1 Preliminaries on Locally Compact Spaces and Radon Measure

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 ∈ 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) = inf{µ(U): U ⊃ E, U open} and inner regular on E if

µ(E) = sup{µ(K): K ⊂ E, K compact}

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 ∈ Cc(X). Moreover, µ satisﬁes the following properties:

µ(U) = sup{I(f): f ∈ Cc(X), 0 ≤ f ≤ 1, supp(f) ⊂ U} for all open U ⊂ X and µ(K) = inf{I(f): f ∈ Cc(X), f ≥ 1K } 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 ∈ 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) = {f ∈ Cc(G) | f ≥ 0, ||f||u > 0}

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 ∈ Cc and all s ∈ 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 ∈ Cc , and taking limits, it then follows for all f ∈ 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) = sup{ f dµ : f ∈ Cc(G), ||f||u ≤ 1, and supp(f) ⊆ U}. 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 existence 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, ϕ ∈ Cc (G), we deﬁne the Haar covering number of f with respect to ϕ by the formula

 n n  X X  (f : ϕ) = inf cj : cj > 0 ∀j ∈ {1, ..., n} and f ≤ cjLsj ϕ for some s1, ..., sn ∈ 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 = {s ∈ G : ϕ(s) > ||ϕ(s)/2||u}, 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 2||f||u X f ≤ L ϕ. ||ϕ|| sj u j=1

2 Haar measure Reeve Garrett

In that case, Pn 2||f||u is a “sum of c ’s” that works. This inequality just stated holds since it holds j=1 ||ϕ||u j trivially outside the support of f and if x ∈ supp(f), then there exists s ∈ U and sj ∈ {s1, ..., sn} such that x = s s, meaning ϕ(s) = L ϕ(x), so f(x) < 2||f||u L ϕ(x), as f(x) ≤ ||f|| and j sj ||ϕ||u sj u

2Lsj ϕ(x) > 1. Then, clearly, by summing over all sj, 1 ≤ j ≤ n, we see the inequality is true for ||ϕ||u all x ∈ G. Notice since ||f||u 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 : ϕ) = (Lsf : ϕ) for all s ∈ G

(ii) (f1 + f2 : ϕ) ≤ (f1 : ϕ) + (f2 : ϕ) (iii) (cf : ϕ) = c(f : ϕ) for any c > 0

(iv) (f1 : ϕ) ≤ (f2 : ϕ) whenever f1 ≤ f2

(v) (f : ϕ) ≥ ||f||u ||ϕ||u

(vi) (f1 : ϕ) ≤ (f1 : f0)(f0 : ϕ)

Proof.

(i) Left multiplication by a ﬁxed element permutes any group, so for all s ∈ G we have the equivalence X X f(t) ≤ cjLsj ϕ(t) ∀t ∈ G ⇐⇒ Lsf(t) ≤ cjLssj ϕ(t) ∀t ∈ G,

which is to say that X X f ≤ cjjLsj ϕ ⇐⇒ Lsf ≤ cjLssj ϕ.

Hence, precisely the same coeﬃcients cj occur in the calculation of (f : ϕ) and (Lsf : ϕ). (ii), (iii), (iv) These are obvious.

(v) Letting cj denote the coeﬃcients in our deﬁnition of (f : ϕ), we see for all s ∈ G that

X −1 X f(s) ≤ cjϕ(sj s) ≤ cj ||ϕ||u, P whence cj ≥ ||f||u/||ϕ||u. P P P (vi) Notice that if f1 ≤ cjLsj f0 and f0 ≤ dkLtk ϕ then f1 ≤ cjdkLsj tk ϕ. As a consequence, X X X (f1 : ϕ) ≤ inf cjdk = inf cj inf dk = (f1 : f0)(f0 : ϕ).

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 ∈ Cc (G). We deﬁne (f : ϕ) Iϕ(f) = (f0 : ϕ)

By applying (vi) in the above lemma, (f : ϕ) ≤ (f : f0)(f0 : ϕ) and (f0 : ϕ) ≤ (f0 : f)(f : ϕ), so we immediately see by dividing appropriately (recall Haar covering numbers are always positive, so we can in fact divide!) that

−1 (f0 : f) ≤ Iϕ(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 Iϕ(f1) + Iϕ(f2) ≤ Iϕ(f1 + f2) + whenever the support of ϕ 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 ∈ 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 ∈ 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 |hi(s) − hi(t)| < δ for each −1 P i whenever t s ∈ U. Assume that supp(ϕ) ⊂ U and h ≤ j cjLsj ϕ. Then, for each i, X X fi(s) = h(s)hi(s) ≤ cjLsj ϕ(s)hi(s) ≤ cjLsj ϕ(s)[hi(sj) + δ] j j

−1 In the last inequality, we applied the fact that for each j, sj s ∈ supp(ϕ) ⊂ U (otherwise, we could just omit those terms from the sum and the original inequality involving h would still hold). By taking the inﬁmum over such sums of cj’s, we thus get the following inequality for each i X (fi : ϕ) ≤ cj[hi(sj) + δ] j

Summing these two inequalities and noting h1 + h2 ≤ 1, we see X (f1 : ϕ) + (f2 : ϕ) ≤ (1 + 2δ) cj j

Since we can take the sum of cj’s arbitrarily close to (h : ϕ), we therefore see that (f1 : ϕ) + (f2 : ϕ) ≤ (1 + 2δ)(h : ϕ). Then, dividing by (f0 : ϕ), we get 1 I (f )+I (f ) ≤ (1+2δ)I (h) ≤ (1+2δ)[I (f +f )+δI (g)] = I (f +f )+2δ[I (f +f )+( +δ)I (g)] ϕ 1 ϕ 2 ϕ ϕ 1 2 ϕ ϕ 1 2 ϕ 1 2 2 ϕ

4 Haar measure Reeve Garrett where we applied (ii) and (iii) of the previous lemma to get the second inequality. But by the inequality stated just before this lemma, we know all the Iϕ terms on the right are bounded independently of ϕ, so for any > 0 we can choose δ > 0 suﬃciently small so that the stated inequality holds. We are now ﬁnally ready to prove the existence of Haar measure.

Theorem 3.6 Left Haar measure exists for any locally compact group G.

Proof. We need a left translation-invariant linear functional so that we can apply the Riesz representation theorem. As we’ve just shown, the Iϕ are close but not quite there, and we’ll use them to construct an actual linear functional. Deﬁne a compact topological space X as follows:

Y −1 X = [(f0 : f) , (f : f0)] + f∈Cc

Recall this is compact by Tychonoff’s theorem. Moreover, every function Iϕ lies in X since for + −1 all f ∈ Cc we have Iϕ(f) ∈ [(f0 : f) , (f : f0)]. Now, for any compact neighborhood U of e n + in G, define K := {I : supp(ϕ) ⊆ U}. Since ∩ K ⊇ K n and there exists ϕ ∈ C U ϕ j=1 Uj ∩j=1Uj c n such that supp(ϕ) ⊆ ∩j=1Uj by Urysohn’s lemma (the intersection is compact since it’s closed in G, which is Hausdorff), we have that {KU : U is a compact neighborhood of e} satisfies the finite intersection property. It’s a fact from point-set topology (stated as Proposition 4.21 in Folland) that a topological space X is compact if and only if for every family {Fα}α∈A of closed sets with the finite intersection property we have ∩α∈AFα 6= ∅. Therefore, ∩KU 6= ∅ where U is indexed over all compact neighborhoods of e. Let’s say I is an element of this set. Since it’s in X, I is not the zero function, so an extended functional obtained from it won’t be trivial either. Since I is in the intersection of the closures of the sets {Iϕ : supp(ϕ) ⊆ U}, it follows that every open neighborhood of I in X intersects each of the sets {Iϕ : supp(ϕ) ⊆ U}. This + means for every > 0, open neighborhood U of e, and finite collection f1, ..., fn ∈ Cc , there + exists ϕ ∈ Cc such that supp(ϕ) ⊆ U and |I(fj) − Iϕ(fj)| < for j = 1, ..., n. From this and the lemma listing properties of the Haar covering number, we show that I is linear and left + translation-invariant. Indeed, given > 0, there exists ϕ ∈ Cc such that |I(cf) − Iϕ(cf)| < 2 and |cIϕ(f) − cI(f)| < 2 , meaning since part (iii) of the lemma shows Iϕ(cf) = cIϕ(f) that |I(cf) − cI(f)| = |I(cf) − Iϕ(cf) + cIϕ(f) − cI(f)| < for all > 0, so I(cf) = cI(f). By similar reasoning, using part (i) of the lemma instead, I is also left translation-invariant: Iϕ(f) = Iϕ(Lsf) + for all s ∈ G and ϕ ∈ Cc , so using the exact same inequality argument we get I(f) = I(Lsf) for all s ∈ G. I is also additive, but first we show subadditivity. By part (ii) of the lemma, Iϕ(f1 + f2) ≤ + + Iϕ(f1) + Iϕ(f2) for all ϕ, f1, f2 ∈ Cc . We also have given > 0 that given f1, f2 ∈ Cc there exists + ϕ ∈ Cc such that |I(f1 + f2) − Iϕ(f1 + f2)| < /3, |I(f1) − Iϕ(f1)| < /3 and |I(f2) − Iϕ(f2)| < /3. As a consequence, I(f1 + f2) ≤ Iϕ(f1 + f2) + /3 ≤ Iϕ(f1) + Iϕ(f2) + /3 ≤ I(f1) + I(f2) + for all > 0. Therefore, I is subadditive. We now use the lemma we proved just before this theorem to show I is in fact additive. By that lemma, we may choose a neighborhood U of e such that Iϕ(f1) + Iϕ(f2) ≤ Iϕ(f1 + f2) + 4 whenever supp(ϕ) ⊆ U. Now, we may choose ϕ such that supp(ϕ) ⊆ U and I(f1), I(f2), and I(f1 + f2) are within /4 of Iϕ(f1), Iϕ(f2), and Iϕ(f1 + f2), respectively, and this gets us what we want: 3 I(f ) + I(f ) ≤ I (f ) + I (f ) + ≤ I (f + f ) + ≤ I(f + f ) + 1 2 ϕ 1 ϕ 2 2 ϕ 1 2 4 1 2

Since > 0 was arbitrary and Iϕ is always sublinear regardless of choice of ϕ, we have that I is in + fact additive. Thus, I is a left translation-invariant linear functional on Cc (G). However, we need

5 Haar measure Reeve Garrett

+ − such a functional on all of Cc(G), and we obtain such a functional by setting I(f) = I(f ) − I(f ) + for an arbitrary f ∈ Cc (G), and in fact we see that I is a positive linear functional (meaning for all f ≥ 0 we have I(f) ≥ 0). Thus, in light of our construction of a positive linear functional I on Cc(G), the ﬁrst proposition we proved, and the Riesz Representation theorem, we know that G admits a left Haar measure µ, and µ is such that for f ∈ Cc(G) Z I(f) = f dµ