4,
Ais heBo elalgeb aBo subse so X,whe easin Sec ion5,Ais
he¾{algeb a gene a ed byS.Assuming, inaddi ion, ha Sand ®
sa is ysome` egula i y'condi ions,weshow ha [R] isanecessa y
and su±cien condi ion o ® obe ex endable o a posi i e egula
measu eonA.
Ou app oach o he ex ension p oblemcomesclose o ha o
B unodeFine i in his P obabili yTheo y"[4](Sec ions9 and 10).
In pa icula ,his con exi ycondi ion"(Sec ion15 inAppendix)is
equi alen ocondi ion[R], al hough deFine i o mula esi in a he
di®e en e ms.
2Condi ion[R]
The ollowinglemmaes ablishesause ulequi alen o mo condi ion
[R].
Lemma1. [R]isequi alen o he ollowing condi ion
X
A2F
c(A)ÂA(s)¸0;8s2X)X
A2F
c(A)®(A)¸0;(1)
Ex ensionso Se Func ions7
o any¯ni e amilyFµS,whe e coe±cien sc(A)'sa ea bi a y eal
numbe s.
P oo . I su±ces oshow ha [R] implies(1).Suppose ha o some
ealcoe±cien sc(A)'s such ha PA2Fc(A)ÂA¸0weha e
PA2Fc(A)®(A)<0.The ea e a ionalnumbe sp(A)'s such ha
PA2Fp(A)®(A)<0 and p(A)¸c(A) o all A2F.Clea ly,
X
A2F
p(A)ÂA¸X
A2F
c(A)ÂA¸0:
Mul iplyingbo hinequali iesPA2Fp(A)ÂA¸0 and PA2Fp(A)®(A)<
0byacommonmul ipleo hedenomina o so nonze ocoe±cien s
p(A)'s,weob ainacon adic ion o[R].
Suppose ha ®isa es ic iono aposi i eaddi i ese unc ion
¹onA.No e ha he¯ s sumin(1)is,byde¯ni ion,asimple
unc iononX.Thencondi ion(1)s a es ha hein eg alo aposi i e
simple unc ionisposi i e([3,III.2.14]).Thusweha e he ollowing
p oposi ion.
P oposi ion1. [R]isanecessa ycondi ion o a unc ion®onS o
be ex endable o a posi i e addi i e se unc iononA.
3Ex ensions oposi i eaddi i ese unc-
ions
Wedeno ebyB0 he ec o space o all simple unc ions(wi h espec
oA)onXand deno ebyB#
0{ healgeb aicdualspace.Thespace
B#
0isisomo phic o he ec o space o all addi i ese unc ions¹on
A.Theisomo phismisgi en by
¹7! ¹whe e ¹(x)=Zx(s)¹(ds):(2)
These Co all posi i esimple unc ionsonXisacon excone
inB0.ThusB0isano de ed ec o space.A unc ional 2B#
0is
8S.O chinniko &J.C.Falmagne
mono onei x¸yimplies (x)¸ (y).A unc ional ismono onei
and onlyi i isposi i e, i.e., x¸0implies (x)¸0.
Weshall use he ollowing gene al ac abou mono onelinea ex-
ensionso linea unc ionalsono de ed ec o spaces([1,Theo em1,
x6,ch.2]).
Theo em 1. Le Lbe a ec o space wi h a coneC.Le L0be a
subspace o Lsuch ha o eachxin L,x+L0mee sCi and only
i ¡x+L0mee sC.Le 0in L#
0be mono one.Then he e exis san
ex ension o 0whichismono oneandin L#.
Now wep o e hemain heo emo his sec ion.
Theo em 2. Le Sbe asubse o Asuch ha Xisa¯ni e uniono
se sin Sand le ®be a unc iononS.Then®canbe ex ended o a
posi i e addi i e unc ion¹onAi and onlyi i sa is¯escondi ion
[R].
P oo . Necessi ywases ablishedinP oposi ion1.
Su±ciency.Le L0be hesubspace o B0gene a ed by he cha ac-
e is ic unc ionso se sinS.Fo x=PA2Fc(A)ÂA2L0whe eFis
a¯ni esubse o S,wede¯ne
0(x)=X
A2F
c(A)®(A):
I ollowsimmedia ely om(1) ha 0iswell{de¯nedand isaposi i e
linea unc ionalonL0.
No e ha o anyx2B0 hese x+L0mee s he coneCo
posi i e unc ionsinB0.Indeed, le X=[n
i=1Ai;Ai2Sand de¯ne
x0=Pn
i=1ÂAi2L0.Then, o m= sups2Xjx(s)j,x+mx02C.
ByTheo em1, 0admi sanex ension o a posi i elinea unc ional
onB0.Byde¯ning¹(A)= (ÂA) o A2A,weob ainanex ension
o ® oaposi i eaddi i e unc iononA.
No e ha heassump ion ha Xisa¯ni euniono se sinSis
essen ial in he heo em.Indeed, le Xbeanin¯ni ese ,A=2X,
and le Sbe he amilyo all single onsinA.Le usde¯ne®( sg)=
1;8s2X.Thusde¯ned®sa is¯escondi ion[R]bu canno be
ex ended oamono oneaddi i e unc iononA.
Ex ensionso Se Func ions9
On heo he hand, in he caseo a¯ni ese Xweha eas onge
esul .
Theo em 3. Le Xbe a¯ni e se ,SµA,and®be a unc ionon
S.Then®canbe ex ended o a posi i e addi i e unc ion¹onAi
and onlyi i sa is¯escondi ion[R]wi hcoe±cien s oma¯ni e se
o in ege s.
P oo . Again,weneed op o esu±ciencyonly.Le X0=[Sand
A0be healgeb a o subse so X0consis ing o se sinA ha a e
subse so X0.ByTheo em2,®can be ex ended oaposi i eaddi i e
se unc ion¹0onA0.Fo anA2A,wede¯ne¹(A)=¹0(A X0).
Clea ly,¹isaposi i eaddi i ese unc iononA.
Le usconside cha ac e is ic unc ionso se sinSasin eg al ec-
o sinRjXjand le Cbe hein e sec iono hesubspace gene a ed by
hese ec o swi h heposi i e coneinRjXj.The coneCisa a ional
polyhed alconeand he e o ehasanin eg alHilbe basis(Theo-
em16.4in[5]).Thuswe can useonly ec o s om hisbasisin he
igh sideo heimplica ionin[R]. I ollows ha in he caseo ¯ni e
se Xcoe±cien sin[R]can be aken oma¯ni ese o in ege s.
Rema k.I wasno ed byJean{PaulDoignon(pe sonalcommuni-
ca ion) ha su±ciencyo condi ion[R] in he¯ni e caseisadi ec
consequence o Fa kas' lemma[5,Co olla y7.1d].
4Ex ensions omeasu esI
The ollowingexampleshows ha , ingene al, condi ion[R] isno
su±cien o a unc ion® obe ex endable oaposi i emeasu e(¾{
addi i ese { unc ion)ona¾{algeb aA.
Example1. Le X=[0;1]and S= [0; ): 2(0;1]g[ [0; ]: 2
[0;1]g.No e ha X2S.Wede¯ne®( 0g)=0 and ®(A)=1i A
is[0; )o [0; ] o 0< ·1.I iseasy o e i y ha husde¯ned®
sa is¯escondi ion[R].

10 S.O chinniko &J.C.Falmagne
Le ¹bea¾{addi i e ex ensiono ® o he¾{algeb aBo Bo el
subse so [0;1]. Weha e
¹(( ;1])=1¡®([0; ])=0; o >0,
¹((0;1])=1¡®( 0g)=1;
¹((s; ])=1¡®([0;s])¡¹(( ;1])=0; o 0<s< :
By¾{addi i i yo ¹,
1=¹((0;1])=¹Ã1
[
1µ1
k+1;1
k¸!=
1
X
1
¹µµ 1
k+1;1
k¸¶=0;
acon adic ion.On heo he hand,bycondi ion[R], he eisan
addi i e ex ensiono ® oB.
Thisexamplesugges s ha ino de okeep[R]asanecessa y
and su±cien condi ion o ex endibili yo ase unc ion o a mea-
su e,some cons ains should beimposedon hequad uple(X;A;S;®).
Namely,weassume ha Xisacompac Hausdo ®space and in oduce
he ollowing` egula i y'condi ionsonSand ®.
De¯ni ion1. (i)A amilySo subse so Xis said obe egula i
(a)Fo eachE2Sand aclosedse FµE he eisE02Ssuch
ha
FµE0µclE0µE:
(b)Fo eachE2Sand anopense G¶E he eisE00 2Ssuch
ha
Eµin E00 µE00 µG:
(ii)A unc ion®ona amilySis said obe egula i o eachE2S
and ">0 he eisase FinSwhose closu eiscon ainedinEand a
se Gwhosein e io con ainsEsuch ha j®(G)¡®(F)j<".
In his sec ion,Ais heBo elalgeb aBo subse so X.
Example2. Since Xisano malspace, he amilieso all opense s
and o all closedse sinXa e exampleso egula amilieso Bo el
se s(c . [2,VII.3.2(2)]).
Ex ensionso Se Func ions11
Example3. Le X=[0;1]and Sbe he amilyo all in e alsin he
o m[a;b).Clea ly,Sisa egula amilyo Bo else s.
Example4. Le S=Band le ®=¹{a egula posi i eaddi i ese
unc iononBin heusualsense(c . [3,III.5.11]).Then®isa egula
unc ionin hesenseo De¯ni ion1.
Lemma2. Le ¹be a egula posi i e measu eon heBo el algeb aB
and le Sbe a egula amilyo Bo else s.Then he es ic iono ¹
oSisa egula unc iononS.
P oo . Le E2Sand ">0.Since ¹is egula and posi i e, he eisa
closedse FµEand anopense G¶Esuch ha ¹(G)¡¹(F)<".
Since Sis egula , he ea eE0;E00 2Ssuch ha FµE0µclE0µ
Eµin E00 µE00 µG.Since ¹isposi i e,¹(E00)¡¹(E0)<".
The e o e he es ic iono ¹ oSisa egula se unc iononS.
Lemma3. Le Sbe a egula amilyo Bo else s such ha Xisa
¯ni e uniono se sin Sand le ®be a egula unc iononSsa is ying
condi ion[R].Then®isex endable o a egula posi i e measu eon
B.
P oo . ByTheo em2,®admi sanex ension o a posi i eaddi i ese
unc ion¹onB.Since ¹isbounded, i de¯nesabounded posi i e
linea unc ional on heBanachspace Bo all uni o mlimi so
unc ionsinB0endowedwi h heno mk¢k1.This unc ional isgi en
by[3,IV.5.1]
(x)=Zx(s)¹(ds);x2B0:
By heRiesz ep esen a ion heo em[3,IV.6.3] he es ic iono his
unc ional(whichwedeno eby hesamesymbol ) o hespace C(X)
o con inuous unc ionsonSisgi en by
(x)=Zx(s)¹¤(ds);x2C(X);
whe e¹¤isa egula posi i emeasu eonB.
12 S.O chinniko &J.C.Falmagne
Nowi su±ces oshow ha ¹¤(E)=¹(E)onS.Le E2Sand
">0.Since ¹¤isposi i eand egula he eisaclosedse Fand an
opense Gsuch ha
FµEµG; ¹¤(F)·¹¤(E)·¹¤(G);and ¹¤(G)¡¹¤(F)<":
Since Sis egula , he ea eE0;E00 2Ssuch ha
FµE0µclE0µEµin E00 µE00 µGand ¹(E00)¡¹(E0)<":
Wedeno eF0=clE0and G0=in E00.Since ¹and ¹¤a eposi i e,
¹(G0)¡¹(F0)<"and ¹¤(G0)¡¹¤(F0)<":(3)
Since Xisano malspace,byU ysohn'slemma, he eisacon in-
uous unc ionxsuch ha
0·x(s)·1; o all s2X;
x(s)=1; o all s2F0;
x(s)=0; o all s=2G0:
Fo ana u alnumbe n,wede¯nea amilyo n+1in e alsin[0;1]
by
Ik=(£k¡1
n;k
n¢; o 1·k·n;
1g; o k=n+1.
The amilyo Bo else sEk=x¡1(Ik);1·k·n+1, o msapa i ion
o X.Clea ly,Sn
k=2EkµG0nF0.The e o e,by he¯ s inequali yin
(3),
n
X
k=2
¹(Ek)·¹(G0)¡¹(F0)<":(4)
Le xnbea unc ion de¯ned byxn(s)=k¡1
n o s2Ek;1·k·n+1.
Thus
j (x)¡ (xn)j· k k¢kx¡xnk<1
nk k(5)
Ex ensionso Se Func ions13
Fu he ,
xn=
n+1
X
k=1
k¡1
nÂEk=
n
X
k=2
k¡1
nÂEk+ÂEk+1:
Thus
(xn)=
n
X
k=2
k¡1
n¹(Ek)+¹(Ek+1);
whichimplies,by(4),
(xn)¡¹(Ek+1)=
n
X
k=2
k¡1
n¹(Ek)<":
Thisinequali y oge he wi honein(5)imply
j (x)¡¹(En+1)j<"+1
nk k:(6)
Clea ly,F0µEn+1µG0,and F0µEµG0.Thus,by(3),
j¹(En+1)¡¹(E)j<":(7)
Since (x)=Rx(s)¹¤(ds),weha e¹¤(F0)· (x)·¹¤(G0).On he
o he hand,¹¤(F0)·¹¤(E)·¹¤(G0).By hesecond inequali yin(3),
j¹¤(E)¡ (x)j<":(8)
Combininginequali ies(6),(7),and (8),weha e
j¹¤(E)¡¹(E)j<3"+1
nk k:
Hence,¹¤(E)=¹(E)=®(E).
Combining he esul so Lemma 2 and Lemma 3,weha e he
ollowing heo em.
Theo em 4. Le Sbe a egula amilyo Bo else s such ha Xisa
¯ni e uniono se sin S.A unc ion®onSisex endable o a egula
posi i e measu eonBi and onlyi i is egula andsa is¯escondi ion
[R].