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TOPOLOGICAL GROUPS - PART 1/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Locally compact second countable spaces 1 2. Measure Theory on LCSC spaces 7 3. Measure Theory and Functional Analysis 12 1. Locally compact second countable spaces All topological spaces are assumed to be Hausdorff, unless stated otherwise. Locally Compact (LC) spaces: The best after compact spaces. A topological space is LC if every x ∈ X has a neighborhood whose closure is compact; equivalently if X has a base B such that B is compact for every B ∈ B. One-point compactification of a LC (non-compact) space (X, T ) is a compact Hausdorff space (X ∪ {∞}, T 0 ) where T 0 is generated by T ∪ {{∞} ∪ (X \ K) : K ⊂ X compact}. Thus (X, T ) is a dense open subspace of (X ∪ {∞}, T 0 ). [101] Let X be LC Hausdorff. Then X is completely regular. Proof. X ∪ {∞} is compact Hausdorff, hence normal, hence completely regular. And complete regularity is passed onto all subspaces. ¤ [102] Let X be LC Hausdorff. If x ∈ X and if U is a neighborhood of x, then there exists open V ⊂ X such that V is compact and x ∈ V ⊂ V ⊂ U . Proof. X is completely regular, hence regular. Let V1 ⊂ X be open such that x ∈ V1 ⊂ V1 ⊂ U . Since X is LC, there is open V2 ⊂ X such that x ∈ V2 and V2 is compact. Let V = V1 ∩ V2 . ¤ [103] Closed subspaces and open subspaces of LC spaces are LC. Proof. If Y ⊂ X is closed and if y ∈ Y , let U ⊂ X be open with y ∈ U and U compact. If V = Y ∩ U , then V is open in Y , y ∈ V and clY [V ] = Y ∩ U is compact since Y is closed. If Z ⊂ X is open, then by [102], Z is LC. ¤ 2 2 Examples: Rn , Cn are LC; discrete spaces are LC; GL(n, R) = Rn \ det−1 ({0}) is open in Rn and hence LC; (a, b) is LC; infinite dimensional Banach spaces are not LC [∵ closed balls are not compact]; Q is not LC [∵ Q ∩ [a, b] is not sequentially compact for reals a < b]. 1 2 T.K.SUBRAHMONIAN MOOTHATHU [104] LC Hausdorff spaces have the Baire property. That is, countable intersection of dense open subsets is nonempty. Proof. Let Un ⊂ X be open and dense for n ∈ N. Choose nonempty open B1 ⊂ X such that B1 is compact B1 ⊂ U1 . At the (n + 1)th step, choose nonempty open Bn+1 ⊂ X such that Bn+1 is T T∞ compact and Bn+1 ⊂ Bn ∩ Un+1 . Then ∅ 6= ∞ ¤ n=1 Bn ⊂ n=1 Un . [105] Let X be a homogeneous LC Hausdorff space. If X is countable, then the topology on X is discrete. Proof. If X is not discrete, then by homogeneity {x} is not open for every x ∈ X. Let X = T {x1 , x2 , . . .} and Un = X \ {xn }. Then Un is open and dense in X but n Un = ∅ contradicting the Baire property. ¤ Remarks: As a corollary we get that Q is not LC since Q is homogeneous. By considering {0}∪{1/n : n ∈ N}, we see that homogeneity of the space is necessary in [105]. [106] Let X be Hausdorff (need not be LC) and Y ⊂ X be dense and LC. Then Y is open in X. Proof. If y ∈ Y , then there is a relatively open V ⊂ Y such that y ∈ Y and clY [V ] is compact. Then V = clY [V ]. Let U ⊂ X be open such that V = Y ∩ U . Since Y ∩ U is dense in U , we have U = V = clY [V ] ⊂ Y . Thus U ⊂ Y . ¤ Corollary: R \ Q is not LC. [107] Let X be LCSC (locally compact, second countable) Hausdorff. Then X is metrizable. Proof. By [101], X is completely regular and hence regular. Uryzohn’s metrization Theorem says that regular SC spaces are metrizable. ¤ [108] Let X be LCSC Hausdorff. Then, (i) X has a countable base {Bn : n ∈ N} such that Bn is compact for every n. (ii) X is σ-compact. In fact, there is a sequence (Kj ) of compact subsets such that Kj ⊂ int[Kj+1 ] S and X = ∞ j=1 Kj . Proof. Let B0 be a countable base of open balls for X and let B = {B(x, ²) ∈ B0 : B(x, 2²) compact}. (The extra factor 2 will be used to prove (ii)). Check that B is also a base for X. To prove (ii), write B = {B(xn , ²n ) : n ∈ N}. Let (αj ) be a strictly increasing sequence in (1, 2) and put S Kj = jn=1 B(xn , αj ²n ). ¤ [109] Let X be LCSC Hausdorff. Then X ∪ {∞} is SC and metrizable. Proof. If D ⊂ X is compact, then D ⊂ X = S∞ j=1 Kj ⊂ S∞ j=1 int[Kj+1 ], so D ⊂ Kj+1 for some j. This means that {{∞} ∪ (X \ Kj ) : j ∈ N} is a countable base at ∞. Combining with a countable TOPOLOGICAL GROUPS - PART 1/3 3 base for X (since X is open in X ∪ {∞}), we get a countable base for X ∪ {∞}. Being compact Hausdorff, X ∪ {∞} is normal and hence regular. Now apply Urysohn’s metrization theorem. ¤ Thus if X is LCSC (Hausdorff), its one-point compactification is a compact metric space. Since compact metric spaces are ‘nice’, we may expect LCSC spaces also to possess many ‘nice’ features. [110] Let (Y, d) be a complete metric space and let X ⊂ Y be a nonempty Gδ set. Then X admits a complete metric. In particular, open subsets of complete metric spaces are completely metrizable. Proof. Suppose X = T∞ n=1 Un where each Un is open in Y . We may assume Un 6= Y . We embed X as a closed subspace of Y × RN to say X is completely metrizable. Let f : X → Y × RN be f (x) = (x, z), where z = (zn ), zn = 1/d(x, Y \ Un ). Easy to see that f is a 1-1 continuous function with f −1 : f (X) → X continuous. To show f (X) is closed, consider (y, w) ∈ Y × RN and let (xm ) be a sequence in X such that (f (xm )) → (y, w). Then (xm ) → y and therefore f (y) = limm→∞ f (xm ) = (y, w) so that it now suffices to show y ∈ X. If y ∈ / X, then y ∈ ∂Un for some n, and then wn = limm→∞ [d(xm , Y \ Un )]−1 = ∞, a contradiction. ¤ Example: If Y = R and X = R \ {0}, then a complete metric on X is d(a, b) = |a − b| + | a1 − 1b |. Try to write down a complete metric on R \ Q. [111] Let X be LCSC. Then X admits a complete metric. Proof. X ∪ {∞} is a compact metric space and X is open in X ∪ {∞}. ¤ Exercise 1 : Let X, Y be non-compact LC Hausdorff spaces and let f : X → Y be a continuous proper map (i.e., f −1 (K) is compact for every compact K ⊂ Y ). Show that (i) f has a continuous extension fe : X ∪ {∞} → Y ∪ {∞0 }. (ii) f is a closed map. [Hint: (i) Put fe(∞) = ∞0 . (ii) If A ⊂ X is closed in X and B is the closure of A in X ∪ {∞}, then B is compact and B = A or A ∪ {∞}. Consider Y ∩ fe(B).] Exercise 2 : Let X be LC Hausdorff and consider K ⊂ U ⊂ X, where K is compact and U is open. Then there exists open V ⊂ X such that V is compact and K ⊂ V ⊂ V ⊂ U . [Hint: X is regular. So for each x ∈ K, there is open Vx such that Vx is compact and x ∈ Vx ⊂ Vx ⊂ U .] Exercise 3 : [Difference between compact and LC spaces] (i) Continuous image of a LC space need Q not be LC. (ii) Infinite product of LC spaces need not be LC; in fact, α Xα is LC iff every Xα is LC and Xα is compact except for finitely may α. (iii) A continuous real valued map on a LC space need not be uniformly continuous. [Hint: (i) Identity map from Q with discrete topology to Q with the usual topology. (ii) RN . (iii) f (x) = x2 on R.] Exercise 4 : Let X be LC Hausdorff. (i) U ⊂ X is open if U ∩ K is open in K for every compact K ⊂ X. (ii) If Y is Hausdorff and f : X → Y is a function such that f |K : K → Y is continuous for every compact K ⊂ X, then f is continuous. [Hint: (i) If x ∈ U , choose open V ⊂ X such that 4 T.K.SUBRAHMONIAN MOOTHATHU x ∈ V and V is compact. Since U ∩ V is open in V , U ∩ V is open in V and hence open in X. (ii) follows from (i).] Space of continuous functions C(X, Y ): We know that pointwise limit of a sequence of continuous functions need not be continuous. But the limit function is continuous if the convergence is locally uniform (i.e., every point has a neighborhood where the convergence is uniform). If the domain X is LC, then it may be seen that locally uniform convergence is equivalent to uniform convergence on compact subsets of X. Thus: [112] Let X be LC and Y be a metric space. If a sequence (fn ) in C(X, Y ) converges to a function f : X → Y uniformly of compact subsets of X, then f ∈ C(X, Y ). Because of this and certain other reasons, the ‘topology of uniform convergence on compact subsets’ is the ‘right’ topology on C(X, Y ). This topology is metrizable when X is LCSC, and this topology can be presented in three ways. Suppose that X is LCSC and let (Y, ρ) be a metric space. Replacing ρ with min{1, ρ}, we may assume ρ ≤ 1. Three (equivalent) topologies on C(X, Y ) are described below. (i) TU C , topology of uniform convergence on compact subsets: For f ∈ C(X, Y ), compact K ⊂ X and ² > 0, let B(f, K, ²) = {g ∈ C(X, Y ) : ρ(f (x), g(x)) < ² ∀x ∈ K}. If g ∈ ∩2i=1 B(fi , Ki , ²i ), note that the continuous function x → ρ(fi (x), g(x)) attains maximum on compact Ki , and therefore there exists αi > 0 such that ρ(fi (x), g(x)) + αi < ²i for every x ∈ Ki , for i = 1, 2. If α = min{α1 , α2 }, then B(g, K1 ∪ K2 , α) ⊂ ∩2i=1 B(fi , Ki , ²i ). It follows that the collection {B(f, K, ²) : f ∈ C(X, Y ), K ⊂ X compact , ² > 0} is a base for a topology on C(X, Y ). This topology is TU C . (ii) TCO , compact-open topology: The subbase {S(K, V ) : K ⊂ X compact ,V ⊂ Y open} generates this topology, where S(K, V ) = {f ∈ C(X, Y ) : f (K) ⊂ V }. [Remark : TP C , the topology of pointwise convergence on C(X, Y ) is the subspace topology coming from the inclusion C(X, Y ) ⊂ Y X . Now, {S(F, V ) : F ⊂ X finite ,V ⊂ Y open} is a subbase for TP C on C(X, Y ); and therefore, TCO = TP C on C(X, Y ) when X is discrete.] Tσ , a topology on C(X, Y ) coming from the σ-compactness of X: Write X = S∞ j=1 Kj , where Kj ’s are compact and Kj ⊂ int[Kj+1 ]. Let dj be the supremum metric on C(Kj , Y ), and define a metric Q d on C(X, Y ) making use of the embedding of C(X, Y ) in ∞ j=1 C(Kj , Y ) by the natural map P∞ dj (f, g) f 7→ (f |K1 , f |K2 , . . .). That is, d(f, g) = , where dj (f, g) = dj (f |Kj , g|Kj ) to be j=1 2j precise. The topology induced by this metric on C(X, Y ) will be denoted by Tσ . Observe that if X is compact, then Tσ is induced by the supremum metric on C(X, Y ). In the general case, the following result says in particular that Tσ is independent of the choice of Kj ’s. [113] Let X be LCSC and (Y, ρ) be a metric space with ρ ≤ 1. Then, TU C = TCO = Tσ on C(X, Y ). Proof. TU C ⊂ TCO : Consider B(f, K, ²). Choose δ > 0 such that ρX (a, b) < δ implies ρ(f (a), f (b)) < ² for every a, b ∈ K. Cover K with finitely many balls B1 , . . . , Bn having compact closures TOPOLOGICAL GROUPS - PART 1/3 5 and diameter < δ. Let Ai = K ∩ Bi , ai ∈ Ai and let Vi = Bρ (f (ai ), ²/2) ⊂ Y . Check that T f ∈ ni=1 S(Ai , Vi ) ⊂ B(f, K, ²). S TCO ⊂ Tσ : Consider S(K, V ) and f ∈ S(K, V ). Since K ⊂ X = ∞ j=1 int[Kj+1 ], there is j with K ⊂ Kj . Also since f (K) is compact and is contained in the open set V , there is ² > 0 such that N² (f (K)) ⊂ V . If g ∈ Bd (f, ²/2j ), then dj (f, g) ≤ 2j d(f, g) < ², and therefore g(K) ⊂ N² (f (K)) ⊂ V . Thus f ∈ Bd (f, ²/2j ) ⊂ S(K, V ). P Tσ ⊂ TU C : Consider Bd (f, ²) and choose k ∈ N so that j>k 2−j < ²/2. If g ∈ B(f, Kk , ²/2), P P then dj (f, g) ≤ ²/2 for j ≤ k so that kj=1 2−j dj (f, g) ≤ ²/2( kj=1 2−j ) < ²/2. Hence d(f, g) ≤ Pk P −j −j < ². Thus, f ∈ B(f, K , ²) ⊂ B (f, ²). ¤ k d j=1 2 dj (f, g) + j>k 2 Exercise-5 : Let X be LCSC and (Y, ρ) be a metric space with ρ ≤ 1. (i) If we consider the topology of pointwise convergence on C(X, Y ), then the evaluation map (x, f ) 7→ f (x) from X × C(X, Y ) to Y need not be continuous. If we consider TU C = TCO = Tσ on C(X, Y ), then the evaluation map (x, f ) 7→ f (x) from X × C(X, Y ) to Y is continuous. [Hint: (i) Consider a sequence (fn ) in C([0, 1], R) such that (fn ) → 0 pointwise and fn (1/n) = 1 for every n ∈ N. (ii) Let (xn ) → x and (fn ) → f . To show (fn (xn )) → f (x), note that (fn ) → f uniformly on the compact set {x} ∪ {xn : n ∈ N}.] From now onwards, when X is LCSC and (Y, ρ) is a metric space (with ρ ≤ 1), the topology on C(X, Y ) will be TU C = TCO = Tσ in our discussions. [114] Let X be LCSC and (Y, ρ) be a metric space. If Y is SC, then C(X, Y ) is SC and hence separable. Proof. It is convenient to work with TCO . Note that it is enough to produce a countable subbase. Let BX be a countable base for X consisting of open balls with compact closures. Let {An : n ∈ N} be the collection of all finite unions of closures of balls from BX . Let BY be a countable base of open balls for Y and let {Vm : m ∈ N} be the collection of all finite unions of balls in BY . We claim that {S(An , Vm ) : n, m ∈ N} is a countable subbase for (C(X, Y ), TCO ). Consider S(K, V ) and f ∈ S(K, V ). Cover f (K) with finitely many balls from BY contained in V and let Vm be their union. Choose ² > 0 so that N² (f (K)) ⊂ Vm . For each x ∈ K choose Bx ∈ BX such that x ∈ Bx and f (Bx ) ⊂ Bρ (f (x), ²) ⊂ N² (f (K), ²). Cover K with finitely many such balls and let An be the union of their closures. Check that f ∈ S(An , Vm ) ⊂ S(K, V ). ¤ [115] Let X be LCSC and (Y, ρ) be a complete metric space. Then (C(X, Y ), d) is a complete metric space. Proof. Let X = S∞ j=1 Kj , where Kj is compact and Kj ⊂ int[Kj+1 ], and let d = P∞ −j j=1 2 dj , where dj is the supremum metric on C(Kj , Y ). Consider a Cauchy sequence (fn ) in (C(X, Y ), d). If x ∈ X, there is j such that x ∈ Kj and then ρ(fn (x), fm (x)) ≤ dj (fn , fm ) ≤ 2j d(fn , fm ) and hence (fn (x)) is Cauchy in the complete space (Y, ρ). Define f : X → Y as the pointwise limit of (fn ). 6 T.K.SUBRAHMONIAN MOOTHATHU Since dj ≤ 2j d, (fn ) is Cauchy with respect to dj . Hence for a fixed j and ² > 0 there is n0 such that dj (fn , fm ) < ²/2 for every n, m ≥ n0 . If n ≥ n0 and x ∈ Kj choose m ≥ n0 with ρ(f (x), fm (x)) < ²/2 by pointwise convergence. Then ρ(f (x), fn (x)) ≤ ρ(f (x), fm (x)) + dj (fm , fn ) < ². Hence S (fn ) → f uniformly on Kj and therefore f |Kj is continuous. Since X = ∞ j=1 int[Kj+1 ], it follows that f ∈ C(X, Y ). To verify that (d(fn , f )) → 0, consider ² > 0 and let k ∈ N be such that P −j < ²/2. Choose n such that d (f, f ) < ²/2 for n ≥ n . Then d(f, f ) < ²/2 for j ≤ k 0 n 0 n k j>k 2 Pk P −j −j and n ≥ n0 so that d(f, fn ) ≤ ²/2( j=1 2 ) + j>k 2 < ² for n ≥ n0 . [Slightly easier argument is to prove the completeness of C(X, Y ) first when X is compact and then to observe in the general Q case that C(X, Y ) is embedded as a closed subspace of ∞ ¤ j=1 C(Kj , Y ).] [116] Corollary: If X is LCSC, then C(X, R) and C(X, C) are complete separable metric spaces. Remark : C(X, C) is not LC in general. For example, since C([0, 1], C) is an infinite dimensional Banach space with respect to the supremum norm and since the supremum metric on C([0, 1], C) comes from this suprem norm, C([0, 1], C) is not LC. Compact subsets of C(X, C) are described by [117] below when X is a compact metric space. Exercise-6 : Let (X, d) be a metric space and A ⊂ X. Then, (i) A is totally bounded iff A is totally bounded, (ii) A is totally bounded iff every sequence in A has a Cauchy subsequence. [Hint: If A is not totally bounded, there is δ > 0 such that A cannot be written as a finite union of open balls of radius δ. Inductively choose x1 , x2 , . . . in A such that d(xj , xn ) ≥ δ for j < n. Then (xn ) has no Cauchy subsequence. Conversely, if (xn ) is a sequence in A, cover A with finitely many balls of radius 1/2. One of them, say B1 , contains infinitely many xn ’s. Let xn1 ∈ B1 . At the (k + 1)th step, cover A ∩ Bk with finitely many balls of radius 2−(k+1) . One of them, say Bk+1 , contains infinitely many xn ’s contained in Bk . Let nk+1 > nk be such that xnk+1 ∈ Bk+1 ∩ Bk . Since d(xnk , xnk+1 ) < diam[Bk ] ≤ 2−k , it follows that (xnk ) is Cauchy.] [117] [Arzela-Ascoli Theorem] Let X be a compact metric space and Γ ⊂ C(X, C). Then Γ is compact iff Γ is pointwise bounded and equicontinuous. Proof. Since Γ is closed in the complete space C(X, C), Γ is always complete. Hence Γ is compact iff Γ is totally bounded iff Γ is totally bounded. We can show that: (i) If Γ is totally bounded, the Γ is equicontinuous. (ii) If Γ is pointwise bounded and equicontinuous, then Γ is uniformly bounded. (iii) If Γ is uniformly bounded and equicontinuous, then Γ is totally bounded (where the crucial point to use is that bounded subsets of C are totally bounded). ¤ Remark : The above result is no longer true if X is only LCSC. If X = R and Γ = {f } ⊂ C(X, C), f (x) = x2 , then Γ is compact but not equicontinuous since f is not uniformly continuous on R. It is not difficult to see that if X is compact and Γ ⊂ C(X, C) is pointwise equicontinuous, then Γ is (uniformly) equicontinuous. The correct generalization of [117] to the case where X is LCSC is: TOPOLOGICAL GROUPS - PART 1/3 7 [1170 ] [Arzela-Ascoli Theorem] Let X be LCSC and Γ ⊂ C(X, C). Then Γ is compact iff Γ is pointwise bounded and pointwise equicontinuous. Proof. Suppose Γ is compact. The set {f (x) : f ∈ Γ} is the continuous image of the compact set {x} × Γ under the evaluation map (x, f ) 7→ f (x) from X × C(X, C) to C [∵ Exercise-5]. So {f (x) : f ∈ Γ} is compact and hence bounded for each x ∈ X. Thus Γ is pointwise bounded. To show Γ is pointwise equicontinuous, consider x ∈ X. Choose compact K ⊂ X with x ∈ int[K]. Since f 7→ f |K from C(X, C) to C(K, C) is continuous, {f |K : f ∈ Γ} is compact in C(K, C), and then by [117], {f |K : f ∈ Γ} is equicontinuous. In particular, Γ is equicontinuous at x. Conversely, suppose that Γ is pointwise bounded and pointwise equicontinuous. Write X = S∞ j=1 Kj , where Kj is compact and Kj ⊂ int[Kj+1 ]. Let Γj = {f |Kj : f ∈ Γ}. Then Γj is pointwise bounded and (uniformly) equicontinuous. Hence by [117], Γj is totally bounded for every j ∈ N. If Γ is not totally bounded, there is δ > 0 and (fn ) in Γ such that d(fn , fm ) ≥ δ for n 6= m [∵ P Exercise-6]. If k ∈ N is such that j>k 2−j < δ/2, then we have dk (fn , fm ) ≥ δ/2 for n 6= m, a contradiction to the total boundedness of Γk . Thus Γ must be totally bounded and therefore Γ is compact. ¤ Example: Let Γ = {fn : n ∈ N} ⊂ C([0, 1], C), where fn (x) = xn . Note that if xn is the positive nth root of 1/n, then (xn ) → 1, fn (1) = 1 and (fn (xn )) = (1/n) → 0. Hence Γ is not equicontinuous and therefore Γ is not compact by [117]. 2. Measure Theory on LCSC spaces We need some abstract measure theory, especially on LCSC spaces. If X is LCSC, the σ-algebra B on X generated by open subsets of X is called the Borel σ-algebra on X and the members of B are called Borel subsets of X. If X is LCSC, let Cc (X, R) = {f ∈ C(X, R) : f has compact support}, where the support of f is defined as supp(f ) = {x ∈ X : f (x) 6= 0} (thus Cc (X, R) = C(X, R) when X is compact). It may be noted that Cc (X, R) is a vector subspace (in fact, a subalgebra) of C(X, R), and the following says that Cc (X, R) has plenty of elements. [118] Let X be LCSC and K ⊂ U ⊂ X, where K is compact and U is open. Then there exists f ∈ Cc (X, R) such that 0 ≤ f ≤ 1, f |K ≡ 1 and supp(f ) ⊂ U . Proof. Choose open W ⊂ X with compact closure such that K ⊂ W ⊂ W ⊂ U . If d is an admissible metric on X, let f : X → [0, 1] be f (x) = d(x, X \ W )/[d(x, K) + d(x, X \ W )]. ¤ Remark : The above result is true, but non-trivial to prove, when X is only LC Hausdorff. Exercise-7 : Let X be LCSC, K ⊂ X be compact and let {Uα : α ∈ I} be an open cover for K. Then there is ² > 0 such that N² (K) is compact and is also covered by {Uα : α ∈ I}. [Hint: If x ∈ K, there is α with x ∈ Uα and then there exists an open ball B with compact closure such that x ∈ B ⊂ B ⊂ Uα .] 8 T.K.SUBRAHMONIAN MOOTHATHU A concept called partition of unity is useful in obtaining global results by patching together various local arguments. Generalizing [118], we have: [119] [Partition of unity] Let X be LCSC, K ⊂ X be compact and let {Uα : α ∈ I} be an open cover for K. Then there exists finitely many functions f1 , . . . , fn ∈ Cc (X, R) such that 0 ≤ fi ≤ 1 P for every i, ni=1 fi (x) = 1 for every x ∈ K, and for each i there is α such that supp(fi ) ⊂ Uα (we say fi ’s are subordinate to {Uα : α ∈ I}). Proof. There is open W ⊂ X with compact closure such that K ⊂ W and {Uα : α ∈ I} is an open cover for W also [∵ Exercise-7]. Since K, X \ W are disjoint closed sets, there is continuous h : X → [0, 1] such that h ≡ 0 on K and h ≡ 1 on X \ W . Now, we can cover W with finitely many open balls B1 , . . . , Bn having compact closures such that for each i, there is αi with Bi ⊂ Uαi . Let gi ∈ Cc (X, [0, 1]) be such that gi ≡ 1 on Bi and P P supp(gi ) ⊂ Uαi for each i. Note that ni=1 gi (x) > 0 for every x ∈ W and hence ni=1 gi (x)+h(x) > P 0 for every x ∈ X. Let fi (x) = gi (x)/[ nj=1 gj (x) + h(x)]. ¤ Coming back to σ-algebras, let X be LCSC and consider the following σ-algebras on X: A1 = the σ-algebra generated by compact subsets of X, A2 = the σ-algebra generated by compact Gδ subsets of X, A3 = the smallest σ-algebra such that every f ∈ C(X, C) is measurable (with respect to the Borel σ-algebra on C), A4 = the smallest σ-algebra such that every f ∈ Cc (X, C) is measurable. Remark : A3 is sometimes called the Baire σ-algebra, and the members of A3 are called Baire subets of X. [120] Let X be LCSC, let B be the Borel σ-algebra on X and let Ai ’s be as given above. Then, B = A1 = A2 = A3 = A4 . Proof. We have A2 ⊂ A1 ⊂ B clearly. In a metric space every closed set and every hence every T compact is Gδ [∵ A = ∞ n=1 N1/n (A)], and closed subsets of LCSC spaces are σ-compact [∵ S∞ S∞ A = j=1 [A∩Kj ] if X = j=1 Kj ]. Thus B = A1 = A2 . If K ⊂ X is compact, then f ∈ Cc (X, R) ⊂ Cc (X, C) given in the proof of [118] satisfies f (x) = 1 iff x ∈ K and thus K = f −1 ({1}) ∈ A4 . So A1 ⊂ A4 . Since A4 ⊂ A3 ⊂ B = A1 , we get B = A3 = A4 . ¤ Exercise-8 : Let (X, A), (Y, A0 ) be measure spaces and suppose that A0 is generated by S ⊂ A0 (for example, the collection of open intervals generates the Borel σ-algebra on R). If f : (X, A) → (Y, A0 ) is a function such that f −1 (S) ∈ A for every S ∈ S, then f is measurable. [Hint: Verify that A00 = {A ∈ A0 : f −1 (A) ∈ A} is a σ-algebra.] TOPOLOGICAL GROUPS - PART 1/3 9 Remark : Let X be LCSC. Any measurable f : (X, BX ) → (C, BC ) is called a (complex) Borel function. The family of all complex Borel functions on X is characterized as the smallest family containing C(X, C) and closed under the operation of taking pointwise limit of sequences of functions. This follows from Exercises 9,10 and 11 below. Exercise-9 : Let X be LCSC and (fn ) be a sequence of complex Borel functions on X converging pointwise to a function f : X → C. Then f is also Borel. [Hint: Enough to show that f −1 (K) ∈ BX T S∞ −1 for every compact K ⊂ C. Let Um = N1/m (K) and show that f −1 (K) = ∞ m=1 n=m fn (Um ). Here the inclusion ⊃ is proved as follows. If x belongs to the right-side, then for every m there is n ≥ m with d(fn (x), K) < 1/m. Since K is compact, there is y ∈ K and a subsequence (nk ) such that limk→∞ fnk (x) = y. Then y = f (x) so that x ∈ f −1 (K).] Exercise-10 : Let X be LCSC and let F be the smallest collection of complex functions on X such that C(X, C) ⊂ F and F is closed under the operation of taking pointwise limits of sequences of functions. Then F contains all complex-valued simple Borel functions (i.e., finite linear combinations of characteristic functions of Borel sets). [Hint: Since F is a vector space, it suffices to show χB ∈ F for every Borel B ⊂ X. Let A = {B ⊂ X : χB ∈ F }. If K ⊂ X is compact, then there is continuous fn : X → [0, 1] such that fn |K ≡ 1 and fn ≡ 0 on X \ N1/n (K). Then (fn ) → χK pointwise, and thus A contains all compact subsets of X. Now it suffices to show that A is a σ-algebra. Observe the following: (i) (fn ) → χB implies (1 − fn ) → χX\B , (ii) Since C(X, R) is closed under the operations of taking maximum and minimum of finitely many functions, the same is true for real valued functions in F and consequently A is closed under finite unions and finite intersections, S (iii) A, B ∈ A implies A\B = A∩(X \B) ∈ A, (iv) If An ∈ A, B1 = A1 and Bn+1 = An+1 \ ni=1 Ai , S∞ P S then Bn ∈ A; and if fn = ni=1 χBi , then (fn ) → χA , where A = ∞ n=1 Bn .] n=1 An = Exercise-11 : Let X be LCSC and f : X → C be Borel. Then f is the pointwise limit of a sequence of simple Borel functions from X to C. [Hint: Assume f is real valued. For −n2n ≤ j < n2n , let P(n−1)2n A(n, j) = f −1 ([ 2jn , j+1 j=−n2n χA(n,j) .] 2n )), which is clearly Borel and define fn : X → R as fn = Exercise-12 : Let f : R → R be differentiable. Then f 0 is the pointwise limit of a sequence of continuous real valued functions on R and hence f 0 is Borel. [Hint: gn (x) = (f (x + 1/n) − f (x))/(1/n).] Aside: BR and R have the same cardinality (proof by transfinite induction), whereas the cardinality of the Lebesgue σ-algebra on R is same as that of P(R) [∵ every subset of the middle-third Cantor set is Lebesgue measurable]. Hence there are non-Borel, Lebesgue measurable sets A ⊂ R. Locally finite regular Borel measure: Let X be LCSC. Any measure on (X, B) is called a Borel measure. A Borel measure is locally finite if µ[K] < ∞ for every compact K ⊂ X (equivalently if every point has a neighborhood with finite measure). Note that every locally finite Borel measure 10 T.K.SUBRAHMONIAN MOOTHATHU µ on a LCSC space X is σ-finite; and µ[X] < ∞ when X is compact. A Borel measure µ on a LCSC space X is regular if for every Borel B ⊂ X, we have: (i) µ[B] = sup{µ[K] : K ⊂ B compact} [inner regularity], (ii) µ[B] = inf{µ[U ] : B ⊂ U, U open in X} [outer regularity]. S Exercise-13 : Let (X, µ) be a measure space, let En , Fn ⊂ X be measurable, and let E = ∞ n=1 En , T∞ F = n=1 Fn . (i) If En ⊂ En+1 and µ[E] < ∞, then for every ² > 0 there is k ∈ N such that µ[E \ Ek ] < ². (ii) If Fn+1 ⊂ Fn and µ[F1 ] < ∞, then for every ² > 0 there is k ∈ N such that µ[Fk \ F ] < ². (iii) The finiteness assumption ‘µ[·] < ∞’ cannot be dropped from (i) and (ii). S P∞ [Hint: (i) Let A1 = E1 and An+1 = En+1 \ En . Then, µ[E] = µ[ ∞ n=1 An ] = n=1 µ[An ] < ∞ S∞ P∞ by assumption. Hence µ[E \ Ek ] = µ[ n=k+1 An ] = n=k+1 µ[An ] → 0 as k → ∞. (ii) Take En = F1 \ Fn and apply (i). For (iii), consider X = R, En = (−n, n) and Fn = (n, ∞).] [121] Let X be a compact metric space and let µ be a Borel measure on X with µ[X] < ∞. Then µ is regular. Proof. Let A be the collection of all B ∈ B such that for every ² > 0, there exist compact K² ⊂ X and open U² ⊂ X with K² ⊂ B ⊂ U² and µ[U² \ K² ] < ². If K ⊂ X is compact, then limn→∞ µ[N1/n (K)] = µ[K] by Exercise-13. So A contains all compact sets. Now we show A is a σ-algebra. If B ∈ A and K² ⊂ B ⊂ U² , then X \ U² ⊂ X \ B ⊂ X \ K² so that X \ B ∈ A. If Bn ∈ A and ² > 0, choose compact Kn,² ⊂ B and open Un,² ⊃ B with µ[Un,² \ Kn,² ] < ²2−(n+1) . S Sk Let U² = ∞ n=1 Un,² . Next, by Exercise-13, there is k ∈ N such that for K² = n=1 Kn,² , we have S∞ S∞ S µ[ n=1 Kn,² \ K² ] < ²/2. If B = n=1 Bn , then K² ⊂ B ⊂ U² and µ[U² \ K² ] ≤ µ[U² \ ∞ n=1 Kn,² ] + S∞ P∞ −(n+1) µ[ n=1 Kn,² \ K² ] < n=1 ²2 + ²/2 = ². Thus B ∈ A. ¤ [122] Let X be LCSC and µ be a locally finite Borel measure on X. Then µ is regular. S Proof. Write X = ∞ j=1 Kj , where Kj is compact and Kj ⊂ int[Kj+1 ]. If B ∈ B, let Bj = B ∩ Kj . Given ² > 0, applying [121] to the restriction of µ to Kj+1 , find compact Aj ⊂ Bj and relatively open Vj ⊂ Kj+1 with Bj ⊂ Vj such that µ[Bj ] − µ[Aj ] < ²2−j and µ[Vj ] − µ[Bj ] < ²2−j . If Uj = Vj ∩ int[Kj+1 ], then Uj is open in int[Kj+1 ] and hence in X. Moreover, Bj ⊂ Uj and S µ[Uj ] − µ[Bj ] < ²2−j . If µ[B] < ∞, then arguing as in the proof of [121], for open U² = ∞ j=1 Uj Sk and compact A² = j=1 Aj (k large), we have A² ⊂ B ⊂ U² and µ[U² \ A² ] < 2². If µ[B] = ∞, S then for compact Dk = kj=1 Aj ⊂ B, we have µ[B] = supk∈N µ[Dk ]; and µ[B] = inf{µ[U ] : U ⊂ X open and B ⊂ U } trivially. ¤ Integration: Recall that Riemann integration: integration of step functions first, then approximation. Lebesgue integration: integration of simple functions first, then approximation. Let X be LCSC and µ be a locally finite Borel measure on X. A real valued simple Borel P P function f = ni=1 ci χBi ≥ 0 is integrable with respect to µ if ni=1 ci µ[Bi ] < ∞. If this happens TOPOLOGICAL GROUPS - PART 1/3 we write R RX f dµ = Pn i=1 ci µ[Bi ]. 11 A non-negative Borel f : X → R is integrable with respect to µ if sup{ X gdµ : g is simple Borel, 0 ≤ g ≤ f } < ∞. If this happens, this finite quantity is the R value of X f dµ. A general real valued Borel function on X is integrable if its positive and negative R R R parts are integrable; and X f dµ = X f+ dµ − X f− dµ. Note that every f ∈ Cc (X, R) is integrable with respect to µ. A complex Borel function on X is integrable if its real and imaginary parts are R R R integrable; and X f dµ = X Re(f )dµ + i X Im(f )dµ. Analogues of standard results of Lebesgue integration theory are true here - we skip the details. [123] Let X be LCSC and let µ, β be locally finite Borel measures on X. (i) If µ[K] = β[K] for every compact K ⊂ X, or if µ[U ] = β[U ] for every open U ⊂ X, then µ = β. R R (ii) If X f dµ = X f dβ for every f ∈ Cc (X, R) with f ≥ 0, then µ = β. Proof. (i) is a corollary to [122]. To prove (ii), consider compact K ⊂ X and choose a sequence (fn ) in Cc (X, R) such that 0 ≤ fn+1 ≤ fn ≤ 1 and (fn ) → χK pointwise. Then by Dominated R R R R Convergence Theorem, µ[K] = X χK dµ = limn→∞ X fn dµ = limn→∞ X fn dβ = X χK dβ = β[K]. Now use (i). ¤ Let L1R (X, µ) be the space of (equivalence classes of) all µ-integrable functions f : X → R (one has to enlarge the class of Borel functions first by completing the Borel σ-algebra, etc.). L1R (X, µ) R is a normed space with respect to the norm kf k1 = X |f |dµ. [124] Let X be LCSC and µ be a locally finite Borel measure on X. Then (L1R (X, µ), k · k1 ) is separable and Cc (X, R) is dense in it. Proof. The collection of all µ-integrable simple Borel functions on X is dense in L1R (X, µ) by the definition of integration. Now let U be a countable base of open balls with compact closures in X and let V be all finite unions of members of U. Consider a µ-integrable simple Borel function P g = ki=1 ci χBi . Since µ is regular by [122], each Bi we can approximate in measure by an open superset W , and then there is V ∈ V such that µ[W ∆V ] is sufficiently small. Also each ci can be P approximated by a rational.. Hence the countable collection F = { ki=1 ci χVi : k ∈ N, ci ∈ Q, Vi ∈ V} is dense in L1R (X, µ). Now to show Cc (X, R) is dense in L1R (X, µ), it is enough to show that χV , V ⊂ X open with µ[V ] < ∞, can be approximated by members of Cc (X, R) since Cc (X, R) is a vector subspace and since F is dense in L1R (X, µ). Let Kn ’s be compact with Kn ⊂ Kn+1 and S V = ∞ n=1 Kn . Let fn ∈ Cc (X, R) be such that 0 ≤ fn ≤ 1, fn ≡ 1 on Kn , and fn ≡ 0 on X \ V . Then, kχV − fn k1 ≤ µ[V \ Kn ] → 0. ¤ Similar argument applies for LpR (X, µ), LpC (X, µ) (1 ≤ p < ∞). It is also true that Lp (X, µ) is complete for 1 ≤ p ≤ ∞. 12 T.K.SUBRAHMONIAN MOOTHATHU [1240 ] Let X be LCSC and µ be a locally finite Borel measure on X. Then for 1 ≤ p < ∞, (LpC (X, µ), k · kp ) is a separable Banach space and Cc (X, C) is dense in LpC (X, µ). Also L2C (X, µ) is R a Hilbert space with respect to the inner product hf, gi = X f gdµ. 3. Measure Theory and Functional Analysis Let µ, β be two measures on an abstract measurable space (X, A). We say β is absolutely continuous with respect to µ if µ[A] = 0 ⇒ β[A] = 0 for every A ∈ A (notation: β << µ); and we say µ, β are singular (or orthogonal ) if there is A ∈ A such that β[A] = 0 = µ[X \ A] (notation: µ ⊥ β). Examples: (i) Let β be the Lebesgue measure on [0, 1], µ[A] = β[A ∩ [0, 1/2]], β2 [A] = [A ∩ [1/2, 1]], and β1 = µ. Then β1 << µ, β2 ⊥ µ and β = β1 + β2 . (ii) Let µ be the Lebesgue measure on [0, 1], and let β1 , β2 be Borel measures on [0, 1] defined by: R β1 [A] = 1 if 0 ∈ A, β1 [A] = 0 if 0 ∈ / A, and β2 [A] = A f dµ where f : [0, 1] → R is µ-integrable and ≥ 0 (here, the countable additivity of β2 is checked using the Monotone Convergence Theorem). Then β1 << µ and β2 ⊥ µ. (iii) Let f : R → R be a Lipschitz continuous homeomorphism, µ be the Lebesgue measure on R and define β[A] = µ[f (A)]. Then β is a Borel measure on R. Let λ > 1 be a Lipschitz constant for f . If B ⊂ R is Borel with µ[B] = 0 and ² > 0, then there are countably many intervals Jn such that S P∞ P∞ B⊂ ∞ n=1 Jn and n=1 µ[Jn ] < ²/λ. If Kn = f (Jn ), then µ[Kn ] ≤ λµ[Jn ] so that n=1 µ[Kn ] < ². S∞ Also f (B) ⊂ n=1 Kn . Hence β[A] = µ[f (A)] = 0. Thus β << µ. Exercise-14 : Let µ, β be finite measures on (X, A). Then β << µ iff for every ² > 0 there is δ > 0 such that β[A] < ² for every A ∈ A with µ[A] < δ. [Hint: Let ² > 0. If there is no δ > 0 with the required property, then there are An ∈ A such that µ[An ] < n−2 and β[An ] ≥ ². If T S∞ B= ∞ k=1 n=k An , then µ[B] = 0 and β[B] ≥ ² > 0. To obtain ‘⇐’, note that if β[A] > ² > 0 and if δ > 0 is chosen for this ², then µ[A] ≥ δ > 0.] Exercise-15 : Let µ, β be measures on (X, A) with β << µ and β ⊥ µ. Then β ≡ 0. Recall the following elementary geometric fact: if u, v are two vectors in the plane, then v can be written uniquely as v = v1 + v2 so that the vector v1 is parallel to u and v2 ⊥ u. A result of the same spirit for measures is: [125] [Lebesgue Decomposition Theorem] (true for σ- finite measures also) Let β, µ be two finite measures on a measurable space (X, A). Then there are unique measures βa , βs on (X, A) such that β = βa + βs , βa << µ and βs ⊥ µ. Proof. Note that beta << β + µ. The idea is to obtain a suitable g such that β[A] = for every A ∈ A. Then βa , βs can be defined with the help of g. R A gd(β + µ) TOPOLOGICAL GROUPS - PART 1/3 13 We have (β + µ)[X] < ∞ and hence L2R (X, β + µ) ⊂ L1R (X, β + µ). Consider the linear functional R φ : L2R (X, β + µ) → R given by φ(h) = X hdβ. By Cauchy-Schwarz,we have Z Z |φ(h)| ≤ |h|dβ ≤ |h|d(β + µ) = khk1 = kh · 1k1 ≤ khk2 k1k2 = khk2 [(β + µ)[X]]1/2 , X X and hence φ is bounded. By Riesz representation theorem, there is g ∈ L2R (X, β + µ) such that φ(h) = hh, gi. That is, R R (i) X hdβ = X hgd(β + µ) ∀h ∈ L2R (X, β + µ), and R (ii) β[A] = A gd(β + µ) ∀A ∈ A (taking h = χA in (i)). We use (ii) for the rest of the proof; (i) will be used to prove the next theorem. First we claim that 0 ≤ g ≤ 1. Let An = {x ∈ X : g(x) ≤ −n−1 } and Bn = {x ∈ X : g(x) ≥ 1 + n−1 }. Using A = An in (ii), β[An ] ≤ −n−1 (β + µ)[An ] and hence (β + µ)[An ] = 0. Similarly, A = Bn in (ii) gives β[Bn ] ≥ (1 + n−1 )(β[Bn ] + µ[Bn ]) which implies β[Bn ] = 0 and then µ[Bn ] = 0. Thus 0 ≤ g ≤ 1 (β + µ)-almost everywhere. Let S = {x ∈ X : g(x) = 1}, T = X \ S. Then, T = S∞ n=1 Tn , Tn = {x ∈ X : g(x) ≤ 1 − n−1 }. Taking A = S in (ii), we get µ[S] = 0. For Y ∈ A with µ[Y ] = 0, taking A = Y ∩ Tn in (ii), we P have β[Y ∩ Tn ] ≥ (1 − n−1 )β[Y ∩ Tn ] implying β[Y ∩ T ] ≤ ∞ n=1 β[Y ∩ Tn ] = 0. Thus if we define βs [Y ] = β[Y ∩ S], βa [Y ] = β[Y ∩ T ], then βs ⊥ µ, βa << µ and β = βa + βs . Uniqueness: Suppose β = αa + αs is another decomposition with respect to µ. Then αa [S] = 0 since µ[S] = 0. Hence βs [Y ] = β[Y ∩ S] = αa [Y ∩ S] + αs [Y ∩ S] = αs [Y ∩ S] ≤ αs [Y ]. Thus βs ≤ αs and hence βa ≥ αa . Then the measure ν = αs − βs = βa − αa satisfies ν ⊥ µ and ν << µ. Therefore ν ≡ 0. ¤ [126] [Radon-Nikodym Theorem for finite measures] Let β, µ be finite measures on a measurable space (X, A) and suppose that β << µ. Then there exists a non-negative µ-integrable function R f : X → R such that β[A] = A f dµ for every A ∈ A. Moreover, f is unique µ-almost everywhere. (Thus β << µ ⇔ β is obtained by integrating against µ.) Proof. Let φ, g, S, T be as in the proof of [125]. Let β = βa + βs be the Lebesgue decomposition of β with respect to µ. Since β << µ, we have βa = β and βs = 0 by the uniqueness of the decomposition. Now, let us try to make a guess about f . Taking h = f χA in (i) of the proof R R R of [125] we see that the required f must satisfy A f dβ = A f gd(β + µ), or (adding A f dµ to R R both sides and rearranging,) A f (1 − g)d(β + µ) = A f dµ. On the other hand we must have R R A f dµ = β[A] = A gd(β + µ) by (ii) from the proof of [125], so that we may expect the identity f (1 − g) = g. R Define f : X → R as f ≡ 0 on S and f = g/(1 − g) on T . Then A f (1 − g)d(β + µ) = R R A∩T f (1 − g)d(β + µ) = A∩T gd(β + µ) = (by (ii) from [125] )β[A ∩ T ] = βa [A] = β[A]. But we R R R have already derived above that A f (1 − g)d(β + µ) = A f dµ. Thus β[A] = A f dµ for every A ∈ A. 14 T.K.SUBRAHMONIAN MOOTHATHU Uniqueness µ-almost everywhere: If h is another such function and if An = {x ∈ X : f (x)−h(x) ≥ R R R then An f dµ = β[An ] = An hdµ and hence 0 = An (f −h)dµ ≥ n−1 µ[An ] implying µ[An ] = 0 n−1 }, and therefore µ[{x ∈ X : f (x) > h(x)}] = 0. Similarly µ[{x ∈ X : h(x) > f (x)}] = 0 also. ¤ [1260 ] [Radon-Nikodym Theorem] Let β, µ be σ-finite measures on a measurable space (X, A) and let β << µ. Then there exists a non-negative measurable function f : X → R such that R β[A] = A f dµ for every A ∈ A. Moreover, f is unique µ-almost everywhere. Proof. Write X = S∞ i=1 Xi , a disjoint union of measurable sets Xi with (β + µ)[Xi ] < ∞. Get fi : Xi → R by [126] and let f : X → R be f |Xi = fi . ¤ We know that locally finite Borel measures on LCSC spaces are σ-finite. Therefore, we have: [12600 ] [Radon-Nikodym Theorem for LCSC spaces] Let β, µ be locally finite Borel measures on a LCSC space X and let β << µ. Then there exists a non-negative Borel function f : X → R such R that β[A] = A f dµ for every Borel A ⊂ X. Moreover, f is unique µ-almost everywhere. Remark : The function f given by the Radon-Nikodym Theorem is called the Radon-Nikodym derivative of β with respect to µ and is denoted by dβ dµ . This notation is justified by the following R R observation. If g : (X, A) → R is a non-negative measurable function, then X gdβ = X g dβ dµ dµ [∵ R R R dβ dβ if g = χA , then X gdβ = β[A] = A dµ µ = X g dµ µ; extend this to nonnegative simple functions by linearity and then use approximation]. Execise-16 : Let µ be the Lebesgue measure on [0, 1], let f : [0, 1] → [0, 1] be a twice differentiable homeomorphism with f 0 > 0 on (0, 1) (example: f (x) = xn ) and define β[A] = µ[f (A)] for Borel A ⊂ [0, 1]. We already know that β << µ (since f is Lipschitz by Mean value theorem). Show that dβ dµ = f 0 . [Hint: Since f 0 > 0, f is increasing on (0, 1) and hence on [0, 1]. If [a, b] ⊂ [0, 1], then R Ra 0 R dβ 0 [a,b] f dµ = b f (t)dt = f (b) − f (a) = µ[f ([a, b])] = β[[a, b]] = [a,b] dµ dµ. Since closed intervals R 0 R dβ generate the Borel sigma algebra on [0, 1], we deduce A f dµ = A dµ dµ for every Borel A ⊂ [0, 1]. By uniqueness, dβ dµ = f 0. ] Exercise-17 : Let Y, X be measurable spaces, g : Y → X be a measurable surjection and let β be a finite measure on Y . Define µ[A] = β[g −1 (A)] for measurable A ⊂ X. Then µ is a R R measure on X and Y f ◦ gdβ = X f dµ for integrable f : X → R. [Hint: If f = χA , then R R R −1 Y f ◦ gdβ = g −1 (A) 1dβ = β[g (A)] = µ[A] = X f dµ.] Locally finite Borel measure = positive linear functional: Let X be LCSC. If µ is a locally finite R Borel measures on X and if we define φ : Cc (X, R) as φ(f ) = X f dµ, then it is easy to see that φ is a positive linear functional (here, φ positive means φ(f ) ≥ 0 for every f ∈ Cc (X, R). Our aim is to prove the converse that every positive linear functional on Cc (X, R) has this form. We need some preparation. TOPOLOGICAL GROUPS - PART 1/3 15 Let X be a set. We say A0 ⊂ P(X) is an algebra if ∅ ∈ A0 and if A0 is closed under finite unions and complementation. For example, if X = [0, 1] and A0 is the collection of all finite unions of subintervals of [0, 1], then A0 is an algebra and the σ-algebra A generated by A0 is the Borel σ-algebra on [0, 1]. By a measure on an algebra A0 we mean a function µ : A0 → [0, ∞] such that µ[∅] = 0 S P∞ 0 and µ[ ∞ n=1 An ] = n=1 µ[An ] whenever A1 , A2 , . . . ∈ A are pairwise disjoint subsets such that S∞ 0 0 n=1 An ∈ A (note that the union need not be in A sometimes, in which case the condition holds automatically). We omit the proof of the following. [127] [Caratheodory’s Extension Theorem (special case)] Let X be a set, A0 be an algebra on X and let µ be a finite measure on A0 . If A is the σ-algebra generated by A0 , then µ extends to a unique finite measure on (X, A). Example: Let X = [0, 1] and A0 be the collection of all finite unions of subintervals of [0, 1]. If S P A ∈ A0 , write A = nn=1 Ai , a disjoint union of intervals, and define µ[A] = ni=1 length[Ai ]. Then µ extends to the Lebesgue measure on the Borel σ-algebra on [0, 1]. Another preparation needed is the following topological fact about the Cantor space {0, 1}N . [128] If X is a compact metric space, then there is a continuous surjection f : {0, 1}N → X. Proof. Let A(1), . . . , A(n1 ) ⊂ X be finitely many nonempty compact sets with diam[A(i)] < 1 and Q S 1 A(i). At the (k+1)th step, for each (i1 , . . . , ik ) ∈ kj=1 {1, 2, . . . , nj }, choose finitely many X = ni=1 nonempty compact sets A(i1 , . . . , ik , 1), . . . , A(i1 , . . . , ik , n(k+1) ) having diameter < 1/(k + 1) such Q nk that their union equals A(i1 , . . . , ik ). Since {0, 1}N = ∞ k=1 {0, 1} , it is easy to find a continuous Q surjection g : {0, 1}N → ∞ k=1 {1, 2, . . . , nk }. Hence it suffices to find a continuous surjection h : Q∞ Q∞ k=1 {1, 2, . . . , nk } k=1 {1, 2, . . . , nk } → X (for then we may take f = h◦g). For y = (y1 , y2 , . . .) ∈ T∞ ¤ define h(y) to be the unique point in k=1 A(y1 , . . . , yk ). Verify that this does the job. Now we prove Riesz Representation Theorem for Measures [RRTM] in three steps (adapting the proof due to V.S. Sunder): first for {0, 1}N , then for compact metric spaces, and then for LCSC spaces. [129] [RRTM for {0, 1}N ] Let X = {0, 1}N and let φ : C(X, R) → R be a positive linear functional. R Then there is a unique finite Borel measure µ on X such that φ(f ) = X f dµ for every f ∈ C(X, R). Proof. For F = {n1 < n2 < · · · < nk } ⊂ N and w = w1 w2 · · · wk ∈ {0, 1}k , let A(F, w) = {x ∈ X : xni = wi for 1 ≤ i ≤ k}. Each A(F, w) is both compact and open. Let A0 be the collection of all finite unions of A(F, w)’s, where F ⊂ N is finite and w ∈ {0, 1}|F | . Then A0 is an algebra and every member of A0 is both compact and open; in particular, χA ∈ C(X, R) for A ∈ A0 . If µ : A0 → [0, ∞) is defined as µ[A] = φ(χA ), then µ[A ∪ B] = φ(χA∪B ) = φ(χA + χB ) = φ(χA ) + φ(χB ) = µ[A] + µ[B] S for disjoint A, B ∈ A0 . Since every member of A0 is both compact and open, ∞ / A0 if n=1 An ∈ 16 T.K.SUBRAHMONIAN MOOTHATHU An ∈ A0 are pairwise disjoint nonempty sets. Hence µ is countably additive on A0 trivially. Also µ[X] = φ(1) ∈ [0, ∞). Thus µ is a finite measure on A0 . Since A0 is also a base for the topology of X, the algebra generated by A0 is the Borel σ-algebra. So µ extends to a finite Borel measure on X R by [127]. Clearly φ(χB ) = µ[B] = X χB dµ for every Borel B ⊂ X. By linearity and approximation, R φ(f ) = X f dµ for every f ∈ C(X, R). Uniqueness: If β is another such measure, then β[A] = φ(χA ) = µ[A] for every A ∈ A0 and then β[U ] = µ[U ] for every open U ⊂ X by approximation. Then β = µ by [123]. ¤ Lemma: Let X be a compact metric space and consider C(X, R) with the supremum norm. If φ : C(X, R) → R is linear, then the following are equivalent: (i) φ is positive. (ii) φ(f ) ≤ φ(g) for f, g ∈ C(X, R) with f ≤ g. (iii) φ is bounded and kφk = φ(1). Proof. (i) ⇔ (ii) is easy. To see (ii) ⇒ (iii), consider f ∈ C(X, R) with |f | ≤ 1. Then f ≤ |f | ≤ 1 and −f ≤ |f | ≤ 1 so that φ(f ) ≤ φ(1) and −φ(f ) = φ(−f ) ≤ φ(1). Hence |φ(f )| ≤ φ(1), and equality holds when f = 1. To prove (iii) ⇒ (i), it suffices to show φ(f ) ≥ 0 for f ∈ C(X, R) with 0 ≤ f ≤ 1. Since 0 ≤ 1 − f ≤ 1, φ(1) − φ(f ) = φ(1 − f ) ≤ kφk ≤ φ(1) and so φ(f ) ≥ 0. ¤ Remarks: (i) The above lemma is true for linear φ : C(X, C) → C, but the proof is more involved. (ii) If X is only LCSC, a positive linear functional φ : (Cc (X, R), sup norm) → R need not be R bounded. For instance, consider φ : Cc (X, R) → R given by φ(f ) = R f (x)dx. [1290 ] [RRTM for compact metric spaces] Let X be a compact metric space and let φ : C(X, R) → R be a positive linear functional. Then there is a unique finite Borel measure µ on X such that R φ(f ) = X f dµ for every f ∈ C(X, R). Proof. Let Y = {0, 1}N and let p : Y → X be a continuous surjection given by [128]. Define T : C(X, R) → C(Y, R) as T (f ) = f ◦ p. Then T is a linear isometry and T (f ) ≥ 0 whenever f ≥ 0. So C(X, R) may be identified with a vector subspace of C(Y, R) and φ may be considered as a positive linear functional on this subspace. By the lemma, φ is bounded and kφk = φ(1). By Hahne (f )) = φ(f ) Banach Theorem, there is a bounded linear functional φe : C(Y, R) → R such that φ(T e = kφk = φ(1X ) = φ(T e (1X )) = φ(1 e Y ). So φe is positive by the for every f ∈ C(X, R) and kφk R e lemma. Therefore, by [129] there is a unique finite Borel measure β on Y such that φ(g) = gdβ Y β[p−1 (A)] for every g ∈ C(Y, R). Define µ[A] = for Borel A ⊂ X. Then µ is a finite Borel measure R R e (f )) = on X and Y f ◦ pdβ = X f dµ for every f ∈ C(X, R) by Exercise-17. Hence φ(f ) = φ(T R R e ◦ p) = φ(f f ◦ pdβ = f dµ for every f ∈ C(X, R). Uniqueness of µ follows from [123]. ¤ Y X TOPOLOGICAL GROUPS - PART 1/3 Remark : It may be noted that kφk = φ(1) = R X 17 1dµ = µ[X], so that positive linear functionals of norm 1 on C(X, R) correspond to Borel probability measures on X, when X is a compact metric space. [12900 ] [Riesz Representation Theorem for Measures for LCSC spaces] Let X be LCSC and let φ : Cc (X, R) → R be a positive linear functional. Then there is a unique locally finite Borel R measure µ on X such that φ(f ) = X f dµ for every f ∈ Cc (X, R). Proof. Step-1: Let K ⊂ X be compact and Un = N1/n (K). Note that Un+1 ⊂ Un for all n and Un is compact for all large n. Let gn ∈ C(X, [0, 1]) be such that gn ≡ 1 on Un+1 and gn ≡ 0 on X \ Un . Then 0 ≤ gn+1 ≤ gn ≤ 1. Let C+ (K, R) = {f ∈ C(K, R) : f ≥ 0}. If f ∈ C+ (K, R), we see that there is fe ∈ Cc (X, R) with fe ≥ 0 and fe|K = f (∵ choose open U with K ⊂ U , let fe = f on K, fe ≡ 0 on X \ U and apply Tietze). Since 0 ≤ fegn+1 ≤ fegn , (φ(fegn )) is a decreasing sequence of nonnegative reals and hence converges. Define φK : C+ (K, R) → R as φK (f ) = limn→∞ φ(fegn ). To verify that φK (f ) is independent of the choice of the extension fe, consider another extension f ∗ ∈ Cc (X, R) of f with f ∗ ≥ 0. Suppose that Un is compact for every n ≥ n0 . Given ² > 0 let An = {x ∈ Un : |fe(x) − f ∗ (x)| ≥ ²} for n ≥ n0 . Then An ’s are compact, An+1 ⊂ An and T T T ∗ n≥n0 An ⊂ n≥n0 Un = K so that n≥n0 An = ∅ (∵ f = f = f on K). So there is n(²) ≥ n0 such that An(²) = ∅. Then |fe(x) − f ∗ (x)| < ² for every x ∈ Un(²) . Now gn ≡ 0 on X \ Un(²) gor n ≥ n(²). Hence |fegn − f ∗ gn | ≤ ²gn ≤ ²gn0 for every n ≥ n(²) ≥ n0 . That is, fegn − f ∗ gn ≤ ²gn0 and f ∗ gn − fegn ≤ ²gn0 for n ≥ n(²). Applying φ, |φ(fegn ) − φ(f ∗ gn )| ≤ ²φ(gn0 ) for every n ≥ n(²) which implies limn→∞ φ(fegn ) = limn→∞ φ(f ∗ gn ). Next note that φK has the following properties: φK (f + h) = φK (f ) + φK (h) and φK (αf ) = αφK (f ) for every f, h ∈ C+ (K, R) and α ≥ 0. Hence by the following result φK extends uniquely to a linear functional φK on C(K, R). Result: Let Y be a compact metric space and ψ : C+ (Y, R) → R be such that ψ(f +h) = ψ(f )+ψ(h) and ψ(αf ) = αψ(f ) for every f, h ∈ C+ (Y, R) and α ≥ 0. Then ψ has a unique extension to a linear functional on C(Y, R). [Proof : Let f1 , f2 , h1 , h2 ∈ C+ (Y, R) and suppose f1 − f2 = h1 − h2 . Then f1 + h2 = h1 + f2 so that by applying ψ and rearranging we obtain ψ(f1 ) − ψ(f2 ) = ψ(h1 ) − ψ(h2 ). For f ∈ C(Y, R), define ψ(f ) = ψ(h1 ) − ψ(h2 ), where h1 , h2 ∈ C+ (Y, R) are such that f = h1 − h2 (for example, f = f+ − f− ). Then this extension ψ is well-defined by the above argument. Check other properties.] Note that the extended linear functional φK : C(K, R) → R is positive (∵ if f ≥ 0, then φK (f ) = limn→∞ φ(fegn ) ≥ 0 since φ is positive). So by [1290 ], there is a unique finite Borel R measure µK on K with φK (f ) = K f dµK . Also observe that φK (f |K ) = φ(f ) for nonnegative f ∈ Cc (X, R) with supp(f ) ⊂ K (∵ taking fe = f , φK (f |K ) = limn→∞ φ(f gn ) = φ(f ) since f gn = f ). 18 T.K.SUBRAHMONIAN MOOTHATHU Step-2: Write X = S∞ j=1 Kj , where Kj ’s are compact and Kj ⊂ int[Kj+1 ]. Step-1 gives a positive linear functional φj : C(Kj , R) → R defined using φ and let µj be the unique finite Borel measure on Kj corresponding to φj . We claim that µj+1 |Kj = µj . Choose gn ’s for Kj as in Step-1 and for f ∈ C+ (Kj , R), let fe ∈ Cc (X, R) be an extension with fe ≥ 0 and supp(fe) ⊂ Kj+1 . Then supp(fegn ) ⊂ Kj+1 and hence φj+1 (fegn |Kj+1 ) = φ(fegn ) for all R large n by the observation at the end of Step-1. Hence φ(fegn ) = Kj+1 fegn dµj+1 for all large n R R and therefore, Kj f dµj = φj (f ) = limn→∞ φ(fegn ) = limn→∞ Kj+1 fegn dµj+1 , which is equal, by R R Dominated convergence theorem, to Kj+1 f χKj dµj+1 = Kj f d(µj+1 |Kj ). Hence the claim holds by the uniqueness part in [1290 ]. For Borel A ⊂ X, define µ[A] = limj→∞ µj [A ∩ Kj ] ∈ [0, ∞]. It may be checked that µ is a Borel measure. For countable additivity, note the fact that if (xn,j ) is a double sequence of reals increasing in each variable, then limn→∞ limj→∞ xn,j = limj→∞ limn→∞ xn,j , and use this with xn,j = µj [An ∩ Kj ]. If K ⊂ X is compact, then K ⊂ Kj for some j and this implies µ[K] ≤ µj [K] < ∞ and so µ is locally finite. Also, if f ∈ Cc (X, R) is nonnegative, then supp(f ) ⊂ Kj for R R R R some j so that φ(f ) = φj (f |Kj ) = Kj f dµj = Kj f dµ = X f dµ, and therefore φ(f ) = K f dµ for every f ∈ Cc (X, R) since f = f+ − f− . Uniqueness: if β is another such measure and if βj = β|Kj , then βj = µj by the uniqueness in [1290 ] and hence β = µ. ¤ Remark : RRTM has other formulations when X is a compact metric space: (i) bounded linear functional on C(X, R) correspond to finite signed real-valued Borel measures, (ii) bounded linear functional on C(X, C) correspond to ‘finite’ complex-valued Borel measures, etc. Remark : Let X be a compact metric space. Then the dual of the Banach space (C(X, C), sup norm) can be identified with M (X) := {all complex-valued ‘finite’ Borel measures on X} by Riesz Theorem. Now recall from Functional Analysis that if Y is a Banach space, then the weak* topology on the dual Y ∗ is the smallest topology such that the evaluation maps φ → φ(y) from Y ∗ to C are continuous for every y ∈ Y ; the closed unit ball of Y ∗ is weak* compact (Alaoglu’s Theorem) and it is weak* metrizable if Y is separable. Hence the closed unit ball of M (X) = C(X, C)∗ is weak* compact and weak* metrizable. For µ, µn ∈ M (X), (µn ) → µ in the weak* topology iff R R R ( X f dµn ) → X f dµ in C for every f ∈ C(X, C), since φ 7→ φ(f ) = X f dµ is the evaluation map if φ ∈ C(X, C)∗ corresponds to µ ∈ M (X). If (µn ) is a bounded sequence in M (X), then it is contained in some closed ball (which is compact and metrizable in the weak* topology) and hence there exist µ ∈ M (X) and a subsequence (µnk ) such that (µnk ) weak* converges to µ, or R R equivalently ( X f dµnk ) → X f dµ for every f ∈ C(X, C). *****