Haar Measure on Locally Compact Topological Groups Abhishek Gupta February 2, 4, 2015 These are notes of my presentation delivered as a part of the course MTH 638: Abstract Harmonic Analysis at IIT Kanpur. Please send errors to [email protected]. The existence and uniqueness (i.e. upto scaling) of the Haar measure is proved. We follow the proof given by Weil and discussed in the book `Abstract Harmonic Analysis' by G.B. Folland. The modular function is also discussed briefly. By a locally compact group we shall mean a topological group which is locally compact and Hausdorff as a topological space. Note that in such a space the Urysohn lemma applies which we shall use, often without explicit comment. 1 Haar Measure: its Existence and Uniqueness Let G be a locally compact group. Cc(G) denotes the space of real valued compactly supported functions + on G and Cc (G) = ff 2 Cc(G): f ≥ 0; f 6= 0g. Recall that a Radon measure on a topological space X is a Borel measure which is finite on compact subsets (compact subsets of Hausdorff spaces are Borel measurable), outer regular on all Borel subsets and inner regular on all σ-finite Borel sets. Definition 1.1. A left(respectively right) Haar measure on a locally compact topological group G is a Radon measure µ such that µ(E) = µ(xE)(respectively µ(E) = µ(Ex)) for every Borel set E and 8x 2 G. Given a left Haar measure µ, one can immediately see thatµ ~ defined byµ ~(E) = µ(E−1) is a right Haar measure. Because of this, one can simply restrict attention to either left or right Haar measure, following the book we choose left. The modular function studies the failure of a left Haar measure to be right invariant and we shall look at this briefly in the next section. We now recall the Reisz representation theorem(see pp. 212 of Folland's Real Analysis for a proof): Theorem 1.1. (Reisz representation theorem) If I is a positive linear functional on Cc(X), there is a R unique Radon measure µ on X such that I(f) = f dµ 8f 2 Cc(G) (in other words µ represents I). 1 R Cc(G) ⊆ L (µ) as continuous functions are Borel measurable and for f 2 Cc(G), f dµ ≤ jjfjj1µ(supp(f)) < 1 as f is compactly supported. Our strategy to prove the existence of a measure will be to utilize this theorem after producing a pos- itive linear functional on Cc(G). To prove the left invariance of the measure we will use the following proposition: R R + Proposition 1.1. A Radon measure µ is a left Haar measure , Ly dµ = f dµ, 8f 2 Cc (G)8y 2 G. 1 Proof. µy(E) = µ(yE) defines another Radon measure. For a characteristic function χ of a compact R −1 R set S we easily see that Lyχ dµ = µ(y S) = χ dµy, which implies this equality for simple functions + R by linearity of the Lebesgue integral. Thus, for any given φ 2 C (G) the two sets f Lyf dµ : R c f is a simple function; f ≤ φg and f f dµy : f is a simple function; f ≤ φg are the same and so are R R + there supremums. Hence Lf dµ = f dµy 8 f 2 Cc (G). Now if µ is a Haar measure then µ = µy thus we have the required equality. Conversely given the R R integral equality, we have f dµ = f dµy, thus we have two measures representing the linear R functional Lf dµ hence by the uniqueness in Reisz representation theorem µ = µy. Thus µ is a left Haar measure. We note the following lemma proved in a previous lecture which we shall use: Lemma 1.1. (Uniform Continuity Lemma) If f 2 Cc(G) then f is both left and right uniformly continuous. Theorem 1.2. Every locally compact group has a left Haar measure. + Pn Pn Proof. For f; φ 2 Cc (G) define (f : φ) = inff j=1 cjjf ≤ j=1 cjLxj φ, for some xj'sg. This makes sense as we can cover support of f with translates of the open set fx 2 G : φ(x) ≥ 1=2jjφjj1g. Because of compactness of support finitely many translates, say N, will suffice. Thus we have (f : φ) ≤ 2Njjfjj1=jjφjj1. We have the following properties: 1.( f : φ) = (Lyf : φ)8 y 2 G 2.( f1 + f2 : φ) ≤ (f1 : φ) + (f2 : φ) 3.( cf : φ) = c(f : φ)8 c > 0 4. f1 ≤ f2 ) (f1 : φ) ≤ (f2 : φ) 5.( f : φ) ≥ jjfjj1=jjφjj1 + 6.( f : φ) ≤ (f : )( : φ)8 2 Cc (G) + (f:φ) + Fix f0 2 C (G) and define Iφ = for f; φ 2 C (G). By the above properties Iφ's are left-invariant, c (f0:φ) c sub-additive, homogeneous of degree 1 and monotone(i.e. order preserving). The last property implies (f : φ) ≤ (f : f0)(f0 : φ) and (f0 : φ) ≤ (f0 : f)(f : φ). In other words, −1 (f0 : f) ≤ Iφ(f) ≤ (f : f0) + Lemma 1.2. 8f1; f2 2 Cc (G); > 0, 9 a neighborhood V of 1, such that supp φ ⊆ V ) Iφ(f1) + Iφ(f2) ≤ Iφ(f1 + f2) + . + Proof. Let g 2 Cc (G) be such that g = 1 on supp f1 + f2. Define h = f1 + f2 + δg where δ is to be chosen later. Define hi = fi=h which is zero whenever the numerator is. Note that because of this definition we have h1 + h2 ≤ 1. By the preceding proposition we have a neighborhood V of 1 such that −1 + P jhi(x) − hi(y)j ≤ whenever y x 2 V . Consider φ 2 Cc (G) with supp φ ⊆ V . If h ≤ j cjLxj φ then: X −1 X −1 fi(x) = h(x)hi(x) ≤ cjφ(xj x)h(x) ≤ cjφ(xj x)(hi(xj) + δ) j j −1 because jhi(x) − hi(xj)j < δ whenever xj x 2 supp φ. Since h1 + h2 ≤ 1 we have: X X X (f1 : φ) + (f2 : φ) ≤ cj(h1(xj) + δ) + cj(h2(xj) + δ) ≤ cj(1 + 2δ) j j j 2 taking the infimum of right hand side and normalizing we get: Iφ(f1) + Iφ(f2) ≤ (1 + 2δ)Iφ(h) ≤ Iφ(f1 + f2) + δ[2(Iφ(f1 + f2)) + (1 + 2δ)Iφ(g)] + Now note that for any k 2 Cc (G) we have (k : φ) ≤ (k : f0)(f0 : φ) in other words Iφ(k) ≤ (k : f0). Thus choosing δ so that δ[2(f1 + f2 : f0) + (1 + 2δ)(g : f0)] < (which we can always do) ensures that δ[2(Iφ(f1 + f2)) + (1 + 2δ)Iφ(g)] < and we are done. −1 Now let X denote the interval [(f : f) ; (f : f )] and X = Π + X . Hence, by Tychonoff f 0 0 f2Cc (G) f theorem it is a compact Hausdorff space. For a neighborhood V of 1 denote by K(V ) the closure of the set fIφ : supp φ ⊆ V g. Then K(V ) are all non-empty closed subsets of X, since given any open V we can construct a continuous function with support in V by the Urysohn Lemma. The family of closed n n subsets fK(V )g has the finite intersection property as K([i=1Vi) ⊆ [i=1K(Vi). Now by the following elementary fact: Proposition 1.2. A topological space is compact if and only if every family of closed subsets having finite intersection property has a nonempty intersection. we conclude that there is an element I in the intersection of all the K(V )'s. In other words given + + an > 0 and any f1; : : : ; fn 2 Cc (G) and any neighborhood V of 1 9φ 2 Cc (G); suppφ ⊆ V such that jI(fi) − Iφ(fi)j < for all i. Then I has the following properties: 1. I(Lyf) = I(f): To see this choose f1 = f; f2 = Lyf. Then given an > 0 and any neighborhood + V of 1 9φ 2 Cc (G) such that jI(fi) − Iφ(fi)j < , in other words jI(f) − Iφ(f)j < and jIφ(Lyf) − I(Lyf)j < . By triangle inequality we have jI(f) − Iφ(f) + Iφ(Lyf) − I(Lyf)j = jI(f) − I(Lyf)j < 2 since Iφ(f) = Iφ(Lyf). 2. I(cf) = cI(f) for c > 0: To see this choose f1 = f; f2 = cf. Then given an > 0 and any + neighborhood V of 1 9φ 2 Cc (G) such that jI(fi) − Iφ(fi)j < , in other words jI(f) − Iφ(f)j < and jIφ(cf) − I(cf)j < . By triangle inequality we have jI(f) − Iφ(f) + Iφ(cf) − I(cf)j = jcI(f) − I(cf)j < 2 since Iφ(cf) = cIφ(f). 3. I(f + g) = I(f) + I(g). To see this choose f1 = f; f2 = g; f3 = f + g. By the lemma we have a neighborhood V of 1 such that jIφ(f + g) − Iφ(f) − Iφ(g)j < for a given . Also for this + neighborhood V of 1, 9φ 2 Cc (G); suppφ ⊆ V such that jIφ(fi) − I(fi)j < , in other words jIφ(f)−I(f)j < , jIφ(g)−I(g)j < and jI(f +g)−Iφ(f +g)j < . Adding these four inequalities we have: jIφ(f + g) − Iφ(f) − Iφ(g)j + jIφ(f) − I(f)j + jIφ(g) − I(g)j + jI(f + g) − Iφ(f + g)j < 4 8 > 0 The triangle inequality then yields jI(f + g) − I(f) − I(g)j < 4 8 > 0, hence giving the desired result.