GEORGIAN MATHEMATICAL JOURNAL : Vol Period 2 Comma No Period 1 Comma 1 9 9 5 Comma\Noindent 45 Hyphengeorgian 5 3 MATHEMATICAL JOURNAL : Vol

$GEORGIAN MATHEMATICAL JOURNAL : Vol Period 2 Comma No Period 1 Comma 1 9 9 5 Comma\Noindent 45 Hyphengeorgian 5 3 MATHEMATICAL JOURNAL : Vol$

GEORGIAN MATHEMATICAL JOURNAL : Vol period 2 comma No period 1 comma 1 9 9 5 comma\noindent 45 hyphenGEORGIAN 5 3 MATHEMATICAL JOURNAL : Vol . 2 , No . 1 , 1 9 9 5 , 45 − 5 3 ON .. SEPARABLE .. SUPPORTS .. OF .. BOREL MEASURES \ centerlineA period KHARAZISHVILI{ON \quad SEPARABLE \quad SUPPORTS \quad OF \quad BOREL MEASURES } Abstract period Some properties of Borel measures with separable sup hyphen \ centerlineports are considered{A . KHARAZISHVILI period .. In particular} comma it is proved that any sigma hyphen finite Borel measure on a Suslin line has a separable supports and from this \ centerline { AbstractGEORGIAN . Some MATHEMATICAL properties JOURNAL of Borel : Vol . 2 , measures No . 1 , 1 9 9 5with , 45 - 5 separable 3 sup − } fact it is deduced commaON using SEPARABLE the continuum SUPPORTS hypothesis comma OF that any BOREL Suslin MEASURES line contains a Luzin subspace with the cardinality of the continuum period A . KHARAZISHVILI \ centerlineLet E be a topological{ ports are space considered period .. We . say\quad that theIn space particular E has the pro , it hyphen is proved that any Abstract . Some properties of Borel measures with separable sup - $ \pertysigma open− parenthesis$ f i n i t S e closing} parenthesis if for every sigma hyphen finite Borel measure mu ports are considered . In particular , it is proved that any σ− finite defined in this space there Borel measure on a Suslin line has a separable supports and from this \ centerlineexists a separable{ Borel support measure comma on i period a Suslin e period line comma has a separable a separable closed set supports F open parenthesis and from this } fact it is deduced , using the continuum hypothesis , that any Suslin mu closing parenthesis subset E such that \ centerline { fact it isline deduced contains a , Luzin using subspace the with continuum the cardinality hypothesis of the continuum , . that any Suslin } mu open parenthesis E backslashLet E be F opena topological parenthesis space mu . closing We say parenthesis that the closingspace E parenthesishas the pro = - 0 period \ centerline { lineperty contains (S) if for a every Luzinσ− subspacefinite Borel with measure theµ defined cardinality in this space of therethe continuum . } Let us consider someexists examples a separable of topological support ,spaces i . e . having , a separable the property closed set F (µ) ⊂ E such that open parenthesis S closing parenthesis period \ hspaceExample∗{\ 1f period i l l } Let .. It is $ obvious E $ bethat a any topological separable topological space space . \quad E hasWe say that the space $E$ has the pro − the property open parenthesis S closing parenthesisµ( periodE \ F (µ)) = 0. Example 2 period .. Let E be an arbitrary metric space whose topological weight \noindent perty $( S )$ if forevery $ \sigma − $ finite Borel measure is not measurable in a wideLet sense us consider periodsome Then examples according of to topological the well hyphen spaces known having result the property $ \frommu $ the topologicaldefined in measure this theory space the there space E has the property open parenthesis S closing parenthesis period \noindentExample 3exists period .. a Let separable E be the Alexandrov support compactification , i . e(S .). , of a some separable discrete closed set $ F ( topological\mu ) space\subset period ThenE the $ following such that statements are equivalent : a closing parenthesis theExample space E has 1 . the propertyIt is obvious open that parenthesis any separable S closing topological parenthesis space semicolonE has \ beginb closing{ a l i parenthesis g n ∗} card open parenthesis E closing parenthesis is not measurable in a wide sense period\mu (E \setminus F( \mu ) ) = 0 . theproperty(S). \endExample{ a l i g n 4∗} period .. Let E be a Hausdorff topological space period We say that E is a Luzin space if every sigma hyphen finite diffused open parenthesis i period e period comma continuous \ hspace ∗{\ f i l l } LetExample us consider 2 . someLet examplesE be an arbitrary of topological metric space spaces whose topological having the property closing parenthesis Borelweight measure is not defined measurable in a wide sense . Then according to the well - known in E is identically zero period The classical Luzin set on the real line R is a Luzin \ begin { a l i g n ∗} result from the topological measure theory the space E has the property (S). topological space openExample parenthesis 3 . aboutLet LuzinE be sets the Alexandrovsee comma forcompactification example comma of some .. brackleft discrete 1 brackright(S). closing parenthesis period .. One can easily \end{ a l i g n ∗} topological space . Then the following statements are equivalent : check that any sigma hyphen finite Borela ) measure the space definedE has in the the property Luzin topological (S); space 1 99 1 Mathematics Subject Classificationb ) card (periodE) is not 54 H measurable 5 period in a wide sense . \ hspaceKey words∗{\ f and i l l } phrasesExample period 1 . Borel\quad measureIt is comma obvious support that comma any Luzin separable space comma topological Suslin space $ E $ has Example 4 . Let E be a Hausdorff topological space . We say that E property comma Suslinis a Luzin space if every σ− finite diffused ( i . e . , continuous ) Borel measure line period \ begin { a l i g n ∗} defined in E is identically zero . The classical Luzin set on the real line R is a 45 Luzin topological space ( about Luzin sets see , for example , [ 1 ] ) . One the1 72 hyphen property 947 X slash ( 9 S5 slash ) 0 1 .0 0 hyphen 0 45 Dollar 7 period 5 0 slash 0 circlecopyrt-c 1 9 \end{ a l i g n ∗} can easily 9 5 P l enum Publishingcheck Corporation that any σ− finite Borel measure defined in the Luzin topological space Example 2 . \quad Let $ E $1 99 be 1 Mathematics an arbitrary Subject Classification metric space. 54 H whose 5 . topological weight is not measurableKey in words a wide and phrases sense . Borel . Then measure according , support , Luzin to space the , well Suslin property− known , Suslin result from the topologicalline . measure theory the space $ E $ has the property $ ( S ) . $ 45 1 72 - 947 X /95/0100 − 045Dollar7.50/0circlecopyrt − c1995 P l enum Example 3 . \quadPublishingLet Corporation$ E $ be the Alexandrov compactification of some discrete topological space . Then the following statements are equivalent :

\ centerline {a) the space $E$ hasthe property $( S ) ; $ }

\ centerline {b ) card $ ( E ) $ is not measurable in a wide sense . }

Example 4 . \quad Let $ E $ be a Hausdorff topological space . We say that $ E $ i s a Luzin space if every $ \sigma − $ finite diffused ( i . e . , continuous ) Borel measure defined in $ E $ is identically zero . The classical Luzin set on the real line $R$ is a Luzin topological space ( about Luzin sets see , for example , \quad [ 1 ] ) . \quad One can easily

\noindent check that any $ \sigma − $ finite Borel measure defined in the Luzin topological space

\ centerline {1 99 1 Mathematics Subject Classification . 54 H 5 . }

Key words and phrases . Borel measure , support , Luzin space , Suslin property , Suslin l i n e .

\ centerline {45 }

\ hspace ∗{\ f i l l }1 72 − 947X$/ 9 5 / 0 1 0 0 − 0 45 Dollar 7 . 5 0 / 0 circlecopyrt −c 1 9 9 5 $ P l enum Publishing Corporation 46 .. A period KHARAZISHVILI \noindentE is concentrated46 \quad on aA countable . KHARAZISHVILI subset of E period The latter fact immediately implies that every Luzin space has the property open parenthesis S closing parenthesis period .. Notice\noindent here that$ the E $ is concentrated on a countable subset of $ E . $ The latter fact immediately impliestopological that space every E from Luzin Example space 3 is a has Luzin the space property if and onlyif $ the ( S ) . $ \quad Notice here that the topologicalcardinal number space card open $ E parenthesis $ from E Example closing parenthesis 3 is a is Luzin not measurable space in if a andwide sense only period if the cardinalWe shall show number that card the property $ ( open E parenthesis ) $ is S not closing measurable parenthesis in is nota wide comma sense generally . 46 A . KHARAZISHVILI speaking comma a hereE hyphenis concentrated on a countable subset of E. The latter fact immediately Weditary shall property show thatin the classthe of property all topological $ ( spaces S period ) $ is not , generally speaking , a here − ditary propertyimplies in the that class every of Luzin all space topological has the property spaces (S). . Notice here that the Example 5 period ..topological Denote .. space by .. theE from .. symbol Example .. R 3 to isthe a Luzin power space of * if .. and the only .. set if .. the of cardinalall .. real .. numbers Example 5 . \quadnumberDenote card\quad (E) is notby measurable\quad the in\ aquad wide sensesymbol . \quad $ R ˆ{ ∗ }$ equipped with the SorgenfreyWe shall topology show that period the ..property Since the (S) topological is not , generally space R speaking to the power , a here of * - \quadis separablethe \quad commas e the t \ productquad o hyphen f a l l space\quad Rr to e a the l \ powerquad ofnumbers * times R to the power of * is equipped with theditary Sorgenfrey property inthe topology class of all . topological\quad Since spaces the . topological space separable too period ThusExample we see that 5 . Denote by the symbol R∗ the set of all $ Rthe ˆ{ space ∗ } R$ to the power of * times R to the power of * has the property open parenthesis S closing is separable , thereal product numbers equipped− space with $ the R ˆSorgenfrey{ ∗ } \ topologytimes .R Since ˆ{ ∗the } topological$ is separable too . Thus we see that parenthesis period .. On the other∗ hand comma the space ∗ ∗ the space $R ˆspace{ ∗R }is \ separabletimes ,R the ˆ{ product ∗ }$ - space hasR the× R propertyis separable $( too . Thus S we ) R to the power of * times R to the power∗ of∗ * contains a closed discrete subspace D with the . $ \quad On thesee other that the hand space ,R the× R spacehas the property (S). On the other hand , the cardinality of the ∗ ∗ $ R ˆ{ ∗ } \timesspace R R× R ˆ{contains ∗ }$ a contains closed discrete a closed subspace discreteD with the subspacecardinality of $ the D $ continuum period ..continuum Assuming . that Assuming the cardinality that the of the cardinality continuum of the is measurable continuum is measurable within a the wide cardinality sense comma we of can the define in D a probability continuous Borel measure period continuum . \quadin aAssuming wide sense ,that we can the define cardinality in D a probability of the continuous continuum Borel is measure measurable . But such a measureBut vanishes such a on measure all separable vanishes subsets on all of separable D period subsets .. Therefore of D. weTherefore we find infind a open wide parenthesis sense , under we can our define assumption in about $ D the $ cardinality a probability of the continuum continuous closing Borel paren- measure . But such a measure( under vanishes our assumption on all about separable the cardinality subsets of the of continuum $ D ) that. $ \quad Therefore we thesis that ∗ ∗ find ( under ourthe assumption topological space aboutD ⊂ theR × cardinalityR does not have of the the property continuum (S). ) that the topological space DExample subset R 6 to . theLet power (E, · of) be * times a topological R to the group power and of let * doesµ be not a σ− havefinite the property open parenthesisBorel S measure closing parenthesis in the group periodE which is quasi - invariant with respect to some \noindentExample 6the period topological .. Let open parenthesis spaceE $ comma D \subset times closingR parenthesis ˆ{ ∗ } be \times a topologicalR ˆ group{ ∗ }$ does not have theeverywhere property dense $ subgroup ( S Γ ) of E. . $ Assume also that there exists a closed and let mu be a sigmaseparable hyphen finite set F ⊂ E with µ(F ) > 0. Then we may assert that the given Borel measure in thetopological group E which group isE quasiis separable hyphen too invariant . Indeed with respect , assume to that some the opposite is Exampleeverywhere 6 . dense\quad subgroupLet Capital$ ( GammaE , of E\cdot period ..) Assume $ be also a thattopological there exists group a closed and let $ \mu $ be a $ true\sigma : E is not− $ a separable f i n i t e topological space . Using the method of transfi - separable set F subsetnite E recursion with mu let open us define parenthesis an ω − Fsequence closing parenthesis (F ) < ω greatersatisfying 0 period the following .. Then weBorel may assert measure that the in given the group $ E $ which1 is quasi ξ−ξ invariant1 with respect to some everywheretopological group dense E is subgroup separable too $ period\Gamma .. Indeed$ o f comma $ E assume . $ that\quad the oppositeAssume is also that there exists a closed separable set $ F \subset E $ with $ \mu (F) > 0 . $ true : E is not a separable topological space period ..conditions Using the : method of transfi hyphen \quadniteThen recursion we let may us assertdefine an that omega the sub given 1 hyphen sequence open parenthesis F sub xi closing topological group $ E $ is separable too . \quad Indeed , assume that the opposite is parenthesis sub xi less omega sub 1 satisfyinga ) the family the following (Fξ)ξ<ω is increasing by inclusion ; true $ : E $ is not a separable topological1 space . \quad Using the method of transfi − conditions : b )(∀ξ)(ξ < ω1 ⇒ Fξ is a closed separable subgroup of E); nite recursion let us define an $ \omega { 1 } − $ sequence $( F {\ xi } a closing parenthesis the family open parenthesisc ) if ζ < ξ F < sub ω1, xithen closingµ(Fξ parenthesis\ Fζ ) > 0. sub xi less omega sub ) {\ xi } < \omega { 1 }$ satisfying the following 1 is increasing by inclusionTake semicolon as F0 the closed subgroup of E generated by the set F. Assume that b closing parenthesisfor open an ordinalparenthesis number forallξ xi(0 closing< ξ < parenthesis ω1) we have open already parenthesis constructed xi less omega a partial sub \ begin { a l i g n ∗} 1 double stroke right arrowfamily F of sub closed xi is separablea closed separable groups (F subgroupζ )ζ < ξ. ofDenote E closing by H parenthesisthe closure semicolon of the conditionsc closing parenthesis : union if zeta of this less partial xi less family omega . sub It 1 is comma obvious then that muH is open also parenthesis a closed F sub xi backslash\end{ a l i F g n sub∗} zeta closingseparable parenthesis subgroup greater of E. 0Since period the group Γ is dense everywhere in E and E Take as F sub 0 theis closed not separable subgroup , there of E generated exists an element by the setg ∈ FΓ period\ H. It .. is Assume clear that \ centerlinethat for an ordinal{a) thenumber family xi open parenthesis$( F 0{\ lessxi xi less} omega) {\ subxi 1 closing< parenthesis\omega { we1 have}}$ alreadyis increasing constructed by a inclusion ; } partial family of closedg·H∩ separable=⇒ ∅notdefnotdefnotdef groups open parenthesis H−notdef F sub− zetanotdef closing−notdefnotdef parenthesis− subparenright zeta −notdefnotdef−notdef−negationslash−greaterbraceleft−zero−universal−notdef notdef g−notdef·notdefnotdefparenright − notdef − notdefgreater−notdefunion−zeronotdefnotdef less\ centerline xi period Denote{b $ by ) H the ( closure\ f o r a l l \ xi )( \ xi < \omega { 1 }\Rightarrow h − e refore w em y − a t ke a F ξ t e c osed s bgroup o E g nerated b F of{\ thexi union}$ of is this a partial closed family separable period .. It subgroup is obvious that of H $E is also )a closed ; $ } yt e separable subgroup of E period .. Since the group Capital Gamma is dense everywhere in E and t H ∪ g · H. W es allth ereby h fine t e u eded ω− s e quence (ξ) < ω1. \ centerlineE is not separable{c ) comma i f $ there\zeta exists< an element\ xi g< ind Capital\omega Gamman { backslash1 } , $H period then Itξ is $ clear\mu m u − l taneousl comma − y w es all o tain a nu countable f mily (ξ+1\F ξ) < that(F {\ xi }\setminus F {\zeta } ) > 0 . $ } ξ ω1 o irwise d s − j s ts h eryone o w i − h ch h s a s rictly p si−t iv e g times H cap arrowdblright-equal varnothingoint notdefe notdef notdef H-notdef-notdef-notdef notdef- µ m a − e sure . o − wv − e er t e e istence o s ch a f mily c ntradicts t e parenright-notdefTake as $ F notdef-notdef-negationslash-greater{ 0 }$ the closed subgroup braceleft-zero-universal-notdef of $ E $ generated notdef by the g-notdef set times o finiteness o t e notdef$ F notdef . $ parenright-notdef-notdef\quad Assumeσ greater-notdef union-zero notdef notdef e − a sure µ T e − h refore t e u ven t pologicao lg oup E i s parable thath-e refore for wan em ordinal y-a t ke a number .. F xi t e $ ..\ cxi osed s( bgroup 0 g ..< o .. E\ gxi nerated< b yt\omega e { 1 } ) $ . B yt e wet have H cup already g times H constructed period W .. es a allth ereby h sub d fine .. t e u sub n eded omega hyphen s e quencepartial open family parenthesis of xi closed closing parenthesis separable sub groups xi less omega $ ( 1 period F {\zeta } ) {\zeta } < m\ u-lxi taneousl. $ comma-y Denote .. byw es all $H$ o tain a the nu countable closure f mily open parenthesis xi plus 1 backslash Fof xi closing the union parenthesis of this sub xi partial less omega family 1 .. o . \quad It is obvious that $H$ is also a closed irwise d s-j sub oint s ts h sub e eryone .. o .. w i-h ch h s a s rictly p s to the power of i-t iv e mu m\noindent a-e sure periodseparable subgroup of $ E . $ \quad Since the group $ \Gamma $ iso-w dense v-e er everywhere t e .. e istence in o .. $ s ch E a $ f mily and c ntradicts t e o sub sigma finiteness o .. t e $e-a E sure $ mu is .. not T .. separable e-h refore t e u, sub there g ven exists t pologica an to element the power of $ o g lg .. oup\ in .. E\ iGamma .. s parable\setminus Hperiod . B $ .. yt It e is clear that \ [ g \cdot H \cap \Longrightarrow \ varnothing notdef notdef notdef H−notdef−notdef−notdef notdef−parenright −notdef notdef−notdef−negationslash −g r e a t e r b r a c e l e f t −zero−u n i v e r s a l −notdef notdef g−notdef \cdot notdef notdef parenright −notdef−notdef greater −notdef union−zero notdef notdef \ ]

\noindent $ h−e $ reforewem $y−a $ t ke a \quad $ F \ xi $ t e \quad c osed s bgroup \quad o \quad $E$ g nerated b yt e

\noindent t $ H \cup g \cdot H . $ W \quad es allth ereby $ h { d }$ f i n e \quad t e $ u { n }$ eded $ \omega − $ s e quence $( \ xi ) {\ xi } < \omega 1 . $

\noindent m $ u−l $ taneousl $ comma−y $ \quad w es all o tain a nu countable f mily $ ( \ xi + 1 \setminus F \ xi ) {\ xi } < \omega 1 $ \quad o i r w i s e d $ s−j { o i n t }$ s t s $ h { e }$ eryone \quad o \quad w $ i−h $ chhsas rictly p $sˆ{ i−t }$ i v e $ \mu $ m $ a−e $ sure . $ o−w v−e $ er t e \quad e i s t e n c e o \quad s ch a f mily c ntradicts t e $ o {\sigma }$ finiteness o \quad t e

\noindent $ e−a $ sure $ \mu $ \quad T \quad $ e−h$ refore te $u { g }$ ven t $ pologica ˆ{ o }$ l g \quad oup \quad $ E $ i \quad s parable . B \quad yt e ON SEPARABLE SUPPORTS OF BOREL MEASURES .. 47 \ hspaceforegoing∗{\ argumentsf i l l }ON weSEPARABLE actually establish SUPPORTS the validity OF BOREL of a more MEASURES general\quad 47 result period .. Namely comma let E be again a topological group comma mu be a sigma hyphen finite\noindent Borel foregoing arguments we actually establish the validity of a more general r emeasure s u l t . in\ Equad sub commaNamely and , Capital let $ Gamma E $ be be some again everywhere a topological dense subset group of E sub $ period , \ Ifmu $ thebe measure a $ \sigma − $ finite Borel measuremu is quasi in hyphen $ E { invariant, }$ with and respect $ \Gamma to Capital$ be Gamma some comma everywhere then at dense least one subset of the of $ E { . }$ If the measure ON SEPARABLE SUPPORTS OF BOREL MEASURES 47 following two foregoing arguments we actually establish the validity of a more general result . $assertions\mu $ is i valid s quasi : − invariant with respect to $ \Gamma , $ then at least one of the following two assertions is validNamely : , let E be again a topological group , µ be a σ− finite Borel measure a closing parenthesisin theE and topological Γ be some group everywhere E is separable dense semicolon subset of E If the measure µ is quasi - b closing parenthesis the, measure mu vanishes on all closed separable. subsets of E period \ centerline {a ) theinvariant topological with respect group to Γ, then $ atE least$ is one separable of the following ; } two assertions is In particular commavalid .. if : the mentioned measure mu is not identically zero and the given topological group E is not separablea ) the topological comma then group E doesE is not separable have the ; \ centerlineproperty open{b parenthesis ) the measure S closing parenthesis$ \mu $ period vanishes on all closed separable subsets of $ E . $ } b ) the measure µ vanishes on all closed separable subsets of E. In connection with theIn property particular open , parenthesis if the mentioned S closing measure parenthesisµ is not notice identically that we zero have and the Proposition 1 periodgiven .. Let topological open parenthesis group E is E not sub separable i closing parenthesis, then E does i in not I .. have be thean arb itrary countableIn particular family of , topolo\quad hyphenif the mentioned measure $ \mu $ is not identically zero and thegical given spaces topological everyone of which group has the $ property E $ isopen not parenthesis separable S closing , then parenthesis $ E sub $ period does .. not have the Then the product hyphen space property(S). \ beginproduct{ a l sub i g n i∗} in I E sub i has th e property open parenthesis S closing parenthesis too period property ( S ) . Proof period .. Let .. mu .. beIn .. connection a .. sigma with hyphen the finite property .. Borel (S) notice .. measure that ..we defined have .. in .. the \end{ a l i g n ∗} .. product hyphen space Proposition 1 . Let (Ei)i ∈ I be an arb itrary countable family product sub i in I Eof sub topolo i period - gical .. It spaces may be everyone assumed of without which has loss the of generality property that(S). muThen is a prob the \ centerline { In connection withQ the property $ ( S ) $ notice that we have } hyphen product - space i∈I Ei has th e property (S) too . ability measure periodProof Fix . an indexLet i inµ I andbe for a anyσ− Borelfinite set X Borel subset E measure sub i write defined in Proposition 1 . \quad Let $ ( E { i } ) i \ in I $ \quad be an arb itrary countable family of topolo − Line 1 mu i openthe parenthesis product X -closing space Q parenthesisE . It = may mu be parenleftbig assumed without X times loss product of generality E sub j gical spaces everyone of which hasi∈I thei property $ ( S ) { . }$ \quad Then the product − space parenrightbig period Linethat 2µ jis in a Ibackslash prob - ability open measure brace i closing . Fix an brace index i ∈ I and for any Borel set $ \prod { i \ in I } E { i }$ hastheproperty $( S )$ too . This formula definesX the⊂ E probabilityi write Borel measure mu i in the topological space E sub i period .. By the condition comma there exists a separable support F sub i subset E \noindent Proof . \quad Let \quad $ \mu $ \quadYbe \quad a \quad $ \sigma sub i for µi(X) = µ(X × Ej). − $mu fi i period n i t e ..\quad Now considerBorel \ thequad productmeasure hyphen\quad space productd e f i n e d sub\quad i in I Fin sub\quad i periodthe ..\ Thisquad product − space j ∈ I \{i} space$ \ isprod certainly{ sepi hyphen\ in I } E { i } . $ \quad It may be assumed without loss of generality that $ \arablemu $ and i s comma a probThis as one− formula can easily defines check the comma probability represents Borel a measure supportµi forin the the original topological ability measure . Fix an index $ i \ in I $ and for any Borel set $X measure mu periodspace Ei. By the condition , there exists a separable support Fi ⊂ Ei for µi. \subset E { i }$ writ e Q Proposition 2 periodNow .. Let consider I be a the s e product t of indices - space whosei∈ cardinalityI Fi. This is spacenot mea is certainly hyphen sep - surable in a wide sarable ense period and , .. as Assume one can open easily parenthesis check , represents E sub i closing a support parenthesis for the original i in I to be a family\ [ \ begin of topo{ a l logicali g n e d spaces}\measuremu µ.i ( X ) = \mu (X \times \prod E everyone{ j } of). which has\\ Proposition th e property2 open . parenthesisLet I be a S s closing e t of indices parenthesis whose period cardinality .. Denote is not b mea y th j \ in I \setminus \{ i \}\end{ a l i g n e d }\ ] e symbol E the topo hyphen- surable in a wide s ense . Assume (Ei)i ∈ I to be a family of topo logical logical sum of the familyspaces open everyone parenthesis of which E sub has i th closing e property parenthesis(S). i inDenote I period b y .. th Then e symbol the spaceE E als o has th e propertythe topo - logical sum of the family (Ei)i ∈ I. Then the space E als o has \noindentopen parenthesisThis formulaSth closing e property parenthesis defines period the probability Borel measure $ \mu i $ inProof the periodtopological .. Let mu be an arbitrary sigma hyphen finite Borel measure defined in the topolo hyphen \noindentgical spacespace E period $ We E write{ i } . $ \quad By the(S). condition , there exists a separable support $ FJ ={ openi }\ bracesubset i in I : muE open{ parenthesisi }$ f o E r sub i closing parenthesis = 0 closing brace period $Since\mu the measurei . $muProof is\quad sigma. LetNow hyphenµ be consider anfinite arbitrary comma theσ we− productfinite have the Borel− inequality measurespace defined $ \prod in the{ topoloi \ in I }cardF open{ parenthesisi } .- $ gical I backslash\quad space E.This JWe closing write space parenthesis is certainly less or equal sep omega− sub 0 period

\noindent arable and , as one can easilyJ = {i check∈ I : µ(E ,i) represents = 0}. a support for the original Since the measure µ is σ− ﬁnite , we have the inequality \noindent measure $ \mu . $ card(I \ J) ≤ ω . Proposition 2 . \quad Let $ I $ be a s e t of indices0 whose cardinality is not mea − surable in a wide s ense . \quad Assume $ ( E { i } ) i \ in I $ to be a family of topo logical spaces everyone of which has th e property $ ( S ) . $ \quad Denote b y th e symbol $ E $ the topo − logical sum of the family $ ( E { i } ) i \ in I . $ \quad Then the space $ E $ als o has th e property

\ begin { a l i g n ∗} (S). \end{ a l i g n ∗}

\noindent Proof . \quad Let $ \mu $ be an arbitrary $ \sigma − $ finite Borel measure defined in the topolo − gical space $E . $ Wewrite

\ [ J = \{ i \ in I: \mu (E { i } ) = 0 \} . \ ]

\noindent Since the measure $ \mu $ i s $ \sigma − $ finite , we have the inequality

\ [ card ( I \setminus J) \ leq \omega { 0 } . \ ] 48 .. A period KHARAZISHVILI \noindentFurther comma48 \quad since JA subset . KHARAZISHVILI I and card open parenthesis I closing parenthesis is not measurable in a wide sense comma we find \noindentthat card openFurther parenthesis , since J closing $ J parenthesis\subset is not measurableI $ and in card a wide $ sense ( too I period ) $ From is not measurable in a wide sense , we find thethat latter card fact we $ ( J ) $ is not measurable in a wide sense too . From the latter fact we readilyreadily conclude conclude that that the equality the equality mu parenleftbig cup in i J E sub i parenrightbig = 0 \ [ \mu ( \cup48{\ A .in KHARAZISHVILI{ i } J } E { i } ) = 0 \ ] is valid period Further , since J ⊂ I and card (I) is not measurable in a wide sense , we find Let us now assumethat that card for any (J) index is not i measurablein I backslash inJ a the wide set sense F sub too i is . aFrom separab the latterhyphen fact we le support for the restrictionreadily conclude of the measurethat the muequality onto the space E sub i period .... Then \noindentit is easy toi s see v a that l i d the . set cup sub i in I backslash J F sub i is a separable support for the mea hyphen µ(∪ E ) = 0 \ hspacesure mu∗{\ periodf i l l } Let us now assume that for any∈iJ i index $ i \ in I \setminus J $Proposition the s e t 3 period $ Fis .. valid{ Leti . E} be$ a istopological a separab space− and open parenthesis E sub i closing parenthesis i in I be a countable Let us now assume that for any index i ∈ I \ J the set Fi is a separab - \noindent le support for the restriction of the measure $ \mu $ onto the space family of subspacesle of support E everyone for the of which restriction has th of e the property measure openµ onto parenthesis the space S closingEi. parenthesisThen $ E { i } . $ \ h f i l l Then period .... Then the it is easy to see that the set ∪i∈I\J Fi is a separable support for the mea - sure space µ. \noindent it is easy to see that the set $ \cup { i \ in I \setminus E to the power of prime = iProposition to the power 3 of . cup inLet I toE thebe a power topological of E sub space i subset and ( EEi)i ∈ I be a J }hasF the{ propertyi }$ open iscountable aparenthesis separable S closing support parenthesis for the too period mea − sureProof period$ \mu .. Let.family mu $ be of an subspaces arbitrary of sigmaE everyone hyphen finite of which Borel has measure th e property in the topological(S). Then space the E to the power of primespace period For any index i in I denote by mu i the trace of the measure mu on the\ hspace set ∗{\ f i l l } Proposition 3 . \quad Let $ E $ be a topological space and $ (E sub E i period{ i } Recall) that i the\ measurein I mu $ i be is defined aE0 countable= i by∪ ∈ theIEi formula⊂ E mu i open parenthesishas the X closing property parenthesis(S) too . = mu to the power of * open parenthesis X closing parenthesis\noindent family of subspaces of $ E $ everyone of which has th e property $ ( S ) . $Proof\ h . f i l l LetThenµ be the an arbitrary σ− finite Borel measure in the topological where X is an arbitraryspace BorelE0. For subset any of index the spacei ∈ I denote E sub i by andµi muthe to trace the power of the of measure * is theµ outeron the measure associatedset withE . muRecall period that .. theBy the measure conditionµi is comma defined .. by for the the formula measure mu i .. there \noindentexists a separablespace supporti F sub i subset E sub i period Hence it is easy to see that the closure of the set cup sub i in I F sub i in the space E to theµi(X power) = µ of∗(X prime) is a separable support for the original\ [ E ˆ{\prime } = i ˆ{\cup }\ in I ˆ{ E { i }}\subset E \ ] ∗ measure mu periodwhere X is an arbitrary Borel subset of the space Ei and µ is the outer measure Example 7 period ..associated Let alpha with be aµ. cardinalBy the number condition strictly , larger for the than measure the cardiµi hyphenthere exists a \noindent has the property $ ( S ) $ too . nality of the continuumseparable and let support T be theFi ⊂ unitEi. circleHence in it the is Euclidean easy to see plane that the closure of the set 0 R to the power of 2∪ periodi∈I Fi in .. the Consider space E T asis a compactseparable commutative support for the group original and denote by mu the \noindentprobabilityProof Haar measure .measure\quad inµ. theLet compact $ \mu commutative$ be an group arbitrary Row 1 alpha $ \ Rowsigma 2 period− .$ is easy finite Borel measure in the topological space $to E check ˆ{\ thatprime the topological} Example. $ group For 7 any . T toLet index theα powerbe a$ of cardinal i alpha\ in is number not separableIstrictly $ denote period larger by than .. According $ the\mu cardi to thei $ the trace of- nality the measureof the continuum $ \mu and$ let onT be the the s unit e t circle in the Euclidean plane $result E { of Examplei } .6 $R the2. Recall HaarConsider measure thatT as mu the a compact is measure not concentrated commutative $ \mu on group a separablei and$denote is defined by µ the by proba- the formula α subset of Row 1 alphability Row Haar 2 periodmeasure . thein the topological compact commutativespace T to the group powerT of alphaIt is easy does to not check have the\ [ property\mu i ( X ) = \mu ˆ{ ∗ } (X) \ ] . α open parenthesis Sthat closing the parenthesis topological period group T .. Weis not thus separable conclude . that According the topological to the product result of of spaces having the Example 6 the Haar measure µ is not concentrated on a separable subset of \noindent where $α X $ is an arbitrary Borel subset of the space $ E { i }$ property open parenthesisT Hence S closingthe topological parenthesis space does notT α does have not comma have generally the property speaking (S). commaWe thisand property $ \mu periodˆ{ ∗ }$. is the outer measureExample associated 8 period ..thus Let concludewith E be a discrete$ that\mu the topological topological. $ \quad space productBy with ofthe the spaces cardinality condition having the , \quad for the measure $ \omegamu subi $ 1 period\quadproperty ..ther According e(S) does to not the have classical , generally result of speaking Ulam open , this parenthesis property . see comma for exampleexists comma a separable .. brackleftExample support 1 brackright 8 . $ closing FLet{E parenthesisibe}\ a discretesubset the topologicalE { spacei } with. $thecardinality Hence it is easy to see that the closure o fcardinal the s number e t $ omega\cupω1. subAccording{ 1i is not\ in to measurable theI classical} F in resulta{ widei of} sense$ Ulam in period ( see the , Therefore for space example E $Eˆ is , a Luzin[ 1{\ ] )prime the }$ istopological a separable space support andcardinal comma number for in the particularω1 is original not comma measurable has in the a propertywide sense open . Therefore parenthesisE is S a Luzinclosing parenthesis period .. Nexttopological consider space and , in particular , has the property (S). Next consider the \noindentthe cardinalmeasure numbercardinal omega $ \mu number sub 1. equippedω $1 equipped with its with order its ordertopology topology period . It It is is obvious obvious that that the the topological spacetopological omega sub space 1 isω lo1 callyis lo cally compact compact and lo and cally lo cally countable countable period . .. Denote Denote Example 7 . \quad Let $ \alpha $ be a cardinal number strictly larger than the cardi − nality of the continuum and let $ T $ be the unit circle in the Euclidean plane

\noindent $ R ˆ{ 2 } . $ \quad Consider $ T $ as a compact commutative group and denote by $ \mu $ the probability Haar measure in the compact commutative group $\ l e f t .T\ begin { array }{ c}\alpha \\ . \end{ array } I t \ right . $ i s easy to check that the topological group $ T ˆ{\alpha }$ is not separable . \quad According to the result of Example 6 the Haar measure $ \mu $ is not concentrated on a separable subset o f $\ l e f t .T\ begin { array }{ c}\alpha \\ . \end{ array }Hence\ right . $ the topological space $ T ˆ{\alpha }$ does not have the property $ ( S ) . $ \quad We thus conclude that the topological product of spaces having the

\noindent property $ ( S ) $ does not have , generally speaking , this property .

Example 8 . \quad Let $ E $ be a discrete topological space with the cardinality $ \omega { 1 } . $ \quad According to the classical result of Ulam ( see , for example , \quad [ 1 ] ) the cardinal number $ \omega { 1 }$ is not measurable in a wide sense . Therefore $E$ is a Luzin topological space and , in particular , has the property $ ( S ) . $ \quad Next consider the cardinal number $ \omega { 1 }$ equipped with its order topology . It is obvious that the topological space $ \omega { 1 }$ is lo cally compact and lo cally countable . \quad Denote ON SEPARABLE SUPPORTS OF BOREL MEASURES .. 49 \ hspaceby the∗{\ symbolf i l l mu}ON the SEPARABLE st andard Dieudonne SUPPORTS two OF hyphen BOREL valued MEASURES probability\quad continuous49 measure defined on the Borel sigma hyphen algebra of the space omega sub 1 period .. Notice that the\noindent by the symbol $ \mu $ the st andard Dieudonne two − valued probability continuous measuremeasure mu defined is not concentrated on the Borel on a separable $ \sigma subspace− $ of omega algebra sub 1 of and the therefore space $ \omega { 1 } . $the\ spacequad omegaNotice sub 1 that does notthe have the property open parenthesis S closing parenthesis period At themeasure same time $ the\mu space$ omega is not sub concentrated 1 on a separable subspace of $ \omega { 1 }$ ON SEPARABLE SUPPORTS OF BOREL MEASURES 49 andis antherefore inj ective continuous image of the discrete space E period .. We have thereby the space $ \omegaby the symbol{ 1 }$µ the does st andard not have Dieudonne the two property - valued probability $ ( S continuous ) . $ shown that the propertymeasure open defined parenthesis on the SBorel closingσ− parenthesisalgebra of the is not space commaω . generallyNotice that speaking the commaAt the invariant same time even under the space $ \omega { 1 }$ 1 is an inj ectivemeasure continuousµ is not concentratedimage of the on a discrete separable subspace space of $ω1 Eand . therefore $ \quad the We have thereby inj ective continuousspace mappingsω does period not have the property (S). At the same time the space ω is an shownThe next that proposition the property indicates1 a $ certain ( S relationship ) $ isbetween not the , generally topo hyphen speaking1 , invariant even under inj ective continuousinj ective mappings continuous image . of the discrete space E. We have thereby shown logical spaces havingthat the the property property open (S) is parenthesis not , generally S closing speaking parenthesis , invariant and even Luzin under subspaces inj ective of these spaces period continuous mappings . TheProposition next proposition 4 period .. Assume indicates that th a e continuum certain hypothesisrelationship holds and between le t E the topo − logical spaces havingThe next the proposition property indicates $ ( a S certain ) $ relationship and Luzin between subspaces the topo of - these spaces . be an arbitrary Hausdorfflogical spaces topological having space thewith property th e ( cardinalityS) and Luzin of the subspaces conti hyphen of these spaces . nuum period If to theProposition power of E has 4 th. e propertyAssume thatopen th parenthesis e continuum S closing hypothesis parenthesis holds and comma le t thProposition en at least one 4 of . the\quad next twoAssume ass ertio that ns th e continuum hypothesis holds and le t $ E $ E be an arbitrary Hausdorff topological space with th e cardinality of the conti is valid : E be an arbitrary- Hausdorff nuum . If topologicalhas th e property space(S) with, th en th at least e cardinality one of the next of two the ass conti − 1 closing parenthesisertio E is ns a s is eparable valid : topological space semicolon nuum2 closing $ parenthesis. I f ˆ{ EE contains}$ has s ome th Luzin e property subspace with $ th ( e cardinality S ) of , the $ continuum th en at period least one of the next two ass ertio ns i s v a l i d : 1)E is a s eparable topological space ; Proof period .. Assume2) thatE contains the given s space ome Luzin E is not subspace separable with period th e cardinality Consider the of the inj continuum ec hyphen tive family of all closed separable subsets of E period It is easy to check that the \ centerline { $ 1. ) E$ is a s eparable topological space ; } cardinality of this familyProof . is equalAssume to the that cardinality the given of space the continuumE is not separable period There . Consider hyphen the inj fore this family can be represented as an omega sub 1 hyphen sequence open parenthesis F sub xi \ hspace ∗{\ f i l l } $ec 2 - tive ) family E of $ all contains closed separable s ome subsets Luzin of E. subspaceIt is easy with to check th that e cardinality the of the continuum . closing parenthesis subcardinality xi less omega of this sub family 1 period is equal .. Further to the comma cardinality of the continuum . There - using the method offore transfinite this family recursion can be comma represented define as an an injω ective− sequence omega ( subF ) 1< hyphen ω . Further sequence \noindentopen parenthesisProof x . sub\quad xi closingAssume parenthesis that the sub xi given less omega space1 sub $ 1 ..E of $ pointsξ isξ not of1 the separable space E . Consider the inj ec − tive family of all, using closed the method separable of transfinite subsets recursion of ,$ define E an . inj $ ective It isω1− easysequence to check that the period .. Assume that( forx ) an ordinalof points number of the space E. Assume that for an ordinal number cardinalityxi less omega ofsub this 1ξ ..ξ<ω we family1 have already is equal defined to a the partial cardinality xi hyphen sequence of the of continuum points .. open . There − fore this familyξ can < be ω1 representedwe have already as defined an a$ partial\omegaξ− sequence{ 1 } of − points$ sequence (xζ )ζ < parenthesis x sub zetaξ. closingConsider parenthesis the set sub zeta less xi period $ (Consider F {\ thexi set } ) {\ xi } < \omega { 1 } . $ \quad Further , using the method of transfinite recursion , define an inj ective $ \omega { 1 } P = open parenthesis zeta to the powerP of= cup (ζ∪ less< ξF xi) F∪ sub(ζ∪ zeta< ξ{x closing}). parenthesis cup open −parenthesis$ sequence zeta to the power of cup less xi open brace xζ sub zeta closingζ brace closing parenthesis $ ( x {\ xi } ) {\ xi < \omega { 1 }}$ \quad of points of the space period It is obvious that the set P is separable . Therefore we have E \ P = ∅. Take $ E . $ \quad Assume that for an ordinal number It is obvious that theas setxξ Pany is separablepoint from period the nonempty Therefore set weE have\ P. E backslashThereby Pwe = shall varnothing construct period

Take the needed ω1− sequence (xξ)ξ<ω1 . After that we write \noindentas x sub xi any$ \ pointxi from< the\omega nonempty{ set1 } E$ backslash\quad Pwe period have .. alreadyThereby we defined shall construct a partial $ \ xi − $ sequence of points \quad $ ( x {{\x } zeta } ) {\zeta } < the needed omega sub 1 hyphen sequence open parenthesisX = ξ ∪<ω x1 subξ. xi closing parenthesis sub xi less omega\ xi sub. $ 1 period After that we write Consider the setClearly , card (X) is equal to the cardinality of the continuum . Now , taking X = xi cup less omegainto sub consideration 1 to the power the offact open that brace our topological x xi to the power space ofE closinghas the brace property sub period Clearly comma card(S open), it is parenthesis not diﬃcult X to closing check parenthesis that X is a is Luzin equal subspace to the cardinality of E. of the continuum\ [ P period = Now ( comma\zeta takingˆ{\cup } < \ xi F {\zeta } ) \cup ( \zeta ˆ{\cup } < \ xi \{ x {\Remarkzeta . }\}Assume). again\ that] the continuum hypothesis holds and let into consideration theE be fact an that arbitrary our topological nonseparable space topological E has the propertyspace with the cardinality of the open parenthesis Scontinuum closing parenthesis . Then comma there exists it is not a subspace diﬃcult toX checkof E satisfying that X is thea Luzin following subspace of E period \noindentRemark periodIt is .. Assume obvious again that that the the continuum set $ P hypothesis $ is separable holds and let . E Therefore we have $ E \setminus P = \ varnothing . $ Take be an arbitrary nonseparable topological space withconditions the cardinality : of the ascontinuum $ x {\ periodxi ..}$ Then any there point exists from a subspace the nonemptyX of E satisfying set the $ followingE \setminus P . $ \quad Thereby we shall construct conditions : a ) card (X) is equal to the cardinality of the continuum ; b ) every uncountable a closing parenthesisset cardY ⊂ openX is parenthesis nonseparable X . closing parenthesis is equal to the cardinality of the continuum\noindent semicolonthe needed $ \omega { 1 } − $ sequence $( x {\ xi } ) {\ xi < b closing\omega parenthesis{ 1 }} every. $ uncountable After that set Y we subset write X is nonseparable period \ [ X = \ xi \cup { < \omega } { 1 }ˆ{\{ x }\ xi ˆ{\}} { . }\ ]

\noindent Clearly , card $ ( X ) $ is equal to the cardinality of the continuum . Now , taking into consideration the fact that our topological space $ E $ has the property

\noindent $ ( S ) , $ it is not difficult to check that $X$ is a Luzin subspace of $ E . $

Remark . \quad Assume again that the continuum hypothesis holds and let $ E $ be an arbitrary nonseparable topological space with the cardinality of the continuum . \quad Then there exists a subspace $ X $ of $ E $ satisfying the following

\ begin { a l i g n ∗} conditions : \end{ a l i g n ∗}

\noindent a ) card $ ( X ) $ is equal to the cardinality of the continuum ; b ) every uncountable set $ Y \subset X $ is nonseparable . 50 .. A period KHARAZISHVILI \noindentIn particular50 comma\quad ..A if the. KHARAZISHVILI given topological space E is metrizable comma .. then the mentioned space X has the property open parenthesis S closing parenthesis although all uncountable subsetsIn particular , \quad if the given topological space $ E $ is metrizable , \quad then the mentionedof X are nonseparable space $ open X $ parenthesis has the moreover property comma $ ( in this S case ) X$ is although a Luzin space all closing uncountable subsets parenthesis period \noindentThe proofof the $ existence X $ are of the nonseparable mentioned space ( X moreover subset E is , quite in similar this case $ X $ is a Luzin space ) . 50 A . KHARAZISHVILI to that of PropositionIn 4 period particular , if the given topological space E is metrizable , then the TheExample proof 9 of period the .. existence Assume that of the thecontinuum mentioned hypothesis space holds $and X consi\subset hyphen E $ is quite similar to that of Propositionmentioned 4 space . X has the property (S) although all uncountable subsets der any Sierpinski setof X Eare on the nonseparable real line R ( open moreover parenthesis , in this about case X Sierpinskiis a Luzin sets space see comma ) . for example comma .... brackleftThe proof 1 of brackright the existence closing of the parenthesis mentioned period space ....X ⊂ SinceE is Equite is a similar separable to metricExample space 9 comma . \quad it hereditarilyAssume that has the the continuum hypothesis holds and consi − der any Sierpinskithat of set Proposition $ E $ 4 on. the real line $R ( $ about Sierpinski sets see , for property open parenthesisExample S closing 9 . parenthesisAssume period that the At continuum the same t hypothesis ime comma holds it is and not considifficult - to show that the space E \noindent exampleder , any\ h Sierpinski f i l l [ 1 set ]E )on . the\ h freal i l l lineSinceR( about $ E Sierpinski $ is setsa separable see , for metric space , it hereditarily has the does not contain anyexample uncountable , [ 1 ] Luzin) . Since subspaceE is a period separable metric space , it hereditarily has the Example 1 0 period .. Let E be an arbitrary nonseparable metric space period .. As is \noindent propertyproperty $ ( (S). SAt the ) same . $ t ime At , itthe is not same difficult t ime to show , it that is the not space difficultE to show that the space well known commadoes E contains not contain a discrete any uncountable subspace D with Luzin the subspace cardinality . omega sub 1 period Using $ Ethe $ result of Ulam mentioned above comma we see that D is an uncountable Luzin does not contain anyExample uncountable 1 0 . Let LuzinE be subspacean arbitrary . nonseparable metric space . As subspace of E periodis well known ,E contains a discrete subspace D with the cardinality ω . Using Notice now that there may also exist nonseparable nonmetrizable topo hyphen 1 Example 1 0 . \quadthe resultLet of $ Ulam E $ mentioned be an arbitrary above , we see nonseparable that D is an uncountable metric space Luzin . \quad As i s logical spaces havingsubspace the property of E. open parenthesis S closing parenthesis period .. We wish to show herewell that known such $ , E $ contains a discrete subspace $ D $ with the cardinality $ \omega { 1 } .Notice $ Using now that there may also exist nonseparable nonmetrizable topo - spaces can be encounteredlogical spaces even among having linearly the property ordered (S) topological. We wish spaces to show period here that such spaces theLet result us recall of a few Ulam simple mentioned notions from above the theory , we of see ordered that sets $ period D $ Let is open an uncountable parenthesis E Luzin subspace of $Ecan be . encountered$ even among linearly ordered topological spaces . Let us comma less or equal closingrecall parenthesisa few simple notions from the theory of ordered sets . Let (E, ≤) be an be an arbitrary linearlyarbitrary ordered linearly set period ordered .. setWe . say Wethat say two that nonempty two nonempty open in openhyphen in - tervals Noticetervals now in E that are disj there oint if may the right also end exist hyphen nonseparable point of one of these nonmetrizable intervals is topo − logical spaces havingin E are disjthe oint property if the right $ end ( -point S )of one . of $ these\quad intervalsWe wish is less to than show here that such less than or equal toor the equal left to end the hyphen left end point - point of the of the other other interval interval period .In .. In particular particular , any comma two spacesany two can disj oint be encounteredintervals do not evenhave the among common linearly points period ordered Further topological comma we say spaces . Let us recall adisj few oint simple intervals notions do not have from the the common theory points of . ordered Further , sets we say . that Let a that a linearly orderedlinearly set open ordered parenthesis set (E, ≤ E) comma has the less Suslin or equal property closing if every parenthesis family ofhas nonempty the Suslin property$ ( E if every , family\ leq ) $ be an arbitraryopen linearly disj oint ordered intervals is set at most . \quad countableWe say. We that wish two to em nonempty - phasize that open in − of nonempty open disjthis oint definition intervals immediately is at most implies countable that period for arbitrary .. We wish linearly to em ordered hyphen sets the tervalsphasize that in this $ E definition $ are immediately disj oint implies if the that rightfor arbitrary end − linearlypoint of one of these intervals is less than or equalSuslin to property the left is the end hereditary− point one ( of in thisthe connection other interval we remark . here\quad that In particular , ordered sets the Suslinthe separability property is the property hereditary and the one Suslin open parenthesis property are in not this the connection hereditary we ones anyremark two here disj that oint the separability intervals property do not and have the Suslin thecommon property are points not . Further , we say that a linearlyfor ordered arbitrary set topological $ ( spaces E ) , . Recall\ leq that) the $ Suslin has line the is Suslin any linearly property if every family the hereditary onesordered for arbitrary set E topologicalhaving the Suslin spaces property closing parenthesis and being non period - separable Recall that with the respect Suslin ofline nonempty is any linearly open ordered disj set oint E having intervals the Suslin is property at most and countable being non hyphen . \quad We wish to em − phasize that thisto the definition order topology immediately in (E, ≤). As implies is well known that , the for existence arbitrary of the linearlySuslin separable with respectline is to not the provable order topology in the modern in open set parenthesis theory and does E comma not contradict less or equal this theory closing parenthesisordered period sets .. the As is Suslin well known property comma is the hereditary one ( in this connection we remark here that( see the , for separability example , [ 2 ] ) property . We shall and see thebelow Suslin property are not the existence of thethat Suslin any line Suslin is not line provable equipped in withthe modern its order set topology theory and has the property (S). In thedoes hereditary not contradict ones this theory for arbitrary open parenthesis topological see comma for spaces example ) comma . Recall brackleft that 2 brack-the Suslin line is any linearlythe first ordered place we need set the $ following E $ having auxiliary the st atement Suslin . property and being non − right closing parenthesis periodLemma .. We . shallLet see(E, below≤) be a lin early o rdered s e t having th e Suslin separablethat any Suslin with line respect equipped to with the its order topology topology has the in property $ ( open E parenthesis , \ leq S closing) . $ \quad As isproperty well known and considered , as a topological space with respect to its order topology parenthesis period . Fur - th er , le t µ be an arbitrary Borel measure in E and le t V be an theIn the existence first place weof need the the Suslin following line auxiliary is not st atement provable period in the modern set theory and does not contradictarbitrary this open theory ( see , for example , [ 2 ] ) . \quad We shall see below Lemma period .. Letsubset open of parenthesisE locally negligibleE comma with lessor respect equal to closing th e measure parenthesisµ( the .. be latter a lin means early o rdered s e t having ththat e Suslin for any property po int x ∈ V there exists an open interval V (x) ⊂ V containing \noindentand consideredthat as any a topological Suslin space line with equipped respect to with its order its topology order topologyperiod .. Fur has hyphen the property $ (th er S comma ) le t . mu $ be an arbitrary Borel measure in E and le t V be an arbitrary open Insubset the of first E locally place negligible we need with respect the following to th e measure auxiliary mu open parenthesisst atement the . latter means that for any po int x in V there exists an open interval V open parenthesis x closing parenthesis Lemmasubset V . containing\quad Let $ ( E , \ leq ) $ \quad be a lin early o rdered s e t having th e Suslin property and considered as a topological space with respect to its order topology . \quad Fur − th er , l e t $ \mu $ be an arbitrary Borel measure in $ E $ and le t $ V $ be an arbitrary open

\noindent subset of $ E $ locally negligible with respect to th e measure $ \mu ( $ the latter means

\noindent that for any po int $ x \ in V $ there exists an open interval $ V ( x ) \subset V $ containing ON SEPARABLE SUPPORTS OF BOREL MEASURES .. 5 1 \ hspacethe point∗{\ xf and i l l } beingON SEPARABLE negligible with SUPPORTS respect to OF mu BOREL closing MEASURES parenthesis period\quad ....5 Then 1 th e whole s e t V is \noindentnegligible withthe respect point to $ mu x comma $ and i period being e period negligible comma the with equality respect mu open to parenthesis $ \mu V) closing. $ \ parenthesish f i l l Then = 0 th is fulfilled e whole period s e t $V$ is Proof period Let us define a binary relation thicksim in the given set V by the formula \noindentx thicksimnegligible y Leftrightarrow with mu open respect parenthesis to brackleft$ \mu x comma, $ i y brackright. e . , theclosing equality parenthesis $ \mu ( V ) = 0$ isfulfilled.ON SEPARABLE SUPPORTS OF BOREL MEASURES 5 1 = 0 and mu open parenthesisthe point brackleftx and beingy comma negligible x brackright with respect closing to parenthesisµ). Then = th 0 e period whole s e t V ProofClearly . comma Let us the define relation a thicksim binary is relation an equivalence $ relation\sim $ in in the the set V given period set .. Let $V$ open by the formula is parenthesis V sub i closingnegligible parenthesis with respect i in I to µ, i . e . , the equality µ(V ) = 0 is fulfilled . Proof be the partition of. V canonicallyLet us define associated a binary withrelation thicksim∼ in the period given Then set V itby is not the difficult formula \ [to x check\sim that y \Leftrightarrow \mu ( [ x , y ] ) = 0 and \mu ( [ y , x ] ) = 0 . \ ] a closing parenthesis every set V subx i is∼ openy ⇔ µ in([x, E y semicolon]) = 0 and µ([y, x]) = 0. b closing parenthesis for any index i in I we have mu open parenthesis V sub i closing parenthesis = 0 semicolon Clearly , the relation ∼ is an equivalence relation in the set V. Let (Vi)i ∈ I \noindentc closing parenthesisClearlybe card , the the partition open relation parenthesis of V canonically $ I closing\sim associatedparenthesis$ is an with less equivalence∼ or. equalThen omega it is relation not sub difficult 0 period in to the set $ VNow .we $ immediately\quad checkLet obtain that $ ( V { i } ) i \ in I $ be the partition of $ V $ canonically associated with $ \sim . $ Then it is not difficult Line 1 mu open parenthesis V closing parenthesisa ) every = set sumVi is mu open open in E parenthesis; V sub i closing to check that parenthesis = 0 period Line 2 i in I b ) for any index i ∈ I we have µ(Vi) = 0; It is easy to see that in the formulation of the above lemma we can \ centerline {a ) every set $V { i }$ isopenin $E ;$ } replace an arbitrary open locally negligible set V subsetc)card( EI by) ≤ anω0 arbitrary. Borel lo cally negligible set X subset E period NoticeNow also we that immediately in the same obtain formulation of the \ centerlinelemma the Suslin{b ) property for any of index the considered $ i linearly\ in orderedI $ set we open have parenthesis $ \mu E comma(V less{ ori } equal) = closing 0 parenthesis ; $ } is X µ(V ) = µ(Vi) = 0. quite essential period Indeed comma let us take the ordinal number omega sub 1 equipped with it s \ [order c topology ) card period ( Denote I by ) mu the\ leq Dieudonne\omega probability{ 0 } continuousi ∈ .I \ ] measure defined on the Borel sigmaIt is easyhyphen to see algebra that of in the the space formulation omega of sub the 1 period above lemma Then it we is canobvious replace that the \ centerline {Now wean arbitrary immediately open locally obtain negligible} set V ⊂ E by an arbitrary Borel lo cally whole space omeganegligible sub 1 is lo set callyX ⊂ negligibleE. Notice with also respect that in to the the same measure formulation mu but of we the lemma the have the equality muSuslin open property parenthesis of the omega considered sub 1 closing linearly parenthesis ordered set = ( 1E, period≤) is quite essential \ [ \Usingbegin the{ a above l i g n e lemmad }\mu comma( we V obtain ) = \sum \mu (V { i } ) = 0 . \\ . Indeed , let us take the ordinal number ω1 equipped with it s order topology Proposition 5 period. Denote .. Let byopenµ the parenthesis Dieudonne E comma probability less orcontinuous equal closing measure parenthesis defined on .. bethe a linearlyi ordered\ in sI e\ tend having{ a li the g n eSuslin d }\ ] Borel σ− algebra of the space ω1. Then it is obvious that the whole space ω1 property and equippedis lo with cally its negligible order topology with respect period to .. theThen measure th e topoµ but logical we space have theE equality has the property open parenthesis S closing parenthesis period It is easy to seeµ(ω that1) = 1 in. the formulation of the above lemma we can Proof period .. If our topological space EUsing is separable the above comma lemma then , we there obtain is nothing to prove period replaceTherefore an let arbitrary us assume that open E is locally nonseparable negligible comma i period set e $ period V comma\subset that theE $ linearly by an arbitrary Borel lo cally negligibleProposition set $ X 5 . \subsetLet (E,E≤) .be $ a linearly Notice ordered also s that e t having in the the same formulation of the ordered Suslin property and equipped with its order topology . Then th e topo logical lemmaset open the parenthesis Suslin E property comma less of or the equal considered closing parenthesis linearly is the Suslin ordered line period set $Let ( mu Ebe , \ leq ) $ ispace s E has the property (S). any sigma hyphen finiteProof Borel . measureIf our in topological E period We space E is separable , then there is nothing to quiteintroduce essential the notation . Indeed , let us take the ordinal number $ \omega { 1 }$ equipped with itprove s . Therefore let us assume that E is nonseparable , i . e . , that the linearly V = open brace x inordered E : E set .. is (E, lo≤ cally) is negligiblethe Suslin at line the . Letpointµ ..be x any .. withσ− finite respect Borel to .. measure mu closing in braceorder period topology . Denote by $ \mu $ the Dieudonne probability continuous measure defined on the BorelE. We introduce $ \sigma the notation− $ algebra of the space $ \omega { 1 } Obviously commaV = is{ anx ∈ openE : subsetE is of lo E cally lo cally negligible negligible at the with point respectx towith respect to µ}. . $the Then measure it mu is period obvious Therefore that according the to the above lemma we have mu open parenthesis V whole space $ \omegaObviously{ 1,V}$isis an open lo cally subset of negligibleE lo cally negligible with withrespect respect to to the the measure closing parenthesis = 0measure period µ. Therefore according to the above lemma we have µ(V ) = 0. $ \Considermu $ but the closed we set F = E backslash V period .. We assert that this set is separable period have the equalityConsider $ \mu the closed( set\Fomega= E\V.{ 1We} assert) that = this 1 set is . separable $ . Assume Assume that the oppositethat the is opposite true : .. is the true set : F is the not set separableF is not periodseparable .. This . This means means that the that the linearly ordered set open parenthesis F comma less or equal closing parenthesis is also the \ centerline { Usinglinearly the above ordered lemma set (F, ≤ ,) we is also obtain the Suslin} line . Let λ be the restriction Suslin line period .. Letof lambda the measure be theµ onto the topological space F. The measure λ is certainly σ− restriction of the measurefinite and mu , onto moreover the topological , has the following space F periodproperty The : the measureλ− measure lambda is Propositioncertainly sigma 5 hyphen . \quad finiteLet and $ comma ( Emoreover , comma\ leq has) the $ following\quad propertybe a linearly : the lambda ordered s e t having the Suslin hyphenproperty measure and equipped with its order topology . \quad Then th e topo logical space $ E $ hastheproperty $( S ) .$

\noindent Proof . \quad If our topological space $ E $ is separable , then there is nothing to prove . Therefore let us assume that $ E $ is nonseparable , i . e . , that the linearly ordered s e t $ ( E , \ leq ) $ is the Suslin line . Let $ \mu $ be any $ \sigma − $ finite Borel measure in $E . $ We introduce the notation

\ centerline { $ V = \{ x \ in E : E $ \quad is lo cally negligible at the point \quad $ x $ \quad with respect to \quad $ \mu \} . $ }

Obviously $ , V $ is an open subset of $ E $ lo cally negligible with respect to the measure $ \mu . $ Therefore according to the above lemma we have $ \mu ( V ) = 0 . $

\noindent Consider the closed set $ F = E \setminus V . $ \quad We assert that this set is separable . Assume that the opposite is true : \quad the set $ F $ is not separable . \quad This means that the linearly ordered set $ ( F , \ leq ) $ is also the Suslin line . \quad Let $ \lambda $ be the restriction of the measure $ \mu $ onto the topological space $ F . $ The measure $ \lambda $ i s c e r t a i n l y $ \sigma − $ finite and , moreover , has the following property : the $ \lambda − $ measure 52 .. A period KHARAZISHVILI \noindentof any nonempty52 \quad openA interval . KHARAZISHVILI in F is strictly positive period Now let us consider the topological product F times F equipped with the product hyphen measure lambda times lambda period\noindent of any nonempty open interval in $ F $ is strictly positive . Now let us consider theAccording topological to the classical product result $ of F Kupera\times open parenthesisF $ equipped see comma with for the example product comma− ..measure brackleft$ \lambda 2 brackright\times closing\ parenthesislambda in. the $ Accordingspace F times to F the there classical exists an uncountable result offamily Kupera of nonempty ( see open , for pairwise example , \quad [ 2 ] ) in the space $ F \times52 A . KHARAZISHVILIF $ there exists an uncountable family of nonempty open pairwise disj oint rectanglesof period any nonempty .. It is obvious open interval that the in openF is parenthesis strictly positive lambda . Nowtimes let lambda us consider closing parenthesisdisj oint hyphen rectangles measure of . everyone\quad ofIt is obvious that the $ ( \lambda \times \lambda ) − $the measure topological of product everyoneF × F ofequipped with the product - measure λ × λ. these rectangles is strictlyAccording positive to the period classical .. But result the product of Kupera hyphen ( see measure , for example lambda , times [ 2 lambda ] ) in isthese sigma hyphen rectangles is strictly positive . \quad But the product − measure $ \lambda \times \lambda the$ space i sF $ ×\Fsigmathere exists− $ an uncountable family of nonempty open pairwise finite and thereforedisj such oint a family rectangles of rectangles . It is cannot obvious exist that period the (λ ..× Theλ)− obtainedmeasure of everyone of finitecontradiction and therefore proves our proposition such a family period of rectangles cannot exist . \quad The obtained contradiction provesthese rectangles our proposition is strictly positive . . But the product - measure λ × λ is σ− Notice that a resultfinite analogous and therefore to Proposition such a 5 family and concerning of rectangles Boolean cannot exist . The obtained algebras with measurescontradiction was obtained proves by our several proposition authors . open parenthesis J period L period Kelley commaNotice D that period a Ma result hyphen analogous to Proposition 5 and concerning Boolean algebras with measuresNotice was that aobtained result analogous by several to Proposition authors 5 and ( J concerning . L . Kelley Boolean , D . Ma − haram comma andalgebras others closing with measures parenthesis was period obtained by several authors ( J . L . Kelley , D . As a consequence of Proposition 5 we have \noindent haramMa , and - others ) . Proposition 6 periodharam .. Assume , and othersthat the ) . continuum hypothesis holds period .. Then any Suslin line contains s ome Luzin subspace with the cardinality of the conti hyphen \ centerline {As a consequenceAs of a Proposition consequence of 5 Proposition we have 5} we have nuum period Proposition 6 . Assume that the continuum hypothesis holds . Then To prove this propositionany Suslin it is enough line contains to apply s ome the Luzin results subspace of Propositions with the cardinality of the conti Proposition4 and 5 with 6 regard . \quad to theAssume fact that thatthe cardinality the continuum of any Suslin hypothesis line is holds . \quad Then any Suslin line contains- s ome Luzin subspace with the cardinality of the conti − less than or equal tonuum the cardinality . of the continuum period .. The latter fact readily follows comma for example comma from the well hyphen known Erd dieresis-o s endash Rado \noindent nuum . To prove this proposition it is enough to apply the results of Propositions 4 combinatorial and 5 with regard to the fact that the cardinality of any Suslin line is less than theorem open parenthesis about this theorem and its various applications see brackleft 2 brackright To prove this propositionor equal to the it cardinality is enough of the to continuum apply the . The results latter fact of Propositionsreadily follows closing parenthesis period, for example , from the well - known Erdo ¨ s – Rado combinatorial theorem ( 4Naturally and 5 with comma regard there arises to the the following fact that question the : cardinality.. does every uncountable of any Suslin line is less than or equalabout to this the theorem cardinality and its various of applicationsthe continuum see [ 2 . ] )\ .quad The latter fact topological space containNaturally an uncountable , there arises subspace the followinghaving the question property : open does parenthesis every uncountable S closing parenthesisreadily ? follows , for example , from the well − known Erd $ \ddot{o} $ s −− Rado combinatorial theorem ( abouttopological this theorem space contain and its an uncountablevarious applications subspace having see the [ property 2 ] ) (S . )? Our next example showsOur next that example the answer shows to that this questionthe answer is negative to this question period is negative . Example 1 1 period .. Assume that the continuum hypothesis holds period .. Let E be \ hspace ∗{\ f i l l } NaturallyExample , 1 there 1 . arisesAssume that the the following continuum hypothesisquestion holds : \quad . LetdoesE every uncountable an arbitrary Sierpinskibe an set arbitrary on the real Sierpinski line R setperiod on the Equip real the line setR. REquip with the the set usualR with the usual density topology open parenthesis about the density topology on R see comma in particular comma \noindent topologicaldensitytopology space contain ( about the an density uncountable topology on subspaceR see , in particular having the, [ 1 property ] ) .. brackleft 1 brackright closing parenthesis period∗ $ ( S ) ? $. Denote by the symbol R the set R equipped with the density topology and Denote by the symbol R to the power of * .. the set R equipped with the density topology∗ Our next exampleconsider shows the that Sierpinski the answer set E as to a subspace this question of the topological is negative space R . Now and consider the Sierpinskiit is not set difficult E asa to subspace check that of the every topological uncountable space subspace of the topological R to the power of *space periodE Nowdoes itnot is have not difficult the property to check (S) that. Notice every hereuncountable that an subspace of the Exampletopological 1 1 space . \quad E doesAssume not have that the property the continuum open parenthesis hypothesis S closing holds parenthesis . \quad periodLet .. $ E $ be analogous example can be constructed in the manner following . Assume Notice here that an again that the continuum hypothesis holds and E is again any Sierpinski set on ananalogous arbitrary example Sierpinski can be constructed set on in the the manner real line following $ Rperiod . .. $ Assume Equip the set $ R $ with the usual the real line R. Consider the family of sets again that the continuum hypothesis{ holdsV \ X and: V Eis is an again open any subset Sierpinski of R and X set on the real line R period Consider the family of sets \noindent density topology ( aboutis atthe most density a countable topology subset of R on}. $ R $ see , in particular , \quad [ 1 ] ) . braceleftbig V backslashIn X the : V set isR anthis open family subset of of sets .. R is and a topology X strictly stronger than the st Denoteis at most by a the countable symbol subset $ of R R ˆ bracerightbig{ ∗ }$ \quad sub periodthe set $ R $ equipped with the density topology and consider theandard Sierpinski Euclidean set topology $ E in R $and as strictly a subspace weaker than of the the density topological topology space In the set R this family of sets is a topology strictly∗∗ stronger than the $ R ˆ{ ∗ } . $in R. NowDenote it is by not the difficultsymbol R the to set checkR equipped thatwith every this uncountable topology and subspace of the st andard Euclideanconsider topology the in Sierpinski R and strictly set E weakeras a subspace than the of .. the density topological topologicaltopology in R periodspace .. Denote $E$ by does the symbol not have R to the the power property of * * the $ set ( R equipped S ) with . this $ \quad Notice here that an topology and consider the Sierpinski set E as a subspace of the topological \noindent analogous example can be constructed in the manner following . \quad Assume again that the continuum hypothesis holds and $ E $ is again any Sierpinski set on the real line $ R . $ Consider the family of sets

\ centerline { $ \{ V \setminus X : V$ is anopen subset of \quad $ R $ and $ X $ }

\ centerline { is at most a countable subset of $ R \} { . }$ }

In the set $ R $ this family of sets is a topology strictly stronger than the st andard Euclidean topology in $ R $ and strictly weaker than the \quad d e n s i t y topology in $R . $ \quad Denote by the symbol $ R ˆ{ ∗ ∗ }$ the s e t $ R $ equipped with this topology and consider the Sierpinski set $ E $ as a subspace of the topological ON SEPARABLE SUPPORTS OF BOREL MEASURES .. 53 \ hspacespace R∗{\ tof the i l l power}ON SEPARABLE of * * period .... SUPPORTS Now it is OF not BORELdiﬃcultMEASURES to prove that\ anyquad uncountable53 subspace of the topological space E does not have the property open parenthesis S closing parenthesis period \noindentIn connectionspace with the $ R last ˆ{ example ∗ ∗ let } us. recall $ \ thath f i a l ltopologicalNow it space is not difficult to prove that any uncountable subspace E is a Sierpinski space if it does not contain a Luzin subspace with the \noindentcardinalityof equal the to topologicalthe cardinality of space E period $ .. E Notice $ does that Sierpinski not have spaces the property $ ( Shave ) some . $ interesting topological properties period For instance comma assume that the ON SEPARABLE SUPPORTS OF BOREL MEASURES 53 continuum hypothesisspace holdsR∗∗ and. takeNow an it arbitrary is not diﬃcult Sierpinski to prove space that E with any uncountable subspace Inthe connection cardinality of with the continuum the last period example .. Then let it can us be recall proved that that any a analytic topological space $ E $ is a Sierpinskiof the topological space space ifE itdoes does not not have contain the property a (LuzinS). subspace with the subset of E is a Borel setIn connection in E period with the last example let us recall that a topological space cardinalityIn conclusion let equal us formulate to the one cardinality unsolved problem of concerning $ E . topolo $ \quad hyphenNotice that Sierpinski spaces have some interestingE is a Sierpinski topological space if itproperties does not contain . For a Luzin instance subspace , with assume the cardi- that the gical spaces havingnality the property equal to open the cardinality parenthesis of SE. closingNotice parenthesis that Sierpinski period spaces have some continuumProblem period hypothesis .. Obtain a holds characterization and take of spaces an arbitrary having the property Sierpinski open parenthesis space $ Sclosing E $ with interesting topological properties . For instance , assume that the continuum parenthesis hypothesis holds and take an arbitrary Sierpinski space E with the cardinality thein s cardinality ome purely topo of logical the t continuum erms period . \quad Then it can be proved that any analytic subset of $E$of the is continuum a Borel . set Then in it $E can be proved . $ that any analytic subset of E is a References Borel set in E. 1 period J period Oxtoby comma Measure and category period .. Springer hyphen Verlag comma \ hspace ∗{\ f i l l } In conclusionIn conclusion let let us us formulate formulate one one unsolved unsolved problem problem concerning concerning topolo - topolo − New York endash Heidelgical hyphen spaces having the property (S). berg endash Berlin comma 1 971 period \noindent gical spacesProblem having . theObtain property a characterization $ ( of S spaces ) having . $ the property (S) 2 period Handbookin of smathematical ome purely topo logic logical period t North erms .hyphen Holland Publishing Company comma Amsterdam endash New York endash Oxford comma 1 977 period Problem . \quad Obtain a characterizationReferences of spaces having the property $ ( open parenthesis Received1 . J 1 . 7 Oxtoby period ,6 Measure period 1 and 993 categoryclosing parenthesis . Springer - Verlag , New York – SAuthor ) $ quoteright s address : in s ome purelyHeidel topo - logical berg – Berlin t erms , 1 971 . . I period Vekua Institute2 . of Handbook Applied Mathematics of mathematical logic . North - Holland Publishing Company of Tbilisi State University \ centerline { References, Amsterdam} – New York – Oxford , 1 977 . 2 comma University St period comma Tbilisi(380043 Received 1 7 . 6 . 1 993 ) Republic of Georgia 1 . J . Oxtoby , Measure and categoryAuthor . \quad ’ s addressSpringer : − Verlag , New York −− Heidel − berg −− Berlin , 1 971 . I . Vekua Institute of Applied Mathematics of Tbilisi State University 2 . Handbook of mathematical logic2 , University . North St− .Holland , Tbilisi 380043 Publishing Company , Amsterdam −− New York −− Oxford , 1 977Republic . of Georgia

\ centerline {( Received 1 7 . 6 . 1 993 ) }

\ centerline {Author ’ s address : }

\ centerline { I . Vekua Institute of Applied Mathematics }

\ centerline { of Tbilisi State University }

\ centerline {2 , University St . , Tbilisi 380043 }

\ centerline { Republic of Georgia }