A
new necessary
and sufficient condition for
the
Egoroff theorem
in
non-additive
measure
theory
MasayukiTakahashi,ToshiakiMurofushi, ShnAsahina
Departmentof Computational Intelligence and SystemsSciences,
TokyoInstituteofTechnology
E-mail:masayuh@&.dis.titech.ac.jp
Abstract: This paper is a briefsummary of“M.Takahashi, T.MuroMhi, S.Asahina, $A$
new
necessary and sufficient condition for the Egorofftheorem in non-additivemeasure
theory,Fuzzy Sets and Systems, to appear.” This
paper
statesthata
newlydefinedcondition, calledcon-dition(M),is
a
necessalyand sufficient condition for the Egorofftheoreminnon-additivemeasure
theory. The existingnecessaryandsufficient conditions for the Egorofftheorem
are
described bya
doubly-indexedsequenceofmeasurablesets,whilecondition(M)is described by
a
singly-indexedsequenceofmeasurablesets.
1 INTRODUCTION
SinceSugeno[5]introduced the concept ofnon-additive measure, which he called
a
$b\eta$measure,non-additive
measure
theoryhasbeen constructedalongthe lines oftheclassicalmeasure
theory[1,4, 7]. Generally,theorems intheclassical
measure
theoryno
longer holdinnon-additivemeasure
theory,
so
that to find necessary and/or sufficient conditions for such theorems to hold is $ve\iota y$importantfor theconstructionof non-additive
measure
theory.TheEgorofftheorem, which assertsthat almost everywhereconvergence implies almost
uni-form convergence, is
one
ofthe most important convergence theorems in the classicalmeasure
theory. Innon-additive
measure
theory,thistheorem does not holdwithoutadditionalconditions.Necessaryand sufficient conditions for the Egorofftheoremtoholdin non-additive
measure
theoryhave beenproposed, forinstance,theEgoroffcondition [3]or
a
certaincondition calledcondition (E) in [2]. Howeverboth conditionsare
complex since theyare
described by a doubly-indexedsequence ofmeasurable sets. This paper gives
a
new
simpler condition (M) which is describedbya singly-indexedsequenceofmeasurablesetsandshowsthatcondition(M)is equivalenttothe
Egoroffcondition.
Thispaperisabriefsummaryofaforthcomingpaper[6].
2 Preliminaries
Througout thepaper, $(X, \mathscr{S})$is assumedtobeameasurablespaceand$\mathbb{N}$denotesthesetofpositive
integers. Inaddition,every measurable function$f$is assumedto befinite realvalued, i.e., $-\infty<$
$f(x)<\infty$for all$x\in X.$
Definition 1 Anon-additivemeasure on$\mathscr{S}$isa
setfunction$\mu:\mathscr{S}arrow[0,\infty]$ satisfyingthe
follow-ingtwoconditions:
(i) $\mu(\emptyset)=0,$
(ii) $A,B\in \mathscr{S},$ $A\subset B\Rightarrow\mu(A)\leq\mu(B)$
.
数理解析研究所講究録
Unless stated otherwise, all subsets
are
supposed to belong to $\mathscr{S}$and$\mu$ is assumedto be a
non-additive
measure
on$\mathscr{S}.$Definition2 Let$\{f_{n}\}$be asequenceof measurablefunctions,and$f$beameasurable function.
(i) $\{f_{n}\}$issaid to convergeto$f$almost everywhere, written$f_{n}arrow fae$, ifthereexists$N$suchthat
$\mu(N)=0$and$\{f_{n}(x)\}$ convergesto$f(x)$ forall$x\in X\backslash N.$
(ii) $\{f_{n}\}$issaid toconvergeto$f$almostuniformly,written$f_{n}arrow fau$,ifforevery$\epsilon>0$thereexists
$N_{\epsilon}$ such that$\mu(N_{\epsilon})<\epsilon$and$\{f_{n}\}$ convergesto$f$uniformlyon$X\backslash N_{\epsilon}.$
Definition3 $\mu$ is said to satisfy the Egoroffconditionif,foreverydoubly-indexedsequence$E_{m,n}$
such that$E_{m,n}\supset E_{m’,n’}$ for$m\geq m’$ and$n\leq n’$ and$\mu(\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n})=0$, and foreverypositive
number$\epsilon$,thereexistsasequence$\{n_{m}\}$ ofpositiveintegerssuchthat$\mu(\bigcup_{m=1}^{\infty}E_{m,n_{m}})<\epsilon.$ $[3]$ Murofushi et al. [3] show that the Egoroffcondition is a necessary and sufficient condition for the Egorofftheorem, i.e.,the Egoroff conditionissatisfiediffalmosteverywhereconvergence
implies almost uniformconvergence.
3 TheEgorofftheorem
Inthis section, wedefine condition(M),whichisequivalent tothe Egoroff condition.
Definition4 $\mu$ issaidtosatisfy condition$(w$if$\mu(\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty}E_{i})=0$implies that forevery
posi-tivenumber$\epsilon$ thereexists
a
sequence $\{m_{n}\}$ ofpositiveintegerssuch that $\mu(\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{m_{n}}E_{i})<\epsilon.$ The following lemma is necessary to show that condition (M) is equivalent to the Egoroffcondition.
Lemma 1 Let $\{E_{m,n}\}$ be a doubly-indexedsequence, $E_{m,n}\supset E_{m’,n’}$
for
$m\geq m’$ and $n\leq n’$, and$\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n}=\emptyset$
.
Thenthere exists a sequence $\{A_{n}\}$such that$\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty}A_{j}=\emptyset$,andfor
everystrictlyincreasingsequence$\{k_{n}\}$ ofpositive integers, there existsanon-decreasing sequence$\{n_{i}\}$
ofpositive integerssuch that$\bigcup_{m=1}^{\infty}\bigcap_{i=m}^{\infty}E_{i,n_{i}}\subset\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{k_{n}}A_{i}.$
We
can
show Lemma 2, whichisageneralization of Lemma 1.Lemma2 Foreverydoubly-indexedsequence$E_{m,n}$suchthat$E_{m,n}\supset E_{m’,n’}$
for
$m\geq m’$and$n\leq n’,$there existsasequence$\{A_{n}\}$
ofsets
such that$\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty}A_{j}=\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n}$,andfor
everystrictlyincreasingsequence$\{k_{n}\}_{J}$ there existsanon-decreasingsequence$\{n_{i}\}$such that$\bigcup_{m=1}^{\infty}\bigcap_{\dot{|}=m}^{\infty}E_{i.n_{i}}\subset$ $\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{k_{n}}A_{j}.$
Itisnotdifficulttoshow that the Egoroffcondition implies condition(M), while it is noteasy
toshow the inverse implication. Ifitholdsthat foreverydoubly-indexedsequence$\{E_{m,n}\}$ such that
$E_{m,n}\supset E_{m’,n’}$for$m\geq m’$and$n\leq n’$, there exists
a
nondecreasingsequence $\{n_{i}\}$ofpositive integerssuchthat$\bigcup_{m=1}^{\infty}\bigcap_{i=m}^{\infty}E_{i,n_{i}}=\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n}$,thenthat condition(M)impliestheEgoroffcondition
is obvious. Assume that the above if-part is true, and condition (M) is satisfied. Let $\{E_{m,n}\}$ be a doubly-indexed sequence, $E_{m,n}\supset E_{\sqrt{},n’}$ for $m\geq m’$ and$n\leq n’$, and $\mu(\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n})=0.$
Sincethere existsa nondecreasingsequence $\{n_{j}\}$ ofpositive integerssuch that $\bigcup_{m=1}^{\infty}\bigcap_{i=m}^{\infty}E_{i,n_{l}}=$
$\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n}$, foreachpositive integer$i$, let$A_{i}=E_{i,n_{j}}$
.
Then, the Egoroff condition is satisfiedobviously. However, it doesnot necessarily hold that forevery doubly-indexed
sequence
$\{E_{m,n}\}$such that$E_{m,n}\supset E$, for$m\geq m’$and$n\leq n’$,thereexists anondecreasingsequence $\{n_{i}\}$ofpositive
integerssuch that$\bigcup_{m}^{\infty}$ $\infty i=mE_{i,n_{j}}=\bigcup_{m=1}^{\infty}\bigcap_{n=1}^{\infty}E_{m,n}.$
ByusingLemma 2,
we
can
showthenexttheorem.Theorem 1 Condition $(w$isequivalenttothe Egorvffcondition.
4 Concluding remarks
In this paper, we have defined condition (M), which is a necessary and sufficient condition for
the Egorofftheorem with respect to non-additive
measure
and is describedby a singly-indexedsequenceofmeasurablesets.
It
seems
thatit is easiertotreat condition(M) than theEgoroffcondition since condition(M)is describedbyasingly-indexed sequenceofmeasurable sets. However, concrete usefulness and
potentialapplicability
are
notknown. Theseare
topics for future research.REFERENCES
[1] D. Denneberg,Non-additiveMeasureandlntegral,2nded., Kluwer, Dordrecht, 1997.
[2] J. Li, A$fi_{1}$rtherinvestigation forEgoroff’s theorem with respect to monotone set functions,
Kybernetika, 39(2003)753-760.
[3] T.Murofushi,K. Uchino, and S.Asahina, Conditions forEgoroff’s theorem in non-additive
measure
theory,Fua
Sets and Systems, 146(2004) 135-146.[4] E. Pap,Null-additiveSetFunctions,Kluwer, Dordrecht, 1995.
[5] M. Sugeno, TheoryofFuzzy Integrals andItsApplications,DoctoralThesis, Tokyolnstitute of Technology, 1974.
[6] M.Takahashi, T.Murofushi, and S.Asahina, Anew necessary and sufficient conditionfor the
Egorofftheorem in non-additive
measure
theory, Fuzzy SetsandSystems,toappear.[7] Z. WangandG.J.Klir,Fuzy MeasureTheory,Plenum,NewYork, 1992.