Garling's Proof of the Riesz Representation Theorem

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 ∀f ∈ 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 ∈ C(X)- measurable. ie. MX is generated by sets of the form f −1(F ) where F ⊂ C is closed and f ∈ 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({0})

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) < ∞ for any compact K ∈ MX . (c) µ is inner regular, in that, for each Baire set E,

µ(E) = sup{µ(K): K ∈ MX compact}

(d) µ is outer regular, in that, for each E ∈ MX ,

µ(E) = inf{µ(U): U ∈ MX open} 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 ∀f ∈ 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 ∀f ∈ C(X) X X

Let K be a compact Gδ-set and write ∞ \ K = Un n=1

For each n ∈ N, by Urysohn’s lemma, ∃fn ∈ C(X) such that 0 ≤ fn ≤ 1 and

fn ≡ 1 on K and fn ≡ 0 on X \ 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 := span{χA : A ∈ A} 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 ∈ G, then f ∈ G since each characteristic function is real-valued. (b) Clearly, G contains the constant function 1 (c) If x, y ∈ X are distinct, then there exist open sets U, V ⊂ Y such that x ∈ U, y ∈ V and U ∩ V = ∅. Then A := U is a cl-open set and y∈ / A. Hence, χA ∈ 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 ∀f ∈ 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, ∞) by

ν(A) := τ(χA) Then ν is well-defined, ν(∅) = 0, and ν is finitely additive. Furthermore, if A ∈ A is expressed as a countable union ∞ 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 ∈ C(X), it suffices to show that, for each c ∈ R, E := {x ∈ X : f(x) ≤ c} = f −1(−∞, c] is in S(A). But then, for each n ∈ N,

En := {x ∈ X : f(x) < c + 1/n}

is open. Hence, En ∈ A ⊂ S(A). Clearly, ∞ \ E = En n=1 so E ∈ 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 ∈ 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 ∀f ∈ 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 ∈ S, define τs : Cb(S) → C by τs(f) := f(s). Then τs ∈ 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 ∈ µ(S) \ µ(F ). Then since µ is injective, s0 ∈/ F . Since {s0} is compact, F is closed and S is locally compact Hausdorﬀ, (a version of) Urysohn’s lemma implies that ∃f ∈ Cb(S) such that

f(s0) = 1 and f|F = 0

Consider the weak-∗ neighbourhood of τs0 given by

U := {τs ∈ µ(S): |τs(f) − τs0 (f)| < 1/2}

Then U ⊂ µ(S) is open in the subspace topology, and if τs ∈ U, then s∈ / F . Hence, τs ∈/ 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, ∃f ∈ C(βS) such that f(µ(s)) = 0 for all s ∈ S but f 6= 0. This implies that f ∈ Cb(S) has the property that

f(s) = τs(f) = f(µ(s)) = 0 ∀s ∈ 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 ∈ Cb(S) = C(βS). Now, f −1({1}) is a closed set containing V . Hence, −1 clβS(V ) ⊂ f ({1}) Similarly, −1 clβS(S \ V ) ⊂ f ({0}) Since f takes values {0, 1}, it follows that

S = clβS(S) = clβS(V ∪ (S \ V ))

= clβS(V ) ∪ clβS(S \ V ) ⊂ f −1({0}) t f −1({1}) = f −1({0, 1}) = S Hence, −1 clβS(V ) = f ({1}) 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 ϕ : C(X) → C(Y ) is a ∗-homomorphism, then ∃ a continuous map p : Y → X such that ϕ = p∗ (3) p is surjective iff p∗ is injective. (4) p is injective iff p∗ is surjective. Proof. (1) Easy exercise. (2) Let ϕ : C(X) → C(Y ), then we get an induced continuous map ϕt : Ω(C(Y )) → Ω(C(X)) given by τ 7→ τ ◦ ϕ Furthermore, the map

µX : X → Ω(C(X)) given by x 7→ τx

(where τx(f) := f(x)) is a homeomorphism. So we get a map

−1 µ ϕt µ p : Y −→Y Ω(C(Y )) −→ Ω(C(X)) −−→X X We claim that p∗ = ϕ, so ﬁx f ∈ C(X) and y ∈ Y , then which is continuous. If y ∈ Y , then ∗ p (f)(y) = f(p(y)) = τp(y)(f)

= µX (p(y))(f) = (µX ◦ p)(y)(f) t = (ϕ ◦ µY )(y)(f) t = (ϕ (τy))(f)

= τy(ϕ(f)) = ϕ(f)(y) Hence, p∗ = ϕ. (3) (a) Suppose p is surjective, WTS: ϕ = p∗ is injective: Suppose f, g ∈ C(X) such that p∗(f) = p∗(g) Then, for any y ∈ Y , f(p(y)) = g(p(y)) Since p is surjective, this implies f = g as required. (b) Suppose p∗ is injective, WTS: p is surjective: Suppose not, then p(Y ) is a proper closed subset in X. By Urysohn’s lemma, ∃f ∈ C(X) such that f 6= 0 but f(p(y)) = 0 ∀y ∈ Y But thus implies that p∗(f) = 0 and f 6= 0, contradicting the injectivity of p∗. (4) The argument that p is injective iff p∗ is surjective is similar [Check!] 5 Lemma 0.11. Let X be a compact Hausdorff space, then there is an extremally disconnected, compact Hausdorff space Y and a continuous surjective map p : Y → X Proof. Let X be a compact Hausdorff space. Let S denote the set X with the discrete topology, then S is locally compact, so by Definition 0.7, we may talk about βS. Fur- thermore, by Lemma 0.9, βS is extremally disconnected, compact and Hausdorff. Let f ∈ C(X), then f is bounded, so the induced map

fe : S → C is both continuous and bounded. This gives a ∗-homomorphism

ϕ : C(X) → Cb(S) given by f 7→ fe ∼ Since Cb(S) = C(βS), the previous lemma gives a continuous function p : βS → X ∗ such that ϕ = p . Furthermore, ϕ is clearly injective, so p is surjective. Theorem 0.12 (Riesz Representation Theorem). Let X be a compact Hausdorﬀ space and τ : C(X) → C be a positive linear functional. Then ∃ a unique Baire measure µ on X such that Z τ(f) = fdµ ∀f ∈ C(X) X Proof. Uniqueness was proved in Lemma 0.3, so we only prove existence. (1) Let τ : C(X) → C be a positive linear functional. Then, let Y be the extremally disconnected, compact Hausdorﬀ space and p : Y → X the surjective map as above. By Lemma 0.10, the map p∗ : C(X) → C(Y ) is an injective ∗-homomorphism. So by [Murphy, Theorem 3.3.8], ∃ a positive linear functional τe : C(Y ) → C such that, for each f ∈ C(X) τe(f) = τ(f ◦ p) By the Special case of the theorem, there is a Baire measure ν on Y such that Z τe(g) = gdν ∀g ∈ C(Y ) Y (2) Since p is continuous, for any open set U ⊂ X, p−1(U) is open in Y . Since

−1 \ \ −1 p ( Un) = p (Un) it follows that, for any Baire set E ⊂ X, p−1(E) is a Baire set in Y . Therefore, we may deﬁne µ : MX → [0, ∞) by µ(A) := ν(p−1(A)) Once again, by properties of p−1, it follows that µ is a measure on X. (3) Furthermore, if A ∈ MX and f = χA, then Z Z Z −1 fdµ = µ(A) = ν(p (A)) = χp−1(A)dν = (f ◦ p)dν X Y Y 6 Hence, it follows (by the dominated convergence theorems) that for any f ∈ C(X), Z Z fdµ = (f ◦ p)dν = τe(f ◦ p) = τ(f) X Y as required. References

[Arveson] W Arveson, https://math.berkeley.edu/~arveson/Dvi/rieszMarkov.pdf [Garling] DJH Garling, Another ‘short’ proof of the Riesz Representation theorem, Proc. Cambridge Philosophical Soc (1973) [Murphy] Gerard Murphy, C* algebras and Operator Theory, Academic Press (1990) [Rudin] W. Rudin, Real and Complex Analysis [Sunder] V. Sunder, https://www.imsc.res.in/~sunder/rrt1.pdf [Varadarajan] Varadarajan, V. On a theorem of F. Riesz concerning the form of linear functionals Fun- damenta Mathematicae 46.2 (1958): 209-220