William Arveson Department of Mathematics University of California Berkeley, CA USA

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1 NOTES ON MEASURE AND INTEGRATION IN LOCALLY COMPACT SPACES William Arveso Departmet of Mathematics Uiversity of Califoria Berkeley, CA USA 25 March 1996 Abstract. This is a set of lecture otes which preset a ecoomical developmet of measure theory ad itegratio i locally compact Hausdorff spaces. We have tried to illumiate the more difficult parts of the subject. The Riesz-Markov theorem is established i a form coveiet for applicatios i moder aalysis, icludig Haar measure o locally compact groups or weights o C -algebras...though applicatios are ot take up here. The reader should have some kowledge of basic measure theory, through outer measures ad Carathéodory s extesio theorem. Cotets Itroductio 1. The trouble with Borel sets 2. How to costruct Rado measures 3. Measures ad liear fuctioals 4. Baire meets Borel 5. The dual of C 0 () The preparatio of these otes was supported i o-egligible ways by Schitzel ad Pretzel. 1 Typeset by AMS-TE

2 2 WILLIAM ARVESON Itroductio At Berkeley the material of the title is taught i Mathematics 202B, ad that discussio ormally culmiates i some form of the Riesz-Markov theorem. The proof of the latter ca be fairly straightforward or fairly difficult, depedig o the geerality i which it is formulated. Oe ca elimiate the most serious difficulties by limitig the discussio to spaces which are compact or σ-compact, but the oe must still deal with differeces betwee Baire sets ad Borel sets; oe ca elimiate all of the difficulty by limitig the discussio to secod coutable spaces. I have take both shortcuts myself, but have ot bee satisfied with the result. These otes preset a approach to the geeral theory of itegratio o locally compact spaces that is based o Rado measures. My ow experiece i presetig alterate approaches has coviced me that Rado measures are the most sesible way to reduce the arbitrariess ad the bother ivolved with doig measure theory i these spaces. We prove the Riesz-Markov theorem i geeral, i a form appropriate for costructig Haar measure o locally compact groups or for dealig with weights o commutative C -algebras. If I have eglected to metio sigificat refereces i the bibliography it is partly because these lecture otes have bee dashed off i haste. I apologize to ay of my colleagues who may have bee abused or offeded, i that order. Fially, I wat to thak Cal Moore for poitig out a error i the proof of Propositio 2.1 (the preset versio has bee fixed) ad Bob Solovay for supplyig the idea behid the example precedig Propositio The trouble with Borel sets Throughout these otes, will deote a locally compact Hausdorff space. A Borel set is a subset of belogig to the σ-algebra geerated by the closed sets of. A Baire set is a elemet belogig to the σ-algebra geerated by the compact G δ s...that is, compact sets K havig the form K = U 1 U 2..., where U 1, U 2,... is a sequece of ope sets i. We will write B (resp. B 0 ) for the σ-algebra of all Borel (resp. Baire) sets. If the topology of has a coutable base the B = B 0. It is a good exercise to prove that assertio. I geeral, however, there is a differece betwee these two σ-algebras, eve whe is compact. At the same time, each of them is more or less ievitable: B is associated with the topology of ad B 0 is associated with the space C 0 () of cotiuous real-valued fuctios o which vaish at (see Propositio 4.1). I these otes we deal maily with Borel sets ad Borel measures. The correspodig results for Baire sets ad Baire measures are treated i 4. The poit we wat to make is that the trouble with measure ad itegratio i locally compact spaces has little to do with the fact that B ad B 0 are differet, ad a lot to do with the fact that ca be very large...i.e., very o-compact. Ad oe eeds the result i geeral if oe wishes to discuss Haar measure o locally compact groups (eve commutative oes), or weights o C -algebras (especially commutative oes). I these otes I have take the approach that I have come to prefer, i which measure meas Rado measure. I have attempted to cast light o the pitfalls that

3 MEASURE AND INTEGRATION 3 ca occur, to avoid verbosity i the mathematics, ad especially I have tried to avoid the pitfalls I have stumbled through i the past. A Rado measure is a positive Borel measure µ : B [0, + ] which is fiite o compact sets ad is ier regular i the sese that for every Borel set E we have µ(e) = sup{µ(k) : K E, K K} K deotig the family of all compact sets. There is a correspodig otio of outer regularity: a Borel measure µ is outer regular o a family F of Borel sets if for every E F we have µ(e) = if{µ(o) : O E, O O}, O deotig the family of all ope sets. The followig result implies that whe is compact (or eve σ-compact) oe has the best of it, i that ier ad outer regularity are equivalet properties. A set is called bouded if it is cotaied i some compact set, ad σ-bouded if it is cotaied i a coutable uio of compact sets. Every σ-bouded Borel set ca obviously be writte as a coutable uio of bouded Borel sets. Propositio 1.1. Let µ be a Borel measure which is fiite o compact sets. The the followig are equivalet. (1) µ is outer regular o σ-bouded sets. (2) µ is ier regular o σ-bouded sets. proof. (1) = (2) Suppose first that E is a bouded Borel set, say E L where L is compact, ad fix ɛ > 0. We have to show that there is a compact set K E with µ(k) µ(e) ɛ. But sice the relative complemet L \ E is bouded, we see by outer regularity that there is a ope set O L \ E such that µ(o) µ(l \ E) + ɛ. It follows that K = L \ O = L O c is a compact subset of E satisfyig µ(k) = µ(l) µ(l O) µ(l) µ(o) µ(l) µ(l \ E) ɛ = µ(e) ɛ, as required. More geerally, suppose that E = E 1 E 2... is a coutable uio of bouded Borel sets E. Without loss of geerality, we may assume that the sets E are disjoit. If some E has ifiite measure, the by the precedig paragraph we have sup{µ(k) : K E, K K} = µ(e ) = +. Hece sup{µ(k) : K E, K K} = µ(e) = +,

4 4 WILLIAM ARVESON ad we are doe. If, o the other had, µ(e ) < for every, the fixig ɛ > 0 we may fid a sequece of compact sets K E with µ(e ) µ(k ) + ɛ/2. Puttig L = K 1 K 2 K, it is clear that L is a compact subset of E for which µ(l ) = µ(k k ) (µ(e k ) ɛ/2 k ) µ(e k ) ɛ. Takig the supremum over all we obtai sup µ(l ) µ(e) ɛ, from which ier regularity follows. (2) = (1) Suppose first that E is a bouded Borel set. The the closure E of E is compact, ad a simple coverig argumet implies that there is a bouded ope set U such that E U. Set L = U, ad fix ɛ > 0. The L \ E is a bouded Borel set, hece by ier regularity there is a compact set K L \ E with µ(k) µ(l \ E) ɛ. Put V = U \ K = U K c. V is a bouded ope set which cotais E, ad we have µ(v ) = µ(u K c ) µ(l K c ) = µ(l) µ(k) µ(l) (µ(l \ E) ɛ) = µ(e) ɛ. Sice ɛ is arbitrary, this shows that µ is outer regular o bouded sets. I geeral, suppose E = E, where each E is a bouded Borel set. Agai, we may assume that the sets E are mutually disjoit. Sice the assertio 1.1 (1) is trivial whe µ(e) = + we may assume µ(e) < +, ad hece µ(e ) < + for every. Fix ɛ > 0. By the precedig paragraph we may fid a sequece of ope sets O E such that µ(o ) µ(e ) + ɛ/2. The E is cotaied i the uio O = O, ad we have µ(o) µ(o ) µ(e ) + ɛ = µ(e) + ɛ as required. We emphasize that ier ad outer regularity are ot equivalet properties whe fails to be σ-compact. I order to discuss this pheomeo, we cosider the family R of all σ-bouded Borel sets. Notice that R cotais if ad oly if is σ-compact; ad i that case we have R = B. But if is ot σ-compact the R is ot a σ-algebra but merely a σ-rig of subsets of. More explicitly, a σ-rig is a ovoid family S of subsets of satisfyig the coditios E, F S = E \ F S E 1, E 2, S = E S.

5 MEASURE AND INTEGRATION 5 There is a theory of measures defied o σ-rigs that is parallel to ad geeralizes the theory of measures defied o σ-algebras. The σ-rig approach to Baire measures was emphasized ad popularized by Paul Halmos [1], who co-iveted the ame itself. Remarks. We are ow able to make some cocrete observatios about the degree of arbitrariess that accompaies measure theory i humogous spaces. Assume is ot σ-compact, let R be the σ-rig of all σ-bouded Borel sets ad let R deote the set of all complemets of sets i R, R = {E c : E R}. The R R = ad R R is the σ-algebra geerated by R. This is a σ-algebra of Borel sets, but it is ot B sice it does ot ecessarily cotai ope sets or closed sets. I ay evet, we have a coveiet partitio of this σ-algebra ito Borel sets which are either σ-bouded or co-σ-bouded. It is atural to ask if a reasoable measure that is iitially defied o R ca be exteded uiquely to a measure o the σ-algebra geerated by R. The aswer is o. Ideed [2, Exercise 9, pp ] shows that a measure o R always has a extesio but that extesios are ot uique. Actually, there is a oe-parameter family of extesios of ay reasoable measure o R. There is a smallest oe (the ier regular extesio) ad a largest oe (the outer regular extesio), ad there is a arbitrary positive costat ivolved with each of the others. A bad apple. Big locally compact spaces ca be pathological i subtle ways. For example, let S be a ucoutable discrete space, let R be the Euclidea real lie, ad let = S R. is a locally compact Hausdorff space, beig the cartesia product of two such. For every subset E ad every s S, let E s R be the sectio of E defied by E s = {x R : (s, x) E}. It is easy to show that that E is ope iff every sectio E s is a ope set i R. Similarly, E is compact iff all but a fiite umber of sectios of E are empty ad all the remaiig sectios are compact subsets of R. It would seem reasoable to guess that if a set E is locally Borel i the sese that its itersectio with every compact set is a Borel set, the it must be a Borel set (may of us have bee so fooled: see [2, Lemma 9, p. 334]). That guess is wrog, as the followig example shows. Sice a complete discussio of the example would require more iformatio about the Baire hierarchy tha we have at had, we merely give eough details for a persistet reader to complete the argumet. We may as well take S to be the set of all coutable ordials. It follows from the above remarks that E K is a Borel set for every compact set K iff every sectio of E is a Borel set i R. Here is a example of a o-borel set E havig the latter property. For every coutable ordial ω S, let E ω be a Borel set i R which belogs to the ω th Baire class but to o properly smaller Baire class. Defie E = {(ω, x) S R : x E ω }.

6 6 WILLIAM ARVESON Clearly every sectio of E is a Borel set. To see that E is ot a Borel set, suppose that it did belog to the σ-algebra B geerated by the family O of all ope sets of. The E would have to belog to some Baire class over O, say to the ω0 th Baire class. It is easy to see that this implies that every sectio E λ must belog to the ω0 th Baire class i the real lie, cotradictig our costructio of the sectios of E. Fially, we poit out that fiite Rado measures behave as well as possible, eve whe the uderlyig space is huge: Propositio 1.2. Every fiite Rado measure is both ier ad outer regular. sketch of proof. Outer regularity of the measure o ay Borel set follows from the ier regularity of the measure o the complemet of the set, because the measure of ay set is fiite. 2. How to costruct Rado measures I this sectio we show how Rado measures ca be costructed from certai simpler etities defied o the family O of all ope subsets of. Let m : O [0, + ] be a fuctio havig the followig properties (A) (B) (C) U K = m(u) < + U V = m(u) m(v ) U 1, U 2, U = m( U ) m(u ), (D) (E) U V = = m(u V ) = m(u) + m(v ) m(u) = sup{m(v ) : V O, V U, V K}. Notice that A ad D together imply that m( ) = 0. If we start with a Rado measure µ o B ad defie m to be the restrictio of µ to O, the such a m obviously has properties A through D, ad a simple argumet establishes E as well. Coversely, we have Propositio 2.1. Ay fuctio m defied o the ope sets which has properties A through E ca be exteded uiquely to a Rado measure defied o all Borel sets. proof. For uiqueess, let µ be a Rado measure which agrees with m o O. Sice Rado measures are obviously determied by their values o compact sets, it suffices to observe that for every compact set K, we have µ(k) = if{m(u) : U K, U O}. Ideed, µ is ier regular by defiitio of Rado measures, ad every compact set is obviously a bouded Borel set. Thus the assertio is a immediate cosequece of the equivalece of coditios (1) ad (2) i propositio 1.1. Turig ow to existece, we cosider the set fuctio µ defied o arbitrary subsets A by µ (A) = if{m(u) : U A, U O}.

7 We claim first that µ is a outer measure, that is MEASURE AND INTEGRATION 7 µ ( ) = 0 A B = µ (A) µ (B) A 1, A 2, = µ ( A ) µ (A ). The first two properties are obvious. To prove the third, it is clear that we eed oly cosider the case i which µ (A ) is fiite for every = 1, 2,.... I that case, fix ɛ > 0 ad choose ope sets U A with the property that µ (U ) µ (A ) + ɛ/2 for every. above, The U is a ope set cotaiig A, hece by property C µ ( A ) m( U ) m(u ) µ (A ) + ɛ, ad the claim follows from the fact that ɛ is arbitrary. It is apparet from the defiitio of µ that µ (U) = m(u) if U is a ope set. We claim ext that every ope set is measurable; that is, for each ope set O we have (2.2) µ (A) = µ (A O) + µ (A O c ), for every subset A. To prove 2.2 it suffices to prove the iequality, sice the opposite oe follows from the subadditivity of µ. For that, fix A ad O. If µ (A) = + the there is othig to prove, so we may assume that µ (A) (ad hece both µ (A O) ad µ (A O c )) is fiite. Fix ɛ > 0 ad choose a ope set U A so that m(u) µ (A) + ɛ. We will prove that (2.3) m(u) m(u O) + µ (U O c ). Note that it suffices to prove 2.3, sice µ (A O) + µ (A O c ) m(u O) + µ (U O c ) m(u) µ (A) + ɛ, ad ɛ is arbitrary. I order to prove 2.3, we use property E above to fid a ope set V whose closure V is a compact subset of U O ad m(u O) m(v ) + ɛ. Notice that V ad U V c are disjoit ope sets which are both cotaied i U. Thus we have by properties B ad D m(v ) + m(u V c ) = m(v (U V c )) m(u).

8 8 WILLIAM ARVESON Fially, sice U O c U V c, m(u O) + µ (U O c ) m(v ) + ɛ + µ (U O c ) m(v ) + m(u V c ) + ɛ m(u) + ɛ, ad 2.3 follows because ɛ is arbitrary. By Carathéodory s extesio theorem [2, Chapter 12, 2] the restrictio of µ to the σ-algebra of µ -measurable sets is a measure; hece the restrictio of µ to the σ-algebra of Borel sets is a measure µ satisfyig µ(o) = m(o) for every ope set O. Notice that µ is fiite o bouded sets by property A, ad µ is outer regular by the defiitio of µ. So if is σ-compact, the µ is already a Rado measure by propositio 1.1. If is ot σ-compact the µ is ot ecessarily a Rado measure, ad we must modify it as follows. Let M deote the σ-rig of all σ-bouded Borel sets. We defie a ew set fuctio µ o Borel sets as follows: µ(e) = sup{µ(b) : B M, B E}. It is a fact that µ is coutably additive. Gratig that for a momet, otice that µ is a Rado measure. Ideed, Propositio 1.1 implies that µ(b) = µ(b) for every B M, that µ is ier regular o M, ad thus by its defiitio µ must be ier regular o all Borel sets. We also have µ(o) = m(o) for every ope set O. To see that, fix O ad choose a ope set V whose closure is a compact subset of O. The we have m(o) = µ(o) µ(o) µ(v ). Sice V is a σ-bouded ope set we have µ(v ) = µ(v ) = m(v ), ad hece m(o) µ(o) m(v ) for all such V. After takig the sup over V ad usig property (E) above, we fid that m(o) = µ(o). We may coclude i this case that µ is a Rado measure which agrees with m o ope sets. It remais to check that µ is coutably additive. For that, let E 1, E 2,... be a sequece of mutually disjoit Borel sets, ad let B be a σ-bouded subset of E. Set B = B E. The B 1, B 2,... are mutually disjoit σ-bouded Borel sets, hece µ(b) = µ(b ) µ(e ). =1 =1 By takig the supremum over all such B we obtai µ( E ) µ(e ). To prove the opposite iequality it suffices to cosider the case i which µ(e ) is fiite for every. I this case, for each positive umber ɛ we ca fid a σ-bouded Borel set B E such that =1 µ(b ) µ(e ) ɛ/2.

9 MEASURE AND INTEGRATION 9 By summig o we obtai µ(b) µ(e ) ɛ where B = B. Sice B is a σ-bouded subset of E we fid that µ( E ) µ(b) µ(e ) ɛ, =1 ad the coutable additivity of µ follows because ɛ is arbitrary. Remark. There are other routes to the costructio of Borel measures which begi with a set fuctio m defied o the family K of all compact sets. Such etities m are called cotets, ad are i a sese dual to set fuctios obeyig the properties A E that we have used. The iterested reader is referred to [1, sectios 53 54] ad [2, chapter 13, sectio 3]. 3. Measures ad liear fuctioals Let C c () be the space of all cotiuous fuctios f : R which have compact support i the sese that the set supp(f) = {x : f(x) 0} is compact. supp(f) is called the support of the fuctio f. C c () is a algebra of fuctios, ad is i fact a ideal i the algebra C() of all cotiuous fuctios f : R, i the sese that f C c (), g C() = fg C c (). If is compact the C c () = C(). If is ot compact, the C c () is sup-orm dese i the algebra C 0 () of all cotiuous real fuctios which vaish at. The Riesz-Markov theorem gives a useful ad cocrete descriptio of positive liear fuctioals Λ : C c () R, that is, liear fuctioals Λ which are positive i the sese that f 0 = Λ(f) 0, f C c (). For example, let µ be a Rado measure o B. Sice compact sets have fiite µ-measure, it follows that every fuctio i C c () belogs to L 1 (, B, µ) ad we ca defie Λ : C c () R by (3.1) Λ(f) = f dµ, f C c (). Λ is a positive liear fuctioal o C c (). The followig lemma shows how certai values of the measure µ ca be recovered directly from Λ. Lemma 3.2. Suppose that µ ad Λ are related by 3.1. The for every ope set U we have µ(u) = sup{λ(f) : 0 f 1, f C c (), supp(f) U}. proof. The iequality is clear from the fact that if 0 f 1 ad supp(f) U the χ U (x) f(x) for every x,

10 10 WILLIAM ARVESON ad after itegratig this iequality we obtai µ(u) f dµ = Λ(f). For the opposite iequality, let K be a arbitrary compact subset of U. We may fid a bouded ope set V satisfyig K V V U. By Tietze s extesio theorem, there is a cotiuous fuctio f satisfyig 0 f 1, f = 1 o K, ad f = 0 o the complemet of V. Thus the support of f is cotaied i V U, ad sice χ K f we may itegrate the latter iequality to obtai µ(k) Λ(f) sup{λ(f) : 0 f 1, f C c (), supp(f) U}. The desired iequality follows from ier regularity after takig the sup over K. The followig theorem of Riesz ad Markov asserts that 3.1 gives the the most geeral example of a positive liear fuctioal o C c (). Theorem 3.3 (Riesz-Markov). Let Λ be a positive liear fuctioal o C c (). The there is a uique Rado measure µ such that Λ(f) = f dµ, f C c (). Remarks. I should poit out that i spite of the fact that this formulatio of the Riesz-Markov theorem is the oe I happe to prefer, it is ot the oly reasoable oe. See [1],[2] for others. The coectio betwee liear fuctioals ad Baire measures will be described i sectio 4 below. proof of Theorem 3.3. The uiqueess of µ is a direct cosequece of Lemma 3.2 ad the results of sectio 1. Ideed, Lemma 3.2 implies that the values of µ o ope sets are determied by the liear fuctioal Λ, ad by propositio 1.1 the value of µ o ay compact set K obeys µ(k) = if{µ(u) : U K, U O}. Fially, sice for a arbitrary Borel set E we have µ(e) = sup{µ(k) : K E, K K}, it follows that the µ(e) is uiquely determied by the liear fuctioal Λ. For existece, we defie a umber m(u) [0, + ] for every ope set U by m(u) = sup{λ(f) : f C c (), 0 f 1, supp(f) U}. We will show first that m satisfies the hypostheses A E of propositio 2.1, ad hece defies a Rado measure µ : B [0, + ] by way of µ(u) = m(u), U O. We the show that Λ is truly itegratio agaist this Rado measure µ. We will require the followig result o the existece of partitios of uity.

11 MEASURE AND INTEGRATION 11 Lemma 3.4. Let {O α : α I} be a ope cover of a compact subset K. The there is a fiite set φ 1, φ 2,..., φ of real cotiuous fuctios o ad there is a fiite subset α 1, α 2..., α I satisfyig (i) (ii) (iii) 0 φ k 1 supp(φ k ) O αk φ k = 1 o K. k This is a stadard result whose proof ca be foud i [2, propositio 9.16]. Let us establish the properties A E of sectio 2. For A, let U be a bouded ope set. Sice the closure of U is compact, a simple coverig argumet shows that we may fid aother bouded ope set V which cotais the closure of U. By Tietze s extesio theorem, there is a cotiuous fuctio g : R such that 0 g 1, g = 1 o U g = 0 o the complemet of V. Ay fuctio f C c () satisfyig 0 f 1 ad supp(f) U must also satisfy 0 f g ad hece Λ(f) Λ(g). It follows that m(u) = sup{λ(f) : 0 f 1, f C c ()} Λ(g) < + ad property A follows. Property B is obvious. For property C, choose ope sets U 1, U 2,..., put U = U, ad choose f C c () with 0 f 1 ad such that f is supported i U. By Lemma 3.4, we may fid a iteger ad cotiuous fuctios φ 1, φ 2,..., φ takig values i the uit iterval, such that supp(φ k ) U k, φ 1 + φ φ = 1 It follows that f = k φ kf, ad hece Λ(f) = Λ(φ k f) ad o supp(f). m(u k ) m(u k ). Property C follows by takig the supremum over all f i the precedig lie. To establish D we prove oly the iequality, sice the opposite oe is a cosequece of C. Let U ad V be disjoit ope sets ad let f ad g be two fuctios i C c () satisfyig 0 f, g 1, supp(f) U, ad supp(g) V. Sice U V = we have 0 f + g 1 ad supp(f + g) U V. Hece Λ(f) + Λ(g) = Λ(f + g) m(u V ), ad the iequality m(u) + m(v ) m(u V ) follows after takig the supremum over f ad g.

12 12 WILLIAM ARVESON For property E, choose f C c () satisfig 0 f 1 ad supp(f) U. Aother simple coverig argumet o the compact subset supp(f) U shows that we ca fid a ope set V havig compact closure such that Sice supp(f) V we have supp(f) V V U. Λ(f) m(v ) sup{m(v ) : V U, V K} ad ow property E follows after takig the sup over f. Usig propositio 2.1 we may coclude that there is a Rado measure µ o B which agrees with m o ope sets. It remais to show that (3.4) Λ(f) = f dµ for every f C c (). Sice both sides of 3.4 are liear i f ad sice C c () is spaed by its oegative fuctios, it suffices to establish 3.4 for the case where 0 f 1. To this ed, fix f ad choose a positive umber ɛ. We will exhibit a pair of simple Borel fuctios u, v havig the followig properties. (3.5) (3.6) (3.7) u f v (v u) dµ ɛ u dµ Λ(f) v dµ + ɛ Note that this will complete the proof. Ideed, we may itegrate 3.5 to obtai u dµ f dµ v dµ ad sice the itegrals of u ad v are withi ɛ of each other by 3.6, we see from 3.7 ad the precedig iequality that Λ(f) f dµ 2ɛ. The result follows sice ɛ is arbitrary. I order to costruct u ad v, fix a positive iteger ad defie a sequece of bouded ope sets O 0 O 1 O by O k = {x : f(x) > k/}, for k = 1, 2,...,, ad let O 0 be ay bouded ope set which cotais supp(f). Notice that O is empty ad that the closure of O k is cotaied i O k 1 for k = 1, 2,...,. Defie u ad v as follows u = 1 v = 1 c k c k 1,

13 MEASURE AND INTEGRATION 13 where c k deotes the characteristic fuctio of O k. We have c = 0 because O is empty, ad it apparet that 0 u v. We will show that if is sufficietly large the the coditios are satisfied. For 3.5, otice that f, u, v all vaish outside O 0, ad that if x O k 1 \ O k for k = 1, 2,..., the u(x) = 1 k 1 1 = k 1 i=1 < f(x) k = 1 k 1 1 = v(x), i=0 which proves 3.5. For 3.6 we have v u = 1 c 0; hece (v u) dµ = 1 µ(o 0) which is smaller tha ɛ provided is sufficietly large. To prove 3.7 we employ a device from [1, 56, p. 246]. Defie a sequece φ 1, φ 2,..., φ of cotiuous fuctios by φ k = [(f k 1 ) 0] 1 = [(f k 1 ) 1 ] 0. Each φ k vaishes o the complemet of O 0, hece φ k C c (). Moreover, we have 0 if x / O k 1 (3.8) φ k (x) = f(x) k 1 if x O k 1 \ O k 1 if x O k. Clearly, 0 φ k 1. Notig that for x O k 1 \ O k, φ 1 (x) = φ 2 (x) = = φ k 1 (x) = 1 φ k (x) = f(x) k 1 φ k+1 (x) = φ k+2 (x) = = φ (x) = 0 it follows that f = φ 1 + φ φ. For coveiece, we defie O 1 = O 0. We claim that for every k = 1, 2,..., we have the iequalities (3.9) 1 µ(o k) Λ(φ k ) 1 µ(o k 2), To prove the first iequality, choose ay fuctio g C c () satisfyig 0 g 1 ad supp(g) O k. Sice φ k (x) = 1 for every x O k we have 0 g φ k, ad hece Λ(g) Λ(φ k ).

14 14 WILLIAM ARVESON Takig the supremum over all such g gives µ(o k ) Λ(φ k ) = Λ(φ k ), ad the iequality follows after divisio by. For the secod iequality, otice that for k 2, the closed support of φ k is cotaied i O k 1 O k 2. So by defiitio of µ(o k 2 ) = m(o k 2 ) we have Λ(φ k ) = Λ(φ k ) µ(o k 2 ) ad we obtai the secod iequality after dividig by. The case k = 1 follows similarly from the fact that the closed support of φ 1 is cotaied i O 0. Notig that u dµ = 1 v dµ = 1 µ(o k ) = 1 1 µ(o k ), µ(o k 1 ) = 1 1 µ(o k ) we may sum the iequalities 3.9 from k = 1 to ad use Λ(f) = k Λ(φ k) to obtai u dµ Λ(f) v dµ + 1 µ(o 0). k=0 ad Sice the last term o the right is less tha ɛ whe is large, 3.7 follows. 4. Baire meets Borel We have already poited out that the Borel σ-algebra B is atural because it is the σ-algebra geerated by the topology of. The followig result implies that the Baire σ algebra B 0 has just as strog a claim to ievitability. Propositio 4.1. B 0 is the smallest σ-algebra with respect to which the fuctios i C c () are measurable. proof. We show first that every fuctio f i C c () is B 0 -measurable. Sice the B 0 -measurable fuctios are a vector space ad sice every fuctio i C c () is a differece of oegatives oes, we may assume that f 0. Fix t R ad cosider the set F t = {x : f(x) t}. If t 0 the F t = belogs to B 0. If t > 0 the F t = =2{x : f(x) > t t/} is exhibited as a compact G δ, hece F t B 0. Coversely, if A is ay σ-algebra with the property that every fuctio i C c () is A-measurable, the we claim that A cotais every compact G δ, ad hece A cotais B 0. For that, let K be a compact set havig the form K = U 1 U 2...

15 MEASURE AND INTEGRATION 15 where the sets U are ope. By replacig each U with a smaller ope set if ecessary, we ca assume that each U is bouded, ad hece has compact closure. By Tietze s extesio theorem there are cotiuous fuctios f : [0, 1] with the properties f = 1 o K ad f = 0 o the complemet of U. f belogs to C c () ad is therefore A-measurable. Fially, sice for each x we have { 1 if x K lim f (x) = 0 if x / K, it follows that the characteristic fuctio of K, beig a poitwise limit of a sequece of A-measurable fuctios, is A-measurable. Hece K A. Remark. Sice the closure of C c () i the sup orm f = sup{ f(x) : x } is the algebra C 0 () of all cotiuous real-valued fuctios which vaish at, we see that B 0 could also have bee defied as the smallest σ-algebra with respect to which the fuctios i C 0 () are measurable. Perhaps the most compellig feature of Baire measures is that regularity comes for free o σ-bouded sets. I order to discuss this result it will be coveiet to itroduce some otatio. K 0 will deote the family of all compact G δ s, ad O 0 will deote the family of all ope Baire sets. A Baire measure µ : B 0 [0, + ] is called ier regular o a family F of Baire sets if for every F F we have µ(f ) = sup{µ(k) : K F, K K 0 }. Similarly, µ is outer regular o F if for every F F we have µ(f ) = if{µ(o) : O F, O O 0 }. 4.2 Separatio properties of K 0 ad O 0. There are eough ope Baire sets to form a base for the topology o ; more geerally, give ay compact set K ad a ope set U cotaiig K, there exist sets K 0 K 0 ad U 0 O 0 such that K K 0 U 0 U, see [1, Theorem D, 50]. Ideed, by replacig U with a smaller ope set havig compact closure, if ecessary, we see that oe ca eve choose U 0 to be a bouded ope Baire set. That is about all oe ca say about ope Baire sets i geeral. O the other had, the oly compact Baire sets are the obvious oes, amely the compact G δ s. The proof of the latter is ot so easy, ad we shall ot require the result the sequel. The reader is referred to [1, Theorem A, 51] for a proof. Propositio 4.3. Let µ be a Baire measure which is fiite o compact G δ s. The µ is both ier regular ad outer regular o σ-bouded Baire sets. proof. We will show that for every σ-bouded Baire set E oe has both properties (4.3.A) (4.3.B) µ(e) = sup{µ(k) : K E, K K 0 }, µ(e) = if{µ(u) : U E, U O 0 }.

16 16 WILLIAM ARVESON To this ed, we claim that for every Baire set E ad every K K 0, the itersectio K E satisfies both 4.3.A ad 4.3.B. Ideed, let A deote the family of all Baire sets E for which this assertio is true. It suffices to show that A cotais K 0 ad is a σ-algebra. To see that A cotais K 0, choose E K 0. The assertio 4.3.A is trivial because E itself belogs to K 0. I order to prove 4.3.B, let K K 0. Sice K E is a compact G δ we may fid bouded ope Baire sets U such that K E = U 1 U By replacig U with U 1 U 2 U we ca assume that U 1 U For each we have µ(u ) < because U is bouded, ad thus 4.3.B follows from upper cotiuity of µ: µ(k E) = lim µ(u ). I order to show that A is a σ-algebra we have to show that A is closed uder complemetatio ad coutable uios. We show first that A is closed uder complemetatio. Choose E A, fix K K 0 ad ɛ > 0. Sice E A there are sets L K 0 ad U, V O 0 such that L K E U, K V ad such that both µ(u \L) ad µ(v \K) are smaller tha ɛ. Usig the remarks 4.2 we may assume that both U ad V are bouded, by replacig them with smaller oes if ecessary. Defie sets A, B by A = K U c, B = V L c. A belogs to K 0, B belogs to O 0, ad we have A K E B. Both A ad B are bouded sets, ad therefore have fiite measure. Moreover, sice µ(b) = µ(v \ L) = µ(v \ K) + µ(k \ L) ad sice we have µ(a) = µ(k \ U) = µ(k) µ(k U) µ(k) µ(u) µ(b) µ(a) µ(v \K)+µ(K \L) µ(k)+µ(u) = µ(v \K)+(µ(U) µ(l)) 2ɛ. Sice ɛ is arbitrary, it follows that E c belogs to A. We claim ow that A is closed uder coutable uios. Choose E 1, E 2, A, fix K K 0 ad ɛ > 0. Because 4.3.A ad 4.3.B are valid for K E for every, we may fid K K 0, ad bouded sets U O 0 such that K K E U

17 ad for which MEASURE AND INTEGRATION 17 µ(u ) µ(k ) ɛ/2. Put U = U, L = K 1 K 2 K. The L K 0, U O 0 ad we have L K E U. Moreover, µ(u) µ( L ) (µ(u ) µ(k )) ɛ. =1 Sice the sets L icrease to L as ad sice L K E has fiite measure, it follows that µ(l ) µ( L ). Hece the differece µ(u) µ(l ) is smaller tha 2ɛ whe is sufficietly large. Sice ɛ is arbitrary, we coclude that E A. Thus, A cotais all Baire sets. Now let E be ay σ-bouded Baire set, say E K where K is compact. The by the remarks 4.2 we may assume that K K 0 by slightly elargig each K. Hece E is itself a coutable uio of sets E = (K E) each of which is, by what has already bee proved, a Baire set of fiite measure which is both ier ad outer regular. It is easy to see that this implies E is both ier ad outer regular. Ideed, choose ɛ > 0. For each we fid L K 0 ad a bouded set U O 0 such that ad for which L K E U µ(u ) µ(k ) ɛ/2. Puttig L = L, ad U = U we have L E U ad by estimatig as we have doe above we also have µ(u \ L) ɛ. If µ(e) is ifiite the so is µ(u) ad the precedig iequality implies µ(l) = +. By lower cotiuity of µ, lim µ(l 1 L 2 L ) = µ(l) = +. Sice L 1 L 2 L is a compact G δ subset of E, this establishes ier ad outer regularity at E i this case. If µ(e) < +, the the precedig iequality implies that both µ(l) ad µ(u) are withi ɛ of µ(e). Sice µ(l) = lim µ(l 1 L 2 L ) ad sice L 1 L 2 L is a K 0 -subset of E for each, we coclude that E is both ier ad outer regular i this case as well.

18 18 WILLIAM ARVESON Corollary. Every fiite Baire measure o a compact Hausdorff space is both ier regular ad outer regular. Fially, it is sigificat that the correspodece betwee Rado measures (defied o B) ad Baire measures (defied o B 0 ) is bijective. However, eve here oe must be careful i the formulatio if is ot σ-compact. More precisely, if oe restricts a Rado measure µ to B 0 the oe obtais a Baire measure which is fiite o compact sets. However, the ier regularity of Rado measures o B does ot immediately imply that their restrictios to B 0 are ier regular Baire measures as defied i the paragraphs precedig Propositio 4.3. The problem is that a give Baire set may have may more compact subsets tha it has compact G δ subsets. Nevertheless, Propositio 4.3 implies that the restrictio of a Rado measure o B to the σ-rig R 0 of σ-bouded Baire sets is ier regular. If is σ-compact, the this restrictio is already a regular Baire measure i both the ier ad outer seses. But if is ot σ-compact the i order to obtai a ier regular Baire measure we must first restrict the Rado measure to the σ-rig R 0 ad the use the latter to defie a ier regular Baire measure o the full Baire σ-algebra much as we did i the proof of Propositio 2.1. After these sheaigas oe ca say that every Rado measure restricts to a ier regular Baire measure i geeral. The followig asserts that this map is a bijectio. Propositio 4.4. Let µ be a ier regular Baire measure which is fiite o compact Baire sets. The µ exteds uiquely to a Rado measure o B. proof. For the existece of a Rado extesio of µ we ote that sice fuctios i C c () are Baire measurable ad µ-itegerable, we may defie a positive liear fuctioal Λ o C c () by Λ(f) = f dµ. By Theorem 3.3 there is a Rado measure ν such that f dν = Λ(f) = f dµ, f C c (). I order to show that µ is the restrictio of ν as described i the precedig discussio, let K be ay compact G δ. Notice that there is a sequece of fuctios f C c () such that { 1 if x K f (x) 0 if x / K. Ideed, we ca write K = U where each U is a ope set havig compact closure. Choosig a cotiuous fuctio g takig values i [0, 1] such that g = 1 o K ad g = 0 o the complemet of U, we may take f = g 1 g 2 g. By the mootoe covergece theorem we have ν(k) = lim f dν = lim f dµ = µ(k), ad the desired coclusio follows from the ier regularity of µ o B 0.

19 MEASURE AND INTEGRATION 19 For uiqueess, suppose that ν 1 ad ν 2 are two Rado measures which exted µ i the sese described above. Notice that for every f C c () satisfyig f 0, the value of the itegral f dν 1 is etirely determied by the values of the fuctio F (t) = ν 1 ({x : f(x) t}) for t > 0; i.e., by the values of ν 1 K0 = µ K0. The same applies to ν 2, hece f dν 1 = f dµ = f dν 2. Hece ν 1 = ν 2 by the uiqueess assertio of 3.3. Corollary. If is compact, the every fiite Baire measure exteds uiquely to a measure o B which is both ier ad outer regular. There are two possible reformulatios of the Riesz-Markov theorem i terms of Baire measures; the proofs follow from the precedig discussio. Theorem 4.5. For every positive liear fuctioal Λ defied o C c () there is a uique measure µ defied o the σ-rig of all σ-bouded Baire sets (resp. a uique ier regular Baire measure µ) such that Λ(f) = f dµ, f C c (). 5. The dual of C 0 () We ow discuss oe of the useful cosequeces of the Riesz-Markov theorem. C 0 () will deote the space of all real-valued cotiuous fuctios f o which vaish at i the sese that for every ɛ > 0 the set {x : f(x) ɛ} is compact. C 0 () is a real algebra i that it is closed uder the usual liear operatios ad poitwise multiplicatio. The orm f = sup f(x) x makes C 0 () ito a Baach space i which fg f g. We will make essetial use of the atural order o elemets of C 0 (), defied by f g f(x) g(x) for every x. This orderig makes C 0 () ito a lattice, ad the lattice operatios ca be defied poitwise by f g (x) = max(f(x), g(x)), f g (x) = mi(f(x), g(x)), x. Fially, P will deote the coe of all positive fuctios, P = {f C 0 () : f 0}. Notice that P P = {0} ad P P = C 0 (). There is a atural orderig iduced o the dual C 0 () by the orderig o C 0 (), amely ρ σ ρ(f) σ(f), for every f P, ad a liear fuctioal ρ is called positive if ρ 0.

20 20 WILLIAM ARVESON Propositio 5.1. Every positive liear fuctioal o C 0 () is bouded. Moreover, for every ρ C 0 () there is a smallest positive liear fuctioal Λ such that ρ Λ. Remarks. The secod assertio meas that if Λ is aother positive liear fuctioal satisfyig ρ Λ, the Λ Λ. After a simple argumet (which we omit), this implies that the dual of C 0 () is itself a lattice with respect to the orderig defied above; ad of course Λ = ρ 0. proof. To establish the first assertio let Λ be a positive liear fuctioal. Sice every elemet i the uit ball of C 0 () is a differece of positive fuctios i the uit ball of C 0 (), to show that Λ is bouded it suffices to show that sup{λ(f) : f P, f 1} < +. But if this supremum is ifiite the we ca fid a sequece f P satisfyig f 1 ad Λ(f ) > 2. Lettig g P be the fuctio defied by the absolutely coverget series g = 2 k f k, we have g 2 k f k for every, hece Λ(g) Λ( 2 k f k ) = 2 k Λ(f k ) >. The latter is absurd for large. To prove the secod assertio choose ρ C 0 (). For every f P we defie a oegative umber Λ 0 (f) by Λ 0 (f) = sup{ρ(u) : 0 u f}. Clearly Λ 0 (αf) = αλ 0 (f) for every oegative scalar α ad every f P. We claim that for all f, g P, Λ 0 (f + g) = Λ 0 (f) + Λ 0 (g). The iequality follows from the fact that if 0 u f ad 0 v g the 0 u + v f + g, hece Λ 0 (f + g) ρ(u + v) = ρ(u) + ρ(v). follows after takig the sup over u ad v. For the opposite iequality, fix f, g P ad choose u satisfyig 0 u f + g. Now for each x, f u (x) is either f(x) or u(x), ad i either case u(x) (f u) (x) + g(x). It follows that 0 u f u g. Therefore ρ(u) ρ(f u) = ρ(u f u) Λ 0 (g),

21 ad hece MEASURE AND INTEGRATION 21 ρ(u) ρ(f u) + Λ 0 (g) Λ 0 (f) + Λ 0 (g). The iequality ow follows after takig the supremum over u. Fially, we claim that Λ 0 exteds uiquely to a liear fuctioal Λ defied o all of C 0 (). Give that, the remaiig assertios of 5.1 follow; for if Λ is aother positive liear fuctioal satisfyig ρ Λ, the for every f P ad every 0 u f we have ρ(u) Λ (u) Λ (f), ad Λ(f) Λ (f) follows from the precedig iequality after takig the sup over u. I order to exted Λ 0, choose a arbitrary f C 0 (), write f = f 1 f 2 i ay way as the differece of two elemets f k P, ad defie Λ(f) = Λ 0 (f 1 ) Λ 0 (f 2 ). The oly questio is whether or ot this is well-defied; but that is clear from the fact that if f k, g k P ad f 1 f 2 = g 1 g 2 the Λ 0 (f 1 ) + Λ 0 (g 2 ) = Λ 0 (f 1 + g 2 ) = Λ 0 (g 1 + f 2 ) = Λ 0 (g 1 ) + Λ 0 (f 2 ), hece Λ 0 (f 1 ) Λ 0 (f 2 ) = Λ 0 (g 1 ) Λ 0 (g 2 ). The liearity of the exteded Λ follows from the restricted liearity of the origial Λ 0 o the positive coe. Uiqueess of the extesio is obvious from the fact that P P = C 0 () Remarks. Let Λ be a positive liear fuctioal o C 0 (), ad let Λ be its orm. By the Riesz-Markov theorem there is a positive Rado measure µ such that Λ(f) = f dµ, f C c (). Notice that if f C c () satisfies 0 f 1, the f dµ Λ(f) Λ. From this, we may deduce that for every compact set K we have µ(k) Λ, ad fially by ier regularity µ() Λ < +. I particular, µ is a fiite measure, ad hece is both outer ad ier regular (Propositio 1.2). It is easy to see that i fact µ() = Λ. More geerally, we may deduce the followig descriptio of the dual of C 0 (); this result (together with myriad other results which bear it o relatio whatsoever) is (are) ofte called the Riesz Represetatio Theorem. Theorem 5.2. For every bouded liear fuctioal ρ o C 0 () there is a fiite siged Borel measure µ such that ρ(f) = f dµ, f C 0 (). proof. Note first that ρ ca be decomposed ito a differece Λ 1 Λ 2 of positive liear fuctioals. Ideed, usig 5.1 we may defie Λ 1 = ρ 0, ad it is clear from

22 22 WILLIAM ARVESON the statemet of 5.1 that the liear fuctioal Λ 2 = Λ 1 ρ is positive. Thus the existece of µ follows from the Riesz-Markov theorem ad 5.1 above. Remarks. If oe stipulates that the total variatio measure µ of µ should be a Rado measure, the µ is uique, ad moreover µ is outer regular as well as ier regular because it is fiite. We omit the argumet [2, Chapter 13, 5]. If we agree to defie a siged Rado measure as a fiite siged measure µ whose variatio µ = µ + + µ is ier regular (ad therefore also outer regular), the the vector space M() of all siged Rado measures becomes a Baach space with orm µ = sup µ(e k ) k the supremum beig exteded over all fiite families E 1, E 2,..., E of mutually disjoit Borel sets. Such a measure gives rise to a liear fuctioal ρ C 0 () as i 5.2, ad it is ot hard to show that ρ = µ, [2, Chapter 13, 5]. Thus the dual of C 0 () is aturally isometrically isomorphic to the Baach space M() i such a way that the order structure o the dual of C 0 () correspods to the atural orderig of siged measures. Refereces 1. Halmos, P. R., Measure Theory, Va Nostrad, Priceto, Royde, H. L., Real Aalysis, third editio, Macmilla, New York, 1988.