GARLING'S PROOF OF THE RIESZ REPRESENTATION THEOREM PRAHLAD VAIDYANATHAN Introduction. The goal of this article is to discuss a proof of the Riesz representation theorem due to Garling [Garling]. While this proof does not provide the most general result - it essentially only works for first countable compact Hausdorff spaces - it does have the important property of being easy to understand. A similar proof to this was proved by Varadarajan [Varadarajan] (See also [Sunder]). Here is the theorem we wish to prove. Originally proved by F. Riesz for the unit interval, it has reached its current form through the work of Markov and Kakutani. Theorem 0.1. Let X be a compact Hausdorff space and τ : C(X) ! C be a positive linear functional. Then there is a unique Borel measure µ such that Z τ(f) = fdµ 8f 2 C(X) X Definition 0.2. Let X be a compact Hausdorff space. (1) The Baire σ-algebra MX is the smallest σ-algebra which makes every f 2 C(X)- measurable. ie. MX is generated by sets of the form f −1(F ) where F ⊂ C is closed and f 2 C(X). (2) We denote the Borel σ-algebra (generated by all open/closed sets) by BX . Note that MX ⊂ BX . However, if X is a metric space, then MX = BX , because if F ⊂ X is closed, then F = f −1(f0g) where f(x) := d(x; F ). More generally, if X is first countable, then MX = BX . (3) A measure µ is called a Baire measure if it satisfies the following conditions: (a) The domain of µ contains MX (b) µ(K) < 1 for any compact K 2 MX . (c) µ is inner regular, in that, for each Baire set E, µ(E) = supfµ(K): K 2 MX compactg (d) µ is outer regular, in that, for each E 2 MX , µ(E) = inffµ(U): U 2 MX openg It is known that a measure that satisfies (a) and (b) automatically satisfies (c) and (d) (See [Arveson, Proposition 4.3]), but we will not worry ourselves with that fact. (4) Note that if µ is a Baire measure, then any continuous function is MX -measurable. Hence, we may define τµ : C(X) ! C by Z f 7! fdµ X This is a positive linear functional. Lemma 0.3. Let µ and ν be two Baire measures such that Z Z fdµ = dν 8f 2 C(X) X X Then µ ≡ ν on MX . Proof. Suppose µ and ν are two Baire measures such that Z Z τ(f) = fdµ = fdν 8f 2 C(X) X X Let K be a compact Gδ-set and write 1 \ K = Un n=1 For each n 2 N, by Urysohn's lemma, 9fn 2 C(X) such that 0 ≤ fn ≤ 1 and fn ≡ 1 on K and fn ≡ 0 on X n Un It follows that fn & χK pointwise. Since µ and ν are both finite measures, it follows by the dominated convergence theorem that Z Z µ(K) = lim fndµ = fndν = ν(K) X X Since such compact Gδ-sets generate the Baire σ-algebra, it follows that µ ≡ ν on MX Definition 0.4. A topological space X is said to be extremally disconnected (or a Stonean space) if the closure of every open set is again open. Note that every discrete space is extremally disconnected. Lemma 0.5. Let X be an extremally disconnected Hausdorff space. Then (1) The collection A of all cl-open sets in X is an algebra. ie. It is closed under finite union, intersection, and taking complements. (2) The set G := spanfχA : A 2 Ag is dense in C(X) Proof. (1) This is obvious. (2) We verify that G satisfies the hypotheses of the Stone-Weierstrass theorem (??) (a) If f 2 G, then f 2 G since each characteristic function is real-valued. (b) Clearly, G contains the constant function 1 (c) If x; y 2 X are distinct, then there exist open sets U; V ⊂ Y such that x 2 U; y 2 V and U \ V = ;. Then A := U is a cl-open set and y2 = A. Hence, χA 2 G separates x from y. Thus, G is dense in C(X). Theorem 0.6 (Riesz representation - Special Case). Let X be an extremally disconnected, compact Hausdorff space. If τ : C(X) ! C is a positive linear functional, then there is a unique Baire measure µ such that Z τ(f) = fdµ 8f 2 C(X) X 2 Proof. Uniqueness was proved in Lemma 0.3, so we only prove existence. (1) Let A be as above. Define a set function ν : A! [0; 1) by ν(A) := τ(χA) Then ν is well-defined, ν(;) = 0, and ν is finitely additive. Furthermore, if A 2 A is expressed as a countable union 1 G A = An n=1 since all the sets are open and compact, it follows that atmost finitely many are non-empty. Hence, ν is automatically countably additive. (2) By Caratheodory's theorem, ν extends to a measure µ : S(A) ! R, where S(A) denotes the σ-algebra generated by A. We claim that M ⊂ S(A): If f 2 C(X), it suffices to show that, for each c 2 R, E := fx 2 X : f(x) ≤ cg = f −1(−∞; c] is in S(A). But then, for each n 2 N, En := fx 2 X : f(x) < c + 1=ng is open. Hence, En 2 A ⊂ S(A). Clearly, 1 \ E = En n=1 so E 2 S(A) as required. (3) Therefore, we may define τµ : C(X) ! C by Z τµ(f) := fdµ X Then τµ is a positive (bounded) linear functional on C(X). (4) Now observe that, for each A 2 A, τµ(χA) = µ(A) = ν(A) = τ(χA) By linearity, τµ and τ agree on G (as defined above). Since both τµ and τ are continuous linear functionals, they must agree on all of C(X). Hence, Z τ(f) = fdµ 8f 2 C(X) X From here on, the proof relies on some facts from the theory of C*-algebras. The reader who is unfamiliar with these matters may consult [Murphy] for the details. Note that if A is a commutative C*-algebra, we write Ω(A) for the space of non-zero multiplicative linear functionals on A, equipped with the weak-∗ topology. Definition 0.7. (1) Let S be a locally compact Hausdorff space, then Cb(S) is a uni- tal commutative C*-algebra. By the Gelfand-Naimark Theorem [Murphy, Theo- rem 2.1.10], ∼ Cb(S) = C(Ω(Cb(S)) and Ω(Cb(S)) is a compact Hausdorff space. This is called the Stone-Cech com- pactification of S, and is denoted by βS 3 (2) For each s 2 S, define τs : Cb(S) ! C by τs(f) := f(s). Then τs 2 βS, and clearly the map µ : S ! βS given by s 7! τs is injective and continuous (because βS is equipped with the weak-∗ topology). Lemma 0.8. (1) µ : S ! µ(S) is a homeomorphism. (2) µ(S) is dense in βS Proof. (1) We show that if F ⊂ S is closed, then µ(F ) is closed in µ(S): Let τs0 2 µ(S) n µ(F ). Then since µ is injective, s0 2= F . Since fs0g is compact, F is closed and S is locally compact Hausdorff, (a version of) Urysohn's lemma implies that 9f 2 Cb(S) such that f(s0) = 1 and fjF = 0 Consider the weak-∗ neighbourhood of τs0 given by U := fτs 2 µ(S): jτs(f) − τs0 (f)j < 1=2g Then U ⊂ µ(S) is open in the subspace topology, and if τs 2 U, then s2 = F . Hence, τs 2= µ(F ). Thus, U \ µ(F ) = ; Hence, µ(F ) is closed, as required. (2) Since βS is a compact Hausdorff space, if µ(S) is not dense in βS, then by Urysohn's lemma, 9f 2 C(βS) such that f(µ(s)) = 0 for all s 2 S but f 6= 0. This implies that f 2 Cb(S) has the property that f(s) = τs(f) = f(µ(s)) = 0 8s 2 S so f = 0 must hold. Lemma 0.9. If S is extremally disconnected, then βS is extremally disconnected. Proof. By Lemma 0.8, we simply identify S with µ(S) ⊂ βS. Let U ⊂ βS be an open set, then we WTS: clβS(U) is open. Assume WLOG that U 6= ;. (1) Since U is open and S is dense in βS, U\S 6= ;. Since S is extremally disconnected V := clS(U \ S) is open in S. Furthermore, it is clear that [Check!] clβS(U) = clβS(V ) ∼ (2) WTS: clβS(V ) is cl-open: Since V is cl-open, f := χV 2 Cb(S) = C(βS). Now, f −1(f1g) is a closed set containing V . Hence, −1 clβS(V ) ⊂ f (f1g) Similarly, −1 clβS(S n V ) ⊂ f (f0g) Since f takes values f0; 1g, it follows that S = clβS(S) = clβS(V [ (S n V )) = clβS(V ) [ clβS(S n V ) ⊂ f −1(f0g) t f −1(f1g) = f −1(f0; 1g) = S Hence, −1 clβS(V ) = f (f1g) 4 which is cl-open. Lemma 0.10. Let X and Y be two compact Hausdorff spaces. (1) If p : Y ! X is continuous, then the map p∗ : C(X) ! C(Y ) given by f 7! f ◦ p is a ∗-homomorphism. (2) If ' : C(X) ! C(Y ) is a ∗-homomorphism, then 9 a continuous map p : Y ! X such that ' = p∗ (3) p is surjective iff p∗ is injective.