Measure theory class notes - 19 November 2010, class 32 1

Haar measure (contd)

Definition. Suppose G is a topological group. A measure λ on its Borel σ-field is said to be a

left-invariant Haar measure on G if

• For K ⊆ G compact, λ(K) <∞.

• For U ⊆ G open, if U 6= ∅ then λ(U) > 0.

• For every Borel set B ⊆ G and g ∈ G, λ(gB) = λ(B).

We will show that every locally compact metrizable topological group has a left-invariant Haar

measure.

How would one construct such a λ? Let us look at the topological group (R,+), for which we know

a Haar measure - the usual Lebesgue measure λ. How can we describe the length of an interval

[a, b] without subtracting a from b? In a general group, we do not have any notion of intervals; and

even if we did manage to “subtract”, we will get only an element of the group, not a real number.

Here is an approach: consider translates of the interval(−1

2, 12

). Some finitely many of them cover

[a, b] (all translates of(−1

2, 12

)certainly cover [a, b]; by compactness, some finitely many of them

do). Let n1 be the minimum number of translates needed to cover [a, b]. Then it can be shown

that

n1 − 1 ≤ λ([a, b]) ≤ n1

Let n2 be the minimum number of translates of(−1

4, 14

)needed to cover [a, b]. Then it can be

shown thatn2 − 1

2≤ λ([a, b]) ≤ n2

2

We may continue this way to get better and better approximations, but the problem here is that

it uses the measures of the sets(−1

2, 12

),(−1

4, 14

), etc. In general we have to define λ for all sets,

nothing is given to us.

To work around this, choose a “reference interval”, say [0, 1] (the exact choice doesn’t matter).

For every k ∈ N, let nk be the minimum number of translates of(− 1

2k, 12k

)needed to cover [a, b],

and let mk be the minimum number needed to cover [0, 1]. Then the measure of [a, b] is roughly

nk/mk. It can be shown that as k goes to infinity, nk/mk converges to λ([a, b]). If we had chosen

some other reference interval, we would have got a scalar multiple of the Lebesgue measure. This

is not a problem - after all, a scalar multiple (by a positive real number) of a left-invariant Haar

measure is also a left-invariant Haar measure.

We can try to extend this to a general metrizable locally compact topological group. We can get

a decreasing sequence of nonempty open sets {Ik}∞k=1 decreasing to the identity, and a reference

nonempty open set U0, such that the closures of all these open sets are compact. As before, for any

open set U with compact closure, we look at nk/mk, where nk and mk are the minimum number

of translates of Ik needed to cover U and U0 respectively. We can show that {nk/mk}∞k=1 is a

bounded sequence. But will it converge? It need not. To deal with this, we generalise our notion

of convergence.

Measure theory class notes - 19 November 2010, class 32 2

Ultrafilters and convergence

Definition. U ⊆ 2N is said to be an ultrafilter if

1. For all i ∈ N, N \ {i} ∈ U .

2. If A ∈ U , then every superset of A is also in U .

3. A,B ∈ U =⇒ A ∩B ∈ U .

4. For all A ⊆ N, exactly one of A and N \ A belongs to U .

If U is an ultrafilter, then it contains all co-finite sets (because it contains complements of singletons,

and is closed under finite intersections). By 4, it does not contain any finite sets. Every element

of U is infinite, and in particular ∅ /∈ U . But does such a U exist at all?

Theorem. There exists U ⊆ 2N which is an ultrafilter.

The proof depends on the axiom of choice and is non-constructive. It is hard to explicitly describe

an ultrafilter. Fortunately for our puposes it does not matter which ultrafilter we use, so we fix

some ultrafilter U arbitrarily. Think of the sets in U as being “large” subsets of N, and the ones

not in U as being “small”. The first three conditions in the definition say that all co-finite sets are

large, every superset of a large set is large, and that the intersection of two large sets is large. The

last condition says that the complement of a large set is small, and vice versa.

How does an ultrafilter help us?

Lemma. Suppose {xn}∞n=1 is a bounded sequence of real numbers. Then⋂U∈U

{xn : n ∈ U}

has exactly one element.

Proof. Since the given sequence is bounded, each set {xn : n ∈ U} is compact. These sets have

the finite intersection property (the intersection of any finitely many of them is nonempty) : If

U1, . . . , Uk ∈ U , we know that their intersection belongs to U and so is nonempty - let p be an

element in U1 ∩ . . . ∩ Uk. Then

xp ∈k⋂i=1

{xn : n ∈ U} ⊆k⋂i=1

{xn : n ∈ U}

Since this collection of compact sets has the finite intersection property, the grand intersection is

nonempty: ⋂U∈U

{xn : n ∈ U} 6= ∅

We now show that this intersection has exactly one element: Let α < β be any two real numbers.

Let

S ={n : xn >

α+β2

}, N \ S =

{n : xn ≤ α+β

2

}Exactly one of S and N \ S belongs to U . If S ∈ U , then α /∈ {xn : n ∈ S}, otherwise β /∈{xn : n ∈ N \ S}. In either case, both α and β cannot belong to

⋂U∈U {xn : n ∈ U}.

Measure theory class notes - 19 November 2010, class 32 3

For a bounded sequence {xn}∞n=1 of reals, let x be the unique element of⋂U∈U {xn : n ∈ U}. We

say that the sequence converges along the ultrafilter U to x. How does this compare with the usual

notion of convergence? Let F ⊆ 2N consist of all co-finite subsets of N. F satisfies the first three

conditions in the definition of an ultrafilter, and F ⊆ U .

Lemma. Suppose {xn}∞n=1 is a bounded sequence of reals, and x a real number.

• {xn}∞n=1 converges to x in the usual analysis sense if and only if for every ε > 0, there exists

A ∈ F such that {xn : n ∈ A} ⊆ (x− ε, x+ ε).

• {xn}∞n=1 converges to x along U if and only if for every ε > 0, there exists A ∈ U such that

{xn : n ∈ A} ⊆ (x− ε, x+ ε).

Proof. The first part is easy. For the second part, assume {xn}∞n=1 converges to x along U . Then

{x} =⋂U∈U

{xn : n ∈ U}

Let ε > 0 be given. Let

A = {n : xn ⊆ (x− ε, x+ ε)}

Exactly one of A and N \ A belongs to U . N \ A cannot belong to U , because

{xn : n ∈ N \ A} ⊆ R \ (x− ε, x+ ε) ⊆ R \ {x}

So A ∈ U , and {xn : n ∈ A} ⊆ (x− ε, x+ ε).

Conversely, suppose for every ε > 0, there exists A ∈ U such that {xn : n ∈ A} ⊆ (x − ε, x + ε).

Let U ∈ U , we have to show that x ∈ {xn : n ∈ U}. This is the same as showing that every

neighbourhood of x intersects {xn : n ∈ U}. Let ε > 0 be given. Let A ∈ U be such that

{xn : n ∈ A} ⊆ (x− ε, x+ ε). A ∩ U ∈ U . For any m ∈ A ∩ U , xm ∈ (x− ε, x+ ε) ∩ {xn : n ∈ U}.Since this holds for every ε > 0, x ∈

⋂U∈U {xn : n ∈ U}. So

{x} =⋂U∈U

{xn : n ∈ U}

Note that F ⊆ U . So if a sequence converges to a number in the usual sense, it converges to the

same number along U . Convergence along U has many nice properties that the usual notion of

convergence has. If {xn}∞n=1 converges to x along U and {yn}∞n=1 converges to y along U , then

{xn + yn}∞n=1 converges to x+ y and {xnyn}∞n=1 converges to xy along U . If xn ≤ yn for all n, then

x ≤ y. We will use these properties in the next class.

Consider the sequence (0, 1, 0, 1, 0, 1, . . .). This does not converge in the usual sense, but an ultra-

filter U forces it to converge: if the set of even numbers belongs to U , it converges to 1; otherwise

the set of odd numbers belongs to U , and it converges to 0.

We now show that an ultrafilter exists1:

Theorem. An ultrafilter U ⊆ 2N exists.

1This wasn’t proved in class.

Measure theory class notes - 19 November 2010, class 32 4

Proof. The idea is simple: start with F and keep adding sets to it till it is saturated.

We call a subset of 2N which satisfies the first three conditions in the definition of an ultrafilter

and which does not have the empty set as an element a filter. In other words, S ⊆ 2N is a filter if

1. For all i ∈ N, N \ {i} ∈ S.

2. If A ∈ S, then every superset of A is also in S.

3. A,B ∈ S =⇒ A ∩B ∈ S.

4. ∅ /∈ S.

The condition ∅ /∈ S only rules out S being equal to 2N. F , the collection of co-finite subsets of

N, is a filter. Let

A = {L : L ⊆ 2N,F ⊆ L,L is a filter}A consists of all possible ways of adding more sets to F while retaining the property of being

a filter. Partially order A by inclusion. F ∈ A . We will use Zorn’s Lemma. Every chain is

bounded: If B ⊆ A is a nonempty chain, then⋃

B is a filter (this is easy to see using B being

totally ordered) and F ⊆⋃

B.⋃

B ∈ A and is an upper bound for B. Every chain is bounded,

so by Zorn’s lemma, A has a maximal element.

We will look at a maximal element, after we examine what happens when we try to extend a filter

by throwing one more set in it: suppose S is a filter, and B ⊆ N, B /∈ S. If B is disjoint from

some set in S, there is no hope of extending S to a filter which has B, because of properties 3 and

4 of a filter. So suppose this is not the case, that is, B intersects every element of S. Let

S ′ = {A ∩ C : A ∈ S, B ⊆ C ⊆ N}

If at all we can extend S to a filter which has B, it must have at least all the sets in S ′. Clearly

S ⊆ S ′, since we can take C = N. So S ′ has the complements of all singletons. Suppose A∩C ∈ S ′,

and D is a superset of it. A ∪D ∈ S and C ∪D is a superset of C, so (A ∪D) ∩ (C ∪D) ∈ S ′.

(A ∪D) ∩ (C ∪D) = (A ∩ C) ∪D = D

So D ∈ S ′. S ′ is closed under taking supersets. If A1 ∩ C1, A2 ∩ C2 ∈ S ′, then

(A1 ∩ C1) ∩ (A2 ∩ C2) = (A1 ∩ A2) ∩ (C1 ∩ C2) ∈ S ′

since A1 ∩ A2 ∈ S and C1 ∩ C2 is a superset of B. S ′ is closed under finite intersections. Since

we have assumed B intersects every element of S, we have that ∅ /∈ S ′. S ′ is a filter, S ⊆ S ′, and

B ∈ S ′ (becasue B = N ∩B). We have shown the following:

Suppose S is a filter, and B ⊆ N with B /∈ S ′. There exists a filter S ′ with S ⊆ S ′ and

B ∈ S ′ if and only if B intersects every element of S.

Now we go back to A . Let U be a maximal element of A . It is a filter, so it satisfies the first

three conditions of being an ultrafilter. We want to show that for every A ⊆ N, exactly one of A

and N \ A belong to U . Assume the contrary: suppose neither of A and N \ A belong to U (of

course both cannot belong). U is maximal, so it cannot be extended using A. As we have shown,

this means there is a D1 ∈ U such that A is disjoint from D1. Similarly, U cannot be extended by

N \ A, so there is a is a D2 ∈ U such that N \ A is disjoint from D2. D1 ⊆ N \ A and D2 ⊆ A, so

D1 ∩D2 = ∅. U cannot have two disjoint sets, so this is a contradiction. U is an ultrafilter.