Download Existence and uniqueness of Haar measure

HAAR MEASURE
MIKE COHEN
Remark. These notes roughly follow the approach of D. Cohn’s ”Measure Theory.”
Many thanks to J. Holshouser for producing the original draft of this document, which
I have only edited.
Definition. If G is a both a topological space and a group, then we say G is a
topological group if both multiplication and inversion are continuous.
Lemma. Let G be a topological group with identity e, and let U ⊆ G be open with
e ∈ U . Then there is an open nhood V ⊆ U such that e ∈ V and V V ⊆ U .
Proof. Since multiplication is continuous, the set O = {(x, y) : xy ∈ U } is open. So
find V1 , V2 ⊆ G open such that (e, e) ∈ V1 × V2 ⊆ O. Let V = V1 ∩ V2 . Then e ∈ V ,
and if x, y ∈ V , then (x, y) ∈ O and thus xy ∈ U .
Lemma. Let G be a topological group, K ⊆ G be compact and U ⊆ G open with
K ⊆ U . Then there is an open nhood V of e such that KV ⊆ U .
Proof. For each x ∈ K, let Wx be an nhood of e such that xWx ⊆ U and let Vx be
an nhood of e such that Vx Vx ⊆ Wx . Then {xVx : x ∈ K} is an open cover of K, so
choose x1 , · · · , xn ∈ K such that
[
K⊆
xi Vxi
i≤n
Set V = Vx1 ∩ · · · ∩ Vxn . If x ∈ K, then there is an i such that x ∈ xi Vi . Thus
xV ⊆ (xi Vxi ) Vxi ⊆ xi Wxi ⊆ U
Since x was arbitrary, KV ⊆ U .
Definition. A topological group G is locally compact if there is a compact set
K ⊆ G which has nonempty interior.
Definition. A Borel measure µ on G is regular if
(1) If K ⊆ G is compact, then µ(K) < ∞.
(2) If U ⊆ G is open, then µ(U ) = sup{µ(K) : K ⊆ U is compact}. This
condition is called inner regularity.
(3) If A ⊆ G is Borel, then µ(A) = inf{µ(U ) : U ⊇ A is open}. This condition is
called outer regularity.
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MIKE COHEN
Definition. A nonzero regular Borel measure on G is a left Haar measure if it is
left translation invariant. That is, if A ⊆ G is Borel and x ∈ G, then µ(xA) = µ(A).
Remark. The following theorem, which is the main focus of our talk, asserts the
existence of a left Haar measure for any locally compact topological group. Though
often called Haar’s theorem, Alfred Haar only proved the theorem for groups which
are second countable, in 1933. The statement in its full generality was first proved by
Andre Weil, using the Axiom of Choice. Later Henri Cartan furnished a proof which
avoided the use of AC. The proof sketch that we present here will utilize choice in
the form of Tychonoff’s theorem.
Theorem. Let G be a locally compact Hausdorff topological group. Then there is a
nonzero left Haar measure µ on G. Moreover, if µ and ν are both left Haar measures
on G, then µ = cν for some c ∈ R.
Proof. If K ⊆ G is compact and K ◦ 6= ∅, then we define #(K : V ) to be the least n
such that there are x1 , · · · , xn ∈ G with K ⊆ x1 V ∪ · · · ∪ xn V . Let K(G) be the set
of all compact subsets of G, and let U be all nhoods of e.
Fix now and forever a compact K0 ⊆ G with K0◦ 6= ∅. For each U ∈ U, we define
hU : K(G) → R by
#(K : U )
hU (K) =
#(K0 : U )
Claim.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Let U ∈ U. Then hU has the following properties:
#(K : U ) ≤ #(K : K0 )#(K0 : U ).
0 ≤ hU (K) ≤ #(K : K0 ).
hU (K0 ) = 1.
hU (xK) = hU (K).
If K ⊆ L, then hU (K) ≤ hU (L).
hU (K ∪ L) ≤ hU (K) + hU (L)
If KU −1 ∩ LU −1 = ∅, then hU (K ∪ L) = hU (K) + hU (L).
Proof of Claim. Part 2 follows from part 1, and everything else other than part 7 is
clear from the definition of hU . So suppose that KU −1 ∩LU −1 = ∅. Let n = #(K ∪L :
U ) and let x1 , · · · , xn ∈ G such that K ∪ L ⊆ x1 U ∪ · · · ∪ xn U . We wish to show
that each xi U meets either K or L, but not both. Suppose toward a contradiction
that there is an i such that xi U meets both K and L. Let y ∈ xi U ∩ K. Then there
is a u ∈ U such that y = xi u. So xi = yu−1 and thus xi ∈ KU −1 . Similarly, if
z ∈ xi U ∩ L, then xi ∈ LU −1 . Thus xi ∈ KU −1 ∩ LU −1 , a contradiction.
We now wish to take
Qa “limit” of the hU ’s. For each k ∈ K(G), set IK = [0, #(K :
K0 )] ⊆ R. Set X = K∈K(G) IK . Then by Tychonoff, X is compact and for each
U ∈ U, hU ∈ X. For each V ∈ U, set S(V ) = {hU : U ⊆ V }.
T Now suppose that
V1 , · · · , Vn ∈ U. Let V = V1 ∩ · · · ∩ Vn . Then V ∈ U, so hV ∈ i≤n S (Vi ). Thus the
HAAR MEASURE
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collection {S(V ) : V ∈ U} of closed sets has
T the finite intersection property. Therefore, as X is compact, we can fix an h ∈ V ∈U S(V ).
Claim.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
h has the following properties:
0 ≤ h(K).
h(∅) = 0.
h (K0 ) = 1.
h(xK) = h(K).
If K ⊆ L, then h(K) ≤ h(L).
h(K ∪ L) ≤ h(K) + h(L).
If K ∩ L = ∅, then h(K ∪ L) = h(K) + h(L).
Proof of Claim. The crucial fact of this proof is that the evaluation maps EK : X →
R, given by EK (f ) = f (K), are continuous since X has the product topology. We
present some of the parts, but leave the rest as exercise.
−1
Let K ⊆ G be compact. Since EK is continuous, EK
([0, ∞]) is closed. For
−1
−1
each U ∈ U, hU ∈ EK ([0, ∞]). Thus for each V ∈ U, S(V ) ⊆ EK
([0, ∞]). So
−1
h ∈ EK ([0, ∞]), and thus h(K) ≥ 0. Since K was arbitrary, this proves part 1.
Again, let K ⊆ G be compact. Then φ = ExK − EK is continuous, where the
operation here is pointwise subtraction. Then φ−1 (0) is closed, and we repeat the argument in the previous paragraph to see that h(xK) = h(K). Since K was arbitrary,
this proves part 4.
We finish by proving part 7. Suppose that K ∩L = ∅. Choose U1 , U2 which separate
K and L. By lemma 2, we can find V1 , V2 open nhoods of e such that KV1 ⊆ U1
and LV2 ⊆ U2 . Let V = V1 ∩ V2 . Then V is still an open nhood of e. Now for any
U ⊆ V −1 , U −1 ⊆ V , and thus KU −1 ∩ LU −1 = ∅. By part 7 of claim 1, we then have
that
hU (K) + hU (L) − hU (K ∪ L) = 0
Consider φ = EK + EL − EK∪L . Then φ is continuous, so φ−1 (0) is a closed set
containing hU for each U ⊆ V −1 . Thus h ∈ S (V −1 ) ⊆ ϕ−1 (0).
Now we define an outer measure µ∗ : P(G) → [0, ∞], If U ⊆ G is open, define
µ∗ (U ) = sup{h(K) : K ⊆ U is compact}
For A ∈ P(G), we define
µ∗ (A) = inf{µ∗ (U ) : U ⊇ A is open}
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MIKE COHEN
We aim to apply the Caratheodory condition to create a Borel measure. We must
first verify that µ∗ is an outer measure, i.e. we must check the following:
(1) µ∗ is non-negative,
(2) µ∗ is monotone,
(3) µ∗ (∅) = 0,
(4) µ∗ is countably subadditive.
Parts one through three are clear from the definition on µ∗ and the properties of
h. We thus need to confirm subadditivity. Because
we have defined µ∗ to be outer
S
regular, it suffices to check open sets. Let U = i∈ω Ui be a countable union of open
sets. Let K ⊆ U be compact. Then by a standard fact about compact sets, there are
compact sets K1 , · · · , Kn such that K = K1 ∪ · · · ∪ Kn and Ki ⊆ Ui for each i ≤ n.
Then
X
X
X
h(K) ≤
h (Ki ) ≤
µ∗ (Ui ) ≤
µ∗ (Ui )
i≤n
i∈ω
i≤n
Since this is true for all compact K ⊆ U , µ (U ) ≤ i∈ω µ∗ (Ui ).
Let µ be µ∗ restricted to the µ∗ -measurable sets. By Caratheodory’s theorem, µ is a
measure on these sets. It remains to show that the Borel sets of G are µ∗ -measurable.
It suffices to show that the open sets of G are µ∗ -measurable as the µ∗ -measurable
sets are a σ-algebra. Moreover, since µ∗ is outer regular, it again suffices to check
only the µ∗ -measurability of open sets. In other words, it suffices to check that
∗
P
µ∗ (V ) ≥ µ∗ (V ∩ U ) + µ∗ (V ∩ U c )
whenever U and V are open in G. To see this, first fix > 0, and use the definition
of µ∗ to find a compact set K ⊆ V ∩ U for which h(K) > µ∗ (V ∩ U ) − . Next choose
a compact set L ⊆ V ∩ K c for which h(L) > µ∗ (V ∩ K c ) − ≥ µ∗ (V ∩ U c ) − . Since
K and L are disjoint, we have
µ∗ (V ∩ U ) + µ∗ (V ∩ U c ) − 2 < h(K) + h(L) = h(K ∪ L) ≤ µ∗ (V ).
Since was chosen arbitrarily, we have the inequality we want. Thus µ is a measure
on a σ-algebra of sets which includes the Borel sets of G.
We will now show that µ is a nonzero left Haar measure. We begin with regularity.
It is clear from definition that µ is outer regular. Moreover if K is compact and U is
open containing K, then h(K) ≤ µ(U ); now taking the infimum over all such U , we
see that h(K) ≤ µ(K). This inequality, together with the definition of µ, show that
µ is also inner regular. So we need to check that compact sets get finite measure. Let
K ⊆ G be compact. Choose U ⊆ G open such that K ⊆ U and U is compact. We
can do this as G is locally compact, K is compact, and a finite union of compact sets
is compact. Then
µ(K) ≤ µ(U ) ≤ h U < ∞
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The second inequality follows from the monotonicity of h. Thus µ is regular. Now
µ is a Haar measure as h is invariant under left translations. Finally µ is nonzero as
1 = h (K0 ) ≤ µ (K0 ).
It remains now to show that µ is unique up to constant multiplication, i.e. if ν is
another left Haar measure, then ν = cµ. To see this, fix any nonzero nonnegative
continuous function g : G → R with compact support, and let f : G → R be an
arbitrary continuous function with compact support. The integrals of both functions
with respect to µ and ν will be finite and the integrals of g will also be nonzero. We
wish to show that the following ratio holds:
R
R
f dν
f dµ
R
= R
gdν
gdµ
R
R
R
R
which will imply that f dν = c f µ for the constant c = gdν/ gdµ. Since this
will hold for each f , then the Riesz representation theorem will imply the result we
want.
So let’s prove that the ratio holds. First consider an arbitrary function h : G×G →
R with compact support. Using Fubini’s theorem to reverse the order of integration
as necessary, and the translation-invariance of µ and ν to make appropriate variable
substitutions, we find that
Z Z
Z Z
Z Z
−1
h(x, y)dν(y)dµ(x) =
h(y x, y)dµ(x)dν(y) =
h(y −1 , xy)dν(y)dµ(x).
(We replace x with y −1 x for the first equality and y with
R xy for the second.) Now
apply this identity to the function h(x, y) = f (x)g(yx)/ g(tx)dν(t) and we get:
Z
Z
Z
f (y −1 )
f (x)dµ(x) = g(x)dµ(x) R
dν(y).
g(ty −1 )dν(t)
R
R
Thus the ratio of f dµ to gdµ depends on f and g, but not on µ, which proves
the claim.