2.1.3 Outer Measures and Construction of Measures Recall the Construction of Lebesgue Measure on the Real Line

CHAPTER 2. THE LEBESGUE INTEGRAL I 11

2.1.3 Outer measures and construction of measures Recall the construction of Lebesgue measure on the real line. There are various approaches to construct the Lebesgue measure on the real line. For example, consider the collection of ﬁnite union of sets of the form

]a, b], ] − ∞, b], ]a, ∞[ R. This collection is an algebra A but not a sigma-algebra (this has been discussed on p.7). A natural measure on A is given by the length of an interval. One attempts to measure the size of any subset S of R covering it by countably many intervals (recall also that R can be covered by countably many intervals) and we deﬁne this quantity by

∞ ∞ X [ m∗(S) = inf{ |Ak| : Ak ∈ A,S ⊂ Ak} (2.7) k=1 k=1 where |Ak| denotes the length of the interval Ak. However, m∗(S) is not a measure since it is not additive, which we do not prove here. It is only sigma- subadditive, that is k k [ X m∗ Sn ≤ m∗(Sn) n=1 n=1 for any countable collection (Sn) of subsets of R. Alternatively one could choose coverings by open intervals (see the exercises, see also B.Dacorogna, Analyse avancéepour mathématiciens)Note that the quantity m∗ coincides with the length for any set in A, that is the measure on A (this is surprisingly the more difficult part!). Also note that the sigma-subadditivity implies that m∗(A) = 0 for any countable set since m∗({p}) = 0 for any p ∈ R. In order to get a measure from m∗ one then restricts m∗(S) to a smaller class of subsets which forms a sigma-algebra and contains A where it has the desired property of sigma-additivity. In the following we describe this procedure in an abstract setting.

Deﬁnition - outer measure. Let Ω be a nonempty set. An outer measure is a set function m∗ : P(Ω) → R+ ∪ {∞} satisfying the following conditions:

m∗(∅) = 0 (2.8)

A ⊂ B ⇒ m∗(A) ≤ m∗(B) (2.9) ∞ ∞ [ X m∗ An ≤ m∗(An). (2.10) n=1 n=1

For example, m∗ given by (2.7) is an outer measure on P(R). We shall show in the theorem below that in an appropriate abstract setting the set functions as defined by (2.7) are always outer measures. Note, that a covering by a finite number of intervals or more generally cubes or rectangles which gives rise to so called outer Jordan content will not suffice in the definition of an outer measure. For example, the rational numbers in [0, 1] have outer Jordan content 1 but the outer measure m∗ is zero since the outer measure of a single point is equal to zero and the rational numbers are countable. CHAPTER 2. THE LEBESGUE INTEGRAL I 12

Construction of an outer measure from a pre-measure deﬁned on an algebra. Let A ⊂ P(Ω) an algebra. Suppose there is a function m0 : A → R ∪ {∞} with the following properties:

1. m0(∅) = 0, S 2. if A = k≥1 Ak ∈ A for some countable collection of disjoint Ak ∈ A, then X m0(A) = m0(Ak). k≥1

Such a function is called a pre-measure. In many applications we also have the S property that Ω = k≥1 Ak for some countable collection of disjoint Ak ∈ A. If m0(Ak) < ∞ for all k we call m0 a sigma-finite pre-measure. If m0(Ω) < ∞ we call m0 a finite pre-measure. For any S ⊂ Ω define

∞ ∞ X [ m∗(S) = inf{ m0(Ak): Ak ∈ A,S ⊂ Ak}. (2.11) k=1 k=1

Theorem. m∗ is an outer measure and m∗(A) = m0(A) for any A ∈ A.

Proof. Obviously m∗(∅) = 0 and if A ⊂ B, then any covering of B is also a covering of A. Hence m∗(A) ≤ m∗(B). In order to prove the sigma-subadditivity let > 0 be arbitrary. We choose for each Ak a covering Ckn ∈ A such that

∞ ∞ [ X −k Ak ⊂ Ckn, m0(Ckn) ≤ m∗(Ak) + 2 . n=1 n=1 S∞ Now {Ckn, k ∈ Z+, n ∈ Z+} is a countable covering of k=1 Ak by elements of A and therefore it follows from the deﬁnition of m∗ that

∞ ∞ ∞ ∞ ∞ [ X X X −k X m∗ Ak ≤ m(Ckn) ≤ m∗(Ak) + 2 ≤ + m∗(Ak). k=1 k=1 n=1 k=1 k=1

To prove that m∗(A) = m0(A) for any A ∈ A we note that for any sequence Cn S∞ S∞ of sets in A such that A ⊂ n=1 Cn we have A = n=1(Cn ∩ A) and therefore using the fact that m0 is also countably subadditive for general unions in A and monotone on A ∞ ∞ X X m0(A) ≤ m0(Cn ∩ A) ≤ m0(Cn) n=1 n=1 proving m0(A) ≤ m∗(A) The reversed inequality is obvious since (A, ∅, ∅,...) is a countable cover of A and hence m∗(A) ≤ m0(A).

A toy example - pre-measure and construction of outer measure. Let Ω = {a, b, c}, A = {∅, {a}, {b, c}, Ω}, α, β ∈ [0, ∞] and deﬁne a pre-measure m0 (prove this fact!) by

m0(∅) = 0, m0({a}) = α, m0({b, c}) = β, m0(Ω) = α + β CHAPTER 2. THE LEBESGUE INTEGRAL I 13

Then m∗(∅) = 0, m∗({a}) = α, m∗({b}) = β, m∗({c}) = β and m∗(Ω) = α + β, m∗({a, b}) = α + β, m∗({a, c}) = α + β, m∗({b, c}) = β

It is easily seen that m∗|A = m0 and that in general m∗ is not additive, that is m∗ is not a measure on P(Ω). Indeed, if β > 0 , then m∗({b, c} ∪ {b}) < m∗({b}) + m∗({c}). On the other hand, when β = 0 or β = ∞ then m∗ deﬁnes a measure on P(Ω).

An alternative setting for the construction of outer measures. Since in some situations it may be difficult to find an appropriate algebra (or a pre- measure) we present another setting for constructing outer measures. Let Ω be a nonempty set and E ⊂ P(Ω) be a collection of subsets of Ω such that ∅, Ω ∈ E. Let m0 : E −→ [0, ∞] a set function such that m0(∅) = 0 (so despite its notation m0 is not necessarily a pre-measure or measure defined on E. As before for any S ⊂ Ω define ∞ ∞ X [ m∗(S) = inf{ m0(Ek): Ek ∈ E,S ⊂ Ek}. (2.12) k=1 k=1

Then m∗ is an outer measure and m∗(E) ≤ m0(E) for all E ∈ E (the same proof as above). However, in order to prove m∗(E) = m0(E) we need more restrictive hypotheses on E and m0.

Example - Outer measure on Rd. Let E the collection of all countable coverings by (closed) cubes Ck with edges parallel to the axes and let |Ck| denote the usual d-dimensional size of a cube. The function ∞ X m∗(S) = inf |Ck| k=1 where the infimum is taken over all sets in is E an outer measure on Rd. This construction differs from the general approach assuming the sigma-additivity on an algebra for countable unions which are in the algebra but corresponds to the alternative setting. We only have finite additivity but the local compactness of d R (or the compactness of closed cubes) guarantees that m∗(A) = |A| for any closed cube A: Indeed, first note that A covers itself hence m∗(A) ≤ |A|. Then consider any countable covering of A. As before we want to prove that ∞ X |A| ≤ |Cn|. n=1

For an arbitrary > 0 we choose for each n an open cube On which contains Cn and such that |On| ≤ (1 + )|Cn|. Since A is compact we may select a ﬁnite subcovering of A. Consequently,

N ∞ X X |A| ≤ (1 + ) |Cnk | ≤ (1 + ) |Cn|. k=1 n=1 The monotonicity and the countable subadditivity are shown as above for an abstract pre-measure. CHAPTER 2. THE LEBESGUE INTEGRAL I 14

Definition - m∗-measurable sets. A set A ⊂ Ω is called m∗-measurable if and only if m∗(E) = m∗(E ∩ A) + m∗(E \ A) for every E ⊂ Ω where E \ A = E ∩ Ac. Note that by the monotonicity of the outer measure we always have the inequality with the ≤ sign. Let Σ∗ denote the family of measurable sets. Sets with outer measure equal to zero are always m∗-measurable. When the outer measure is constructed from a pre-measure defined on an algebra A, then all sets A ∈ A are measurable. The notion of m∗- measurable sets filters out sets for which the outer measure m∗ is not additive (for disjoint unions) as we see from our previous toy-example.

A toy example - m∗-measurable sets. Recall that Ω = {a, b, c}, A = {∅, {a}, {b, c}, Ω}, α, β ∈ [0, ∞] and a pre-measure m0 deﬁned by

m0(∅) = 0, m0({a}) = α, m0({b, c}) = β, m0(Ω) = α + β

We had seen that when β > 0 , m∗({b, c} ∪ {b}) < m∗({b}) + m∗({c}). Hence {b, c} is not m∗-measurable. In this case a set is m∗-measurable if and only if it is in the algebra A.

Carath´eodory’s theorem. With the above notations Σ∗is a sigma algebra and the restriction of m∗ to Σ∗ is a measure µ.

Proof. See exercise 4.1.

Theorem - Further results from the construction via an outer measure.

• When m∗ is constructed form a pre-measure m0 deﬁned on an algebra A

then A ⊂ Σ∗ and m0 = m∗|Σ∗ = µ|Σ∗

• When m∗ is constructed form a pre-measure m0 deﬁned on an algebra A then µ is maximal in the following sense: For any measure ν on Σ∗

such that m0 = ν|Σ∗ we have ν(S) ≤ µ(S) for any S ∈ Σ∗. In addition, if µ(S) < ∞ then ν(S) = µ(S). If m0 is sigma-ﬁnite (that is Ω is a countable union of sets in A of ﬁnite pre-measure) then ν(S) = µ(S) for all S ∈ Σ∗, that is ν = µ.

Proof. The first statement follows directly from the construction of the outer measure defined in (2.11) and Carathéodory’s theorem. For the second statement we only show the maximality property. Let ν be a measure that extends m0 to Σ∗ and S ∈ Σ∗. For any countable cover of S by Ak ∈ A using that ν is a measure on Σ∗ (hence countably sub-additive) we have:

∞ ∞ X X ν(S) ≤ ν(Ak) = m0(Ak) k=1 k=1 which implies ν(S) ≤ m∗(S) = µ(S). CHAPTER 2. THE LEBESGUE INTEGRAL I 15

Lebesgue and Borel measurable sets in Rd. For Rd the measure µ and the sigma algebra Σ constructed from an algebra of rectangles and the usual volume as a pre-measure with the help of Carath´eodory’s theorem are called the Lebesgue measure and the sigma algebra of Lebesgue measurable functions (of- ten denoted by L(Rd)), respectively. This sigma algebra is larger than the Borel sets and is the completion of the Borel sets by adjoining all subsets of Borel sets of Lebesgue measure zero. This annoyance is caused by the fact that a subset of a set of measure zero might not be Borel measurable. We shall mainly use the Borel sets which has advantages for the deﬁnition of product measures (see 2.2.5). Further consequences as regularity properties of the Lebesgue measure will be discussed in Chapter 3.

Properties of the Lebesgue measure on Rd. We list a couple of properties without proof of the Lebesgue measure on Rd which follow from the construction via the outer measure given above. See also Chapter 3 for further properties. Let µ denote the Lebesgue measure on Rd and let S ⊂ Rd be Lebesgue measurable. Then

1. µ is translation invariant: In particular, a set S ⊂ Rd is Lebesgue measur- d able if and only if all translates are Lebesgue measurable. For x0 ∈ R let d S + x0 = {x ∈ R : x + x0 ∈ S} then µ(S) = µ(S + x0) for any Lebesgue measurable set.

2. µ is invariant under orthogonal transformations: Let A : Rd −→ Rd be an orthogonal linear mapping and let A(S) = {A(x) = Ax ∈ Rd : x ∈ S}. Then A(S) is Lebesgue measurable and µ(A(S)) = µ(S).

3. Dilations/scaling: Let λ > 0 and denote λS = {λx ∈ Rd : x ∈ S}. Then λS is Lebesgue measurable and µ(λS) = λdµ(S).

4. Finite or countable sets are Lebesgue measurable and have Lebesgue measure 0. 5. There exist uncountable Lebesgue measurable sets of measure 0 as (when d = 1) the Cantor set.

6. There exist sets which are not Lebesgue measurable. Hence L(Rd) is smaller than P(Rd). Now we can illustrate the annoyance described above. From the outer measure m∗ it follows that each m∗-measurable set of outer measure zero is Lebesgue measurable and has Lebesgue measure zero. In particular, each subset of a set with Lebesgue measure zero is Lebesgue measurable (and has Lebesgue measure zero, too). Now, an interval or a line in R2 has Lebesgue measure zero (in R2). On the other hand a Cantor set is a subset of an interval hence Lebesgue measurable in R2. Therefore we have to take some care when using the product structure R2 = R × R.