Haar measure Reeve Garrett 1 Preliminaries on locally compact spaces and Radon measure Definition 1.1 A topological space is locally compact if every point has a compact neighborhood. The spaces we are concerned with in this presentation are locally compact Hausdorff spaces. Results about these spaces in general are proved in Section 4.5 of Folland, starting on page 131. One result about them we will make use of is the following: Theorem 1.2 (Urysohn's Lemma) If X is a locally compact Hausdorff space and K ⊂ U ⊂ X where K is compact and U is open, there exists f 2 C(X; [0; 1]) such that f = 1 on K and f = 0 outside U. Now, we develop the notion of Radon measure. Definition 1.3 Given a Borel measure µ on a locally compact Hausdorff space X and a Borel subset E of X, we say µ is outer regular on E if µ(E) = inffµ(U): U ⊃ E; U openg and inner regular on E if µ(E) = supfµ(K): K ⊂ E; K compactg Definition 1.4 A Radon measure on a locally compact Hausdorff space X is a Borel measure which is finite on compact sets, outer regular on all Borel sets, and inner regular on all open sets. It is shown in Proposition 7.5 of Folland that Radon measures are also inner regular on all σ-finite sets of X. Since one of the main objectives of this presentation is to prove Haar measure exists, and Haar measure is, as we will see, a special kind of Radon measure, it will be helpful to establish the existence of Radon measures for locally compact Hausdorff spaces first and let it do most of the work for us in establishing Haar measure's existence. The result is as follows: Theorem 1.5 (The Riesz Representation Theorem) Given a locally compact Hausdorff space X, if I is a positive linear functional on Cc(X) (the space of continuous functions from X to R with compact support), there is a unique Radon measure µ on X such that I(f) = R f dµ for all f 2 Cc(X). Moreover, µ satisfies the following properties: µ(U) = supfI(f): f 2 Cc(X); 0 ≤ f ≤ 1; supp(f) ⊂ Ug for all open U ⊂ X and µ(K) = inffI(f): f 2 Cc(X); f ≥ 1K g for all compact K ⊂ X: The proof as presented in Folland is quite long, and as the theorem is not the object of this presentation, we omit the proof. Also, note the result given here (taken verbatim from Folland) gives a Radon measure, not just an outer measure as proved in class. 1 Haar measure Reeve Garrett 2 Preliminaries on Haar measure and locally compact groups Definition 2.1 A left Haar measure on a locally compact (Hausdorff) group G is a nonzero left invariant Radon measure on G. Notation 2.2 Given a continuous function f on a topological group G and y 2 G, we define left and right translates of f through y by −1 Lyf(x) = f(y x);Ryf(x) = f(xy) + Notation 2.3 We define Cc (G) = ff 2 Cc(G) j f ≥ 0; jjfjju > 0g 3 Proof of the existence of Haar measure Proposition 3.1 Let G be a locally compact (Hausdorff) group with nonzero Radon measure µ. Then, the measure µ is a left Haar measure on G if and only if Z Z Lsf dµ = f dµ G G + for all f 2 Cc and all s 2 G. Proof. If µ is a Haar measure on G, then the equality of integrals stated above follows by + + definition for all simple functions f 2 Cc , and taking limits, it then follows for all f 2 Cc . For the converse, we put the Riesz representation theorem to work: from the positive linear R functional G · dµ on Cc(G) we can explicitly recover the Radon measure µ of any open subset U ⊆ G as follows: Z µ(U) = supf f dµ : f 2 Cc(G); jjfjju ≤ 1; and supp(f) ⊆ Ug: G Since supp(f) ⊆ U if and only if supp(Lsf) ⊆ sU, we see if the integral is left translation invariant, then µ(sU) = µ(U) for all open U ⊆ G. However, a Radon measure such as µ is outer regular, so the result extends to all Borel subsets of G. Thus, via the Riesz representation theorem and the proposition just proved, the proof of exis- tence of Haar measure reduces to the construction of a left translation invariant linear functional on Cc(G). The key idea needed to do this is the introduction of a translation-invariant device for + comparing functions on Cc (G), which we call the Haar covering number. + Definition 3.2 Given f; ' 2 Cc (G), we define the Haar covering number of f with respect to ' by the formula 8 n n 9 <X X = (f : ') = inf cj : cj > 0 8j 2 f1; :::; ng and f ≤ cjLsj ' for some s1; :::; sn 2 G :j=1 j=1 ; Why is this set of sums we're taking an infimum over nonempty? Let's see why. If we set U = fs 2 G : '(s) > jj'(s)=2jjug, then since supp(f) is compact it can be covered by a finite number of translates of U, say s1U; :::; snU, and the following holds: n 2jjfjju X f ≤ L ': jj'jj sj u j=1 2 Haar measure Reeve Garrett In that case, Pn 2jjfjju is a \sum of c 's" that works. This inequality just stated holds since it holds j=1 jj'jju j trivially outside the support of f and if x 2 supp(f), then there exists s 2 U and sj 2 fs1; :::; sng such that x = s s, meaning '(s) = L '(x), so f(x) < 2jjfjju L '(x), as f(x) ≤ jjfjj and j sj jj'jju sj u 2Lsj '(x) > 1. Then, clearly, by summing over all sj, 1 ≤ j ≤ n, we see the inequality is true for jj'jju all x 2 G. Notice since jjfjju is assumed positive we have the Haar covering number is never 0. Of course, we want to prove that we can get the linear functional we want from this, and this requires establishing some properties of the Haar covering number, which we'll do right now. Lemma 3.3 The Haar covering number has the following properties: (i) (f : ') = (Lsf : ') for all s 2 G (ii) (f1 + f2 : ') ≤ (f1 : ') + (f2 : ') (iii) (cf : ') = c(f : ') for any c > 0 (iv) (f1 : ') ≤ (f2 : ') whenever f1 ≤ f2 (v) (f : ') ≥ jjfjju jj'jju (vi) (f1 : ') ≤ (f1 : f0)(f0 : ') Proof. (i) Left multiplication by a fixed element permutes any group, so for all s 2 G we have the equivalence X X f(t) ≤ cjLsj '(t) 8t 2 G () Lsf(t) ≤ cjLssj '(t) 8t 2 G; which is to say that X X f ≤ cjjLsj ' () Lsf ≤ cjLssj ': Hence, precisely the same coefficients cj occur in the calculation of (f : ') and (Lsf : '). (ii), (iii), (iv) These are obvious. (v) Letting cj denote the coefficients in our definition of (f : '), we see for all s 2 G that X −1 X f(s) ≤ cj'(sj s) ≤ cj jj'jju; P whence cj ≥ jjfjju=jj'jju. P P P (vi) Notice that if f1 ≤ cjLsj f0 and f0 ≤ dkLtk ' then f1 ≤ cjdkLsj tk '. As a consequence, X X X (f1 : ') ≤ inf cjdk = inf cj inf dk = (f1 : f0)(f0 : '): The question still remains as to how we get a linear functional from Haar covering numbers. We'll address that question now. 3 Haar measure Reeve Garrett + Notation 3.4 Fix f0 2 Cc (G). We define (f : ') I'(f) = (f0 : ') By applying (vi) in the above lemma, (f : ') ≤ (f : f0)(f0 : ') and (f0 : ') ≤ (f0 : f)(f : '), so we immediately see by dividing appropriately (recall Haar covering numbers are always positive, so we can in fact divide!) that −1 (f0 : f) ≤ I'(f) ≤ (f : f0): This bound is crucial to our proof for the existence of Haar measure for G. But first we need one more lemma. + Lemma 3.5 Given f1 and f2 in Cc , for every > 0 there is a neighborhood V of the identity e such that I'(f1) + I'(f2) ≤ I'(f1 + f2) + whenever the support of ' lies in V . Proof. Recall that Urysohn's lemma for locally compact Hausdorff spaces states that if X is a locally compact Hausdorff space with K ⊂ U ⊂ X where K is compact and U is open, there exists f 2 C(X; [0; 1]) such that f = 1 on K and f = 0 outside a compact subset of U. So, by Urysohn's + lemma, there exists a function g 2 Cc that takes the value 1 on supp(f1+f2) = supp(f1)[ supp(f2). Let δ > 0 and h = f1 + f2 + δg. h is continuous. Next, let hi = fi=h for i = 1; 2, with the + understanding that hi is 0 outside the support of fi. We see both hi lie in Cc and their sum approaches 1 from below as δ tends towards 0. Since each hi is in Cc, they are uniformly continuous (as shown in Prop 11.2 of Folland, any function in Cc(G) is uniformly continuous, but we don't have time to prove this), so there exists a neighborhood U of e such that jhi(s) − hi(t)j < δ for each −1 P i whenever t s 2 U. Assume that supp(') ⊂ U and h ≤ j cjLsj '.