The Concept of Measure

4. The concept of measure

Definition. Let X be some set, and let E be an arbitrary collection of subsets of X. Assume ∅ ∈ E.A measure on E is a map µ : E → [0, 1] with the following properties (0) µ(∅) = 0. ∞ S∞ (addσ) Whenever (En)n=1 ⊂ E is a pair-wise disjoint sequence, with n=1 En ∈ E, it follows that we have the equality ∞ ∞ [ X µ En = µ(En). n=1 n=1

Property (addσ) is called σ-additivity. ∞ Convention. For a sequence (αn)n=1 ⊂ [0, ∞] we define  ∞  X ∞  α if α ∈ [0, ∞), ∀ n ∈ X  n n N αn = n=1  n=1   ∞ if there exists n ∈ N with αn = ∞. P∞ (Of course, in the first case, it is still possible to have n=1 αn = ∞.) Remark 4.1. If µ is a measure on E, then µ is additive, i.e. N (add) Whenever (En)n=1 ⊂ E is a finite pair-wise disjoint system, such that E1 ∪ · · · ∪ EN ∈ E, it follows that we have the equality µ E1 ∪ · · · ∪ EN = µ(E1) + ··· + µ(EN ).

This follows from (addσ) (0), after completing the sequence E1,...,EN to an infinite sequence by taking En = ∅, ∀ n > N. Comment. The most natural setting for measures is the one when E is a σ-ring. S∞ In this case, the stipulation that n=1 En ∈ E, which appears in the definition, is superfluous. The purpose of this section is to study measures on more rudimentary collections. Examples 4.1. Let X be a non-empty set. A. If we take E = {∅,X} and we define µ(∅) = 0 and µ(X) to be any element in [0, ∞], then µ is obviously a measure on {∅,X}. B. If we take E = P(X) and we define 0 if E = µ(E) = ∅ ∞ if E 6= ∅ then µ is a measure on P(X). C. If we take E = P(X) and we define card E if E is finite µ(E) = ∞ if E is infinite then µ is a measure on P(X). This is called the counting measure.

200 §4. The concept of measure 201

Exercise 1. Let X1, X2 be non-empty sets, let Ek ⊂ P(Xk) be arbitrary collections with ∅ ∈ Ek, k = 1, 2. Let µ1 be a measure on E1 and µ2 be a measure on E2. Consider the collections −1 f∗E1 = {A ⊂ X2 : f (A) ∈ E1} ⊂ P(X2); ∗ −1 f E2 = {f (A): A ∈ E2} ⊂ P(X1).

A. Prove that the map f∗µ1 : f∗E1 → [0, ∞], defined by −1 (f∗µ1)(A) = µ1 f (A) , ∀ A ∈ f∗E1, is a measure on f∗E1. ∗ ∗ B. If f is surjective, prove that the map f µ2 : f E2 → [0, ∞], defined by ∗ ∗ (f µ2) ) = µ2 f(B) , ∀ B ∈ f E2, ∗ is a measure on f E2. We now concentrate on the most rudimentary types of collections E on which measures can be somehow easily defined. Actually, what we have in mind is a set of easy conditions on a map µ : E → [0, ∞] which would guarrantee that µ is a measure. Definition. Let X be some set. A collection J ⊂ P(X) is called a semiring, if it satisfies the following properties: (i) ∅ ∈ J; (ii) if A, B ∈ J, then A ∩ B ∈ J; (iii) if A, B ∈ J and A ⊂ B, then there exists an integer n ≥ 1, and sets D0,D1,...,Dn ∈ J, such that A = D0 ⊂ D1 ⊂ · · · ⊂ Dn = B, and Dk r Dk−1 ∈ J, ∀ k ∈ {1, . . . , n}. Remark that every ring is a semiring. Remark also that, if J contains at least one set, then conditions (ii) and (iii) imply (i). Exercise 2. Prove that the semiring type is not consistent. Give an example of two semirings J1, J2 ⊂ P(X), such that J1 ∩ J2 is not a semiring. Hint: Use the set X = {1, 2, 3}. The following construction is very useful.

Proposition 4.1. Let X1,...,Xn be some sets, and let Jk ⊂ P(Xk), k = 1, . . . , n, be semirings. Then the collection J = A1 × · · · × An : A1 ∈ J1,...,An ∈ Jn ⊂ P(X1 × · · · × Xn) is a semiring. Proof. An obvious induction argument shows that it suﬃces to prove only the case n = 2. The fact that J contains the empty set is trivial. To check the second condition, we start with two sets A, B ∈ J and we show that A ∩ B ∈ J. This is however trivial, because if we write A = A1 × A2 and B = B1 × B2, with A1,B1 ∈ J1 and A2,B2 ∈ J2, then we clearly have

A ∩ B = (A1 ∩ B1) × (A2 ∩ B2), with A1 ∩ B1 ∈ J1 and A2 ∩ B2 ∈ J2, so A ∩ B indeed belongs to J. Finally, we check the third condition. Start with two sets A, B ∈ J with A ⊂ B, and let us construct the desired sequence D0,D1,...,Dn ∈ J, satisfying (iii). Write 202 CHAPTER III: MEASURE THEORY

A = A1 × A2 and B = B1 × B2, with A1,B1 ∈ J1 and A2,B2 ∈ J2, so that we have the inclusions A1 ⊂ B1 and A2 ⊂ B2. Use condition (iii) for the semiring J1 to ﬁnd sets E0,E1,...,Ek ∈ J1, such that A1 = E0 ⊂ E1 ⊂ · · · ⊂ Ek = B1, and Ei r Ei−1 ∈ J1, ∀ i = 1, . . . , k. Also use condition (iii) for the semiring J2 to ﬁnd sets F0,F1,...,F` ∈ J2, such that A2 = F0 ⊂ F1 ⊂ · · · ⊂ F` = B2, and Fj r Fj−1 ∈ J2, ∀ j = 1, . . . , `. Notice that we have

A1 ×A2 = E0 ×A2 ⊂ · · · ⊂ Ek ×A2 = B1 ×A2 = B1 ×F0 ⊂ · · · ⊂ B1 ×F` = B1 ×B2, n so if we put n = k + ` and deﬁne the sets (Dp)p=0 by Ep × A2 if p ≤ k Dp = B1 × Fp−k if p > k then the sets D0,D1,...,Dn all belong to J, and also satisfy

A = D0 ⊂ · · · ⊂ Dn = B. Finally, it is also clear that, for every p ∈ {1, 2, . . . , n} one has (Ep r Ep−1) × A2 if p ≤ k Dp r Dp−1 = B1 × (Fp−k r Fp−k−1) if p > k so we clearly have Dp r Dp−1 ∈ J. Example 4.2. Take X = R. The collection J = {∅} ∪ [a, b): a, b ∈ R, a < b ⊂ P(R) is a semiring. Indeed, the first two axioms are pretty clear. To prove the third axiom, we start with two intervals A = [a, b) and B = [c, d) with A ⊂ B. This means that a ≥ c and b ≤ d. If a = c or b = d, we set D0 = A and D1 = B. If a > c and b < d, we set D0 = A, D1 = [a, d) and D2 = B. More generally, by Proposition 4.1, the collection of ”half-open boxes” n Y n Jn = {∅} ∪ [aj, bj): a1 < b1, . . . , an < bn ⊂ P(R ) j=1 is a semiring. n Exercise 3. Let Jn ⊂ P(R ) be the semiring defined above. Prove that the n σ-ring S(J) generated by Jn coincides with Bor(R ). The ring generated by a semiring has a particularly nice description (compare to Proposition 2.1): Proposition 4.2. Let J be a semiring on X. For a subset A ⊂ X, the following are equivalent: (i) A belongs to R(J), the ring generated by J; n (ii) There exists an integer n ≥ 1, and a pair-wise disjoint system (Aj)j=1 ⊂ J, such that A = A1 ∪ · · · ∪ An. Proof. Denote by R the collection of all subsets A ⊂ X that satisfy condition (ii). It is obvious that J ⊂ R ⊂ R(J), so (see Section III.2) we only need to prove that R is a ring. Let us first remark that we obviously have the property: §4. The concept of measure 203

(i) if A, B ∈ R, and A ∩ B = ∅, then A ∪ B ∈ R. Secondly, we remark that we have have the implication: (ii) A, B ∈ J ⇒ A r B ∈ R. Indeed, since A∩B ∈ J, by the definition of a semiring, there exist D0,D1,...,Dn ∈ J with A ∩ B = D0 ⊂ D1 ⊂ · · · ⊂ Dn = A, and Dk r Dk−1 ∈ J, ∀ k ∈ {1, . . . , n}. Then the equality n [ Ar = (Dk r Dk−1) k=1 shows that A r B indeed belongs to R. Thirdly, we prove the implication: (iii) A, B ∈ R ⇒ A ∩ B ∈ R. m n Write A = A1 ∪ · · · ∪ Am and B = B1 ∪ · · · ∪ Bn, with (Ai)i=1, (Bk)k=1 ⊂ J pair-wise disjoint systems. If we define the sets Dik = Aj ∩ Bk ∈ J,(i, k) ∈ {1, . . . , m} × {1, . . . , n} then it is obvious that m n [ [ A ∩ B = Dik, i=1 k=1 and (Dik)1≤i≤m ⊂ J is a pair-wise disjoint system, therefore A ∩ B indeed belongs 1≤j≤n to R. Finally, we show the implication: (iv) if A, B ∈ R and A ⊃ B, then A r B ∈ R. m Write A = A1 ∪ · · · ∪ Am, with (Ai)i=1 ⊂ J a pair-wise disjoint system. Notice that m [ A r B = (Ai r B), i=1 m with (Ai r B)i=1 a pair-wise disjoint system, so by (i) it suffices to show that Ai rB ∈ R, ∀ i ∈ {1, . . . , m}. To prove this, we fix i and we write B = B1 ∪· · ·∪Bn, n with (Bk)k=1 ⊂ J a pair-wise disjoint system. Then

Ai r B = (Ai r B1) ∩ · · · ∩ (Ai r Bn), and the fact that Ai r B belongs to R follows from (ii) and (iii). Having proven (i)-(iv), it we now prove that R is a ring. By (iii), we only need to prove the implication (∗) A, B ∈ R ⇒ A4B ∈ R. On the one hand, using (iv), it follows that the sets A r B = A r (A ∩ B) and B r A = B r (A ∩ B) both belong to R. Since A4B = (A r B) ∪ (B r A), and (A r B) ∩ (B r A) = ∅, by (i) is follows that A4B indeed belongs to R. Theorem 4.1 (Semiring-to-ring extension). Let J be a semiring on X, and let µ : J → [0, ∞] be an additive map with µ(∅) = 0. (i) There exists a unique additive map µ¯ : R(J) → [0, ∞], such that µ¯ = µ. J (ii) If µ is σ-additive, then so is µ¯. Proof. The key step is contained in the following 204 CHAPTER III: MEASURE THEORY

m n Claim: If (Ai)i=1 ⊂ J and (Bj)j=1 ⊂ J are pair-wise disjoint systems, with

A1 ∪ · · · ∪ Am = B1 ∪ · · · ∪ Bn,

then µ(A1) + ··· + µ(Am) = µ(B1) + ··· + µ(Bn).

To prove this fact, we define the pair-wise disjoint system (Dij)1≤i≤m by Dij = 1≤j≤n Ai ∩ Bj, ∀ (i, j) ∈ {1, . . . , m} × {1, . . . , n}. Since n [ Dij = Ai, ∀ i ∈ {1, . . . , m}, j=1 m [ Dij = Bj, ∀ j ∈ {1, . . . , n}, i=1 using additivity, we have the equalities n X µ(Dij) = µ(Ai), ∀ i ∈ {1, . . . , m}, j=1 m X µ(Dij) = µ(Bj), ∀ j ∈ {1, . . . , n}, i=1 and then we get m m m n n n X X X X X X µ(Ai) = µ(Dij) = µ(Dij) = µ(Bj). i=1 i=1 j=1 j=1 i=1 j=1 To prove (i), for any set A ∈ R(J) we choose (use Proposition 4.2) a finite n pair-wise disjoint system (Ai)i=1 ⊂ J, with A = A1 ∪ · · · ∪ An, and we define

(1)µ ¯(A) = µ(A1) + ··· + µ(An). By the above Claim, the number µ¯(A) is independent of the particular choice of the pair-wise disjoint system (A )n . Also, it is clear thatµ ¯ = µ, andµ ¯ is additive. i i=1 J The uniqueness is clear, because the equalityµ ¯ = µ and additivity ofµ ¯ force (1) J (ii). Assume now that µ is σ-additive, and let us prove thatµ ¯ is again σ- ∞ S∞ additive. Start with a pair-wise disjoint sequence (An)n=1 ⊂ R(J), with n=1 An ∈ R(J), and let us prove the equality ∞ ∞ [ X (2)µ ¯ An = µ¯(An). n=1 n=1 S∞ p Since n=1 An ∈ R, there exists a ﬁnite pair-wise disjoint system (Bi)i=1 ⊂ J, such S∞ that n=1 An = B1 ∪ · · · ∪ Bp. With this choice we have ∞ p [ X (3)µ ¯ An = µ(Bi). n=1 i=1 S∞ For each i ∈ {1, . . . , p}, we have Bi = n=1(Bi ∩ An). Fix for the moment a pair (n, i) ∈ N × {1, . . . , p}. Since Bi ∩ An ∈ R(J), it follows that there exist an ni Nni integer Nni ≥ 1 and a ﬁnite pair-wise disjoint system (Ck )k=1 ⊂ J, such that SNni ni Bi ∩ An = k=1 Ck . §4. The concept of measure 205

ni Since, for each i ∈ {1, . . . , p}, the countable system (Ck ) n∈N ⊂ J is pair- 1≤k≤Nni wise disjoint, and we have the equality

∞ Nni ∞ [ [ ni [ Ck = (Bi ∩ An) = Bi ∈ J, n=1 k=1 n=1 by the σ-additivity of µ, we have

∞ N ni X X ni (4) µ(Bi) = µ(Ck ), ∀ i ∈ {1, . . . , p}. n=1 k=1

ni Since, for each n ∈ N, the ﬁnite system (Ck ) 1≤i≤p ⊂ J is pair-wise disjoint, 1≤k≤Nni and we have the equality

p Nni ∞ [ [ ni [ Ck = (Bi ∩ An) = An ∈ J, i=1 k=1 i=1 by the deﬁnition ofµ ¯, we have

p N ni X X ni µ¯(An) = µ(Ck ), ∀ i ∈ {1, . . . , p}. i=1 k=1 Combining this with (4) yields

∞ ∞ p Nni p X X X X ni X µ¯(An) = µ(Ck ) = µ(Bi), n=1 n=1 i=1 k=1 i=1 and the equality (2) follows from (3).

Definition. Let X be some set, and let E ⊂ P(X) be a collection of sets. We say that a map µ : E → [0, ∞] is sub-additive, if − n Sn (add ) whenever A ∈ E, and (An)k=1 is a ﬁnite sequence in E with A ⊂ k=1 Ak, Pn it follows that µ(A) ≤ k=1 µ(Ak).

Note that we do not require the Ak’s to be pair-wise disjoint. With this terminology, Theorem 4.1 has the following.

Corollary 4.1. Let X be some set, and let J ⊂ P(X) be a semiring. Then any additive map µ : J → [0, ∞] is sub-additive.

Proof. Letµ ¯ : R(J) → [0, ∞] be the additive extension of µ to the ring generated by J. It suﬃces to prove thatµ ¯ is sub-additive. Start with sets A, A1,...,An ∈ R(J) such that A ⊂ A1 ∪ ...An. Deﬁne the sets B1 = A1, and

Bk = Ak r (A1 ∪ · · · ∪ Ak−1), forall k ∈ {1, . . . , n}, k ≥ 2.

Since we work in a ring, the sets Bk, Bk ∩A, Bk rA, and An rBn, n ∈ N, all belong n to R(J). Moreover, the sequence (Bk)k=1 is pair-wise disjoint and it satisﬁes

• Bk ⊂ Ak, ∀ k ∈ {1, . . . , n}, Sn Sn • k=1 Bk = k=1 Ak ⊃ A, 206 CHAPTER III: MEASURE THEORY so by the additivity ofµ ¯, we get n n n X X X µ¯(Ak) = µ¯ (Ak r Bk) ∪ Bk = µ¯(Ak r Bk) +µ ¯(Bk) ≥ k=1 k=1 k=1 n n n X X X ≥ µ¯(Bk) = µ¯ (Bk r A) ∪ (Bk ∩ A) = µ¯(Bk r A) +µ ¯(Bk ∩ A) ≥ k=1 k=1 k=1 n n X [ ≥ µ¯(Bk ∩ A) =µ ¯ [Bk ∩ A] =µ ¯(A). k=1 k=1 In connection with Proposition 4.1, one has the following result.

Proposition 4.3. Let X1,...,Xn be some sets, let Jk ⊂ P(Xk), k = 1, . . . , n be semirings, and let µk : Jk → [0, ∞], k = 1, . . . , n, be additive maps. Consider the semiring (by Proposition 4.1) J = A1 × · · · × An : A1 ∈ J1,...,An ∈ Jn ⊂ P(X1 × · · · × Xn). Then the map µ : J → [0, ∞] deﬁned by

µ(A1 × · · · × An) = µ1(A)1 ··· µn(An) is additive. (Use the convention 0 · ∞ = 0.) Proof. An easy induction argument shows that it is suﬃcient to prove only the case n = 2. Consider for the moment a larger collection L = A1 × A2 : A1 ∈ R(J1),A2 ∈ R(J2) ⊂ P(X1 × X2), and the map ν : L → [0, ∞] deﬁned by

ν(A1 × A2) =µ ¯1(A)1 · µ¯2(A2), whereµ ¯k : R(Jk) → [0, ∞] are the additive maps given by Theorem 4.1. Obviously L is a semiring (by Proposition 4.1) which contains the semiring J, and we have ν = µ. It then suﬃces to prove that ν is additive. This argument shows that, J upon replacing Jk with R(Jk) and µk withµ ¯k, k = 1, 2, we can (and will) assume that both J1 and J2 are rings. We need to prove the property

(∗) whenever B1,B2,...,Bp ∈ J are pairwise disjoint, and B1 ∪ · · · ∪ Bp ∈ J, one has the equality µ(B1 ∪ · · · ∪ Bp) = µ(B1) + ··· + µ(Bp). We prove property (∗) by induction on p. We start with the case p = 2, so we have two disjoint sets B1,B2 ∈ J, such that B1 ∪ B2 ∈ L, and we need to prove that

(5) µ(B1 ∪ B2) = µ(B1) + µ(B2).

If one of the sets B1, B2 is empty, there is nothing to prove. Assume both B1 and B2 are non-empty. Since the sets B1 and B2 belong to J, we can write them as k k k k Bk = A1 × A2 , with A1 ∈ J1 and A2 ∈ J2, k = 1, 2. Since the set B1 ∪ B2 also belongs to J, we can write it as B1 ∪ B2 = A1 × A2 with A1 ∈ J1 and A2 ∈ J2. With these notations we have 1 1 2 2 (6) A1 × A2 = (A1 × A2) ∪ (A1 × A2). Claim 1: The equality (6) forces either 1 2 (a) A1 = A1 = A1, or §4. The concept of measure 207

2 2 (b) A1 = A2 = A2. 1 1 2 2 Indeed, on the one hand the fact that (A1 ∩ A2) ∩ (A1 × A2) = ∅, forces either 1 2 (i) A1 ∩ A1 = ∅, or 1 2 (ii) A2 ∩ A2 = ∅. 1 2 In case (i) we can then form two disjoint subsets D1 = A1 × A2 and D2 = A1 × A2, which will satisfy the inclusions Bk ⊂ Dk, k = 1, 2, as well as

1 2 A1 × A2 = B1 ∪ B2 ⊂ D1 ∪ D2 = (A1 ∪ A1) × A2 ⊂ A1 × A2.

This forces D1 ∪ D2 = A1 × A2 = B1 ∪ B2, so in fact we have Bk = Dk, k = 1, 2. In other words, we have

k k k A1 × A2 = A1 × A2, k = 1, 2,

1 2 and the fact that A1 and A1 are non-empty will force (b). In the exact same way, one can see that in case (ii) we get the equalities (a). Having proven the Claim, the desired equality (5) is now pretty clear. In case 1 2 1 2 (a) the equality (6) forces A2 ∩ A2 = ∅ and A2 ∪ A2 = A2, so we immediately have 1 2 µ(B1 ∪ B2) = µ1(A1) · µ2(A2) = µ1(A1) · µ2(A2) + µ2(A2) = 1 1 1 2 = µ1(A1) · µ2(A2) + µ1(A2) · µ2(A2) = µ(B1) + µ(B2). A similar computation proves that (5) also holds in case (b). Having proven the case p = 2, we now proceed with the inductive step. Assume (∗) holds for p = N and let us prove that it also holds for p = N + 1. Start with some pairwise disjoint sets B1,B2,...,BN ,BN+1 ∈ J, such that the union B = B1 ∪ · · · ∪ BN ∪ BN+1 belongs to J, and let us prove the equality

(7) µ(B) = µ(B1) + ··· + µ(BN ) + µ(BN+1).

Let us write B = T1 × T2, and BN+1 = A1 × A2, with A1,T1 ∈ J1 and A2,T2 ∈ J2. It is pretty obvious that one has the equality

T1 × T2 = (A1 × A2) ∪ (A1 × [T2 r A2]) ∪ ([T1 r A1] × T2).

0 00 Note that the sets B = A1 ×[T2 rA2] and B = [T1 rA1]×T2 both belong to J (it is here that the assumption that J1 and J2 are rings is used, so that T1 r A1 ∈ J1 0 00 and T2 rA2 ∈ J2). It is also clear that, and the sets BN+1, B , and B are pairwise disjoint. 0 00 Claim 2: µ(B) = µ(BN+1) + µ(B ) + µ(B ).

This eqaulity is proven by direct computation, using additivity of µ1 and µ2. If we start with the right hand side, we have

0 00 µ(BN+1) + µ(B ) + µ(B ) =

= µ1(A1) · µ2(A2) + µ1(A1) · µ2(T2 r A2) + µ1(T1 r A1) · µ2(T2) = = µ1(A1) · µ2(A2) + µ2(T2 r A2) + µ1(T1 r A1) · µ2(T2) = = µ1(A1) · µ2(T2) + µ1(T1 r A1) · µ2(T2) = = µ1(A1) + µ1(T1 r A1) · µ2(T2) = µ1(T1) · µ2(T2) = µ(B). 208 CHAPTER III: MEASURE THEORY

0 00 Notice that by construction we have B ∪ B = B r BN+1 = B1 ∪ · · · ∪ BN , so we have the equalities 0 0 0 B = [B ∩ B1] ∪ · · · ∪ [B ∩ BN ], 00 00 00 B = [B ∩ B1] ∪ · · · ∪ [B ∩ BN ]. 0 00 Since all sets B ∩ Bk and B ∩ Bk, k = 1,...,N belong to J, by the inductive hypothesis we have 0 0 0 µ(B ) = µ(B ∩ B1) + ··· + µ(B ∩ BN ), 00 00 00 µ(B ) = µ(B ∩ B1) + ··· + µ(B ∩ BN ), so using Claim 2, we get N X 0 00 µ(B) = µ(BN+1) + µ(B ∩ Bk) + µ(B ∩ Bk) . k=1 0 00 Using the case n = 2, combined with the equality [B ∩ Bk] ∪ [B ∩ Bk] = Bk, we 0 00 have µ(B ∩ Bk) + µ(B ∩ Bk) = µ(Bk), ∀ k = 1,...,N, so we above computation PN yields µ(B) = µ(BN+1) + k=1 µ(Bk), and we are done. Comment. In connection with the above result, one can ask the following

Question: With the notations above, is it true that, if both µ1, . . . , µn are measures, then µ is also a measure? As we shall see a bit later in the course, that the answer is is “yes.” Definition. Let X be a some set, and let E ⊂ P(X) be a collection of sets. We say that a map µ : E → [0, ∞] is σ-sub-additive, if − ∞ S∞ (addσ ) whenever A ∈ E, and (An)n=1 is a sequence in E with A ⊂ n=1 An, it P∞ follows that µ(A) ≤ n=1 µ(An). Note that we do not require the An’s to be pair-wise disjoint. Proposition 4.4 (characterization of semiring measures). Let X be a non- empty set, let J ⊂ P(X) be a semiring, and let µ : J → [0, ∞] be a map with µ(∅) = 0. The following are equivalent: (i) µ is a measure on J; (ii) µ is additive, and σ-sub-additive. Proof. (i) ⇒ (ii). Assume µ is a measure on J. It is clear that µ is additive, so we only need to prove σ-sub-additivity. Use Theorem 4.1 to find a measureµ ¯ on the ring R(J) generated by J, such that µ¯(A) = µ(A), ∀ A ∈ J. Then it suffices to show thatµ ¯ is σ-sub-additive. Start with a set A ∈ R(J), and a ∞ S∞ sequence (An)n=1 ⊂ R(J), such that A ⊂ n=1 An. Define the sets B1 = A1, and

Bn = An r (A1 ∪ · · · ∪ An−1), ∀ n ≥ 2.

Since we work in a ring, the sets Bn, Bn ∩A, Bn rA, and An rBn, n ∈ N, all belong ∞ to R(J). Moreover, the sequence (Bn)n=1 is pair-wise disjoint and it satisﬁes • Bn ⊂ An, ∀ n ∈ N, S∞ S∞ • n=1 Bn = n=1 An ⊃ A, §4. The concept of measure 209 so by σ-additivity ofµ ¯, we get

∞ ∞ ∞ X X X µ¯(An) = µ¯ (An r Bn) ∪ Bn = µ¯(An r Bn) +µ ¯(Bn) ≥ n=1 n=1 n=1 ∞ ∞ ∞ X X X ≥ µ¯(Bn) = µ¯ (Bn r A) ∪ (Bn ∩ A) = µ¯(Bn r A) +µ ¯(Bn ∩ A) ≥ n=1 n=1 n=1 ∞ ∞ X [ ≥ µ¯(Bn ∩ A) =µ ¯ [Bn ∩ A] =µ ¯(A). n=1 n=1 (ii) ⇒ (i). Assume µ : J → [0, ∞] is additive and σ-sub-additive, and let us show that µ is σ-additive. We again use Theorem 4.1, to ﬁnd an additive map µ¯ : R(J) → [0, ∞], such thatµ ¯ = µ. Start with a pair-wise disjoint sequence J ∞ S∞ (An)n=1 ⊂ J, such that the union A = n=1 An belongs to J. On the one hand, by σ-sub-additivity, we have the inequality

∞ X (8) µ(A) ≤ µ(An). n=1 On the other hand, for any integer N ≥ 1, we have N N [ [ µ(A) =µ ¯(A) =µ ¯ An ∪ A r An ≥ n=1 n=1 N N N [ X X ≥ µ¯ An = µ¯(An) = µ(An), n=1 n=1 n=1 which then gives N ∞ X X µ(A) ≥ sup µ(An) = µ(An), N∈N n=1 n=1 P∞ so using (8) we immediately get µ(A) = n=1 µ(An).

The following technical result will be often employed in subsequent sections.

Lemma 4.1 (Continuity). Let J be a semiring, and let µ be a measure on J. ∞ S∞ (i) If (An)n=1 ⊂ J is a sequence of sets, with A1 ⊂ A2 ⊂ ... , and n=1 An ∈ J, then ∞ [ µ An = lim µ(An). n→∞ n=1 ∞ T∞ (ii) If (Bn)n=1 ⊂ J is a sequence of sets, with B1 ⊃ B2 ⊃ ... , and n=1 Bn ∈ J, and µ(B1) < ∞, then ∞ \ µ Bn = lim µ(Bn). n→∞ n=1 Proof. Using Theorem 4.1, we can assume that J is already a ring. (Otherwise we replace J by R(J), and µ by its extensionµ ¯.) 210 CHAPTER III: MEASURE THEORY

(i). Consider the sets D1 = A1, and Dk = An r Ak−1, ∀ k ≥ 2. It is clear that ∞ (Dk)k=1 is a pairwise disjoint sequence in J, and we have the equality n [ (9) Dk = An, ∀ n ≥ 1. k=1 This gives of course the equality ∞ ∞ [ [ Dk = An ∈ J. k=1 n=1 Using this equality, combined with the (σ-)additivity of µ, and with (9), we get

∞ ∞ n n [ X X [ µ An = µ(Dk) = lim µ(Dk) = lim µ Dk = lim µ(An). n→∞ n→∞ n→∞ n=1 k=1 k=1 k=1 T∞ (ii). Consider the sets B = n=1 Bn, and An = B1rBn, ∀ n ≥ 1. It is clear that ∞ S∞ (An)n=1 ⊂ J, and we have A1 ⊂ A2 ⊂ ... . Moreover, we have n=1 An = B1 r B, so by part (i), we get

(10) µ(B1 B) = lim µ(B1 Bn). r n→∞ r

Using the fact that µ(B1) < ∞, it follows that

µ(B) ≤ µ(Bn) ≤ µ(B1) < ∞, ∀ n ≥ 1. This gives then the equalities

µ(B1 r B) = µ(B1) − µ(B) and µ(B1 r Bn) = µ(B1) − µ(Bn), ∀ n ≥ 1, so the equality (10) immediately gives µ(B) = limn→∞ µ(Bn). The above result has a (minor) generalization, which we record for future use. To formulate it we introduce the following.

Notation. Let R be a ring, and let µ be a measure on R. For two sets A, B ∈ R, we write A ⊂ B, if µ(A r B) = 0. µ Using this notation, we have the following generalization of Lemma 4.1.

Proposition 4.5. Let R be a ring, and let µ be a measure on R. ∞ S∞ (i) If (An)n=1 ⊂ R is a sequence of sets, with A1 ⊂ A2 ⊂ ... , and An ∈ µ µ n=1 R, then ∞ [ µ An = lim µ(An). n→∞ n=1 ∞ T∞ (ii) If (Bn)n=1 ⊂ R is a sequence of sets, with B1 ⊃ B2 ⊃ ... , and Bn ∈ µ µ n=1 J, and µ(B1) < ∞, then ∞ \ µ Bn = lim µ(Bn). n→∞ n=1 §4. The concept of measure 211

∞ Sn Proof. (i). Deﬁne the sequence of sets (En)n=1 ⊂ R, by En = k=1 Ak, ∀ n ≥ 1. Notice that, A1 = E1, and for each n ≥ 2, we have An ⊂ En, as well as the equality n−1 [ En r An = [An r Ak]. k=1 Using sub-additivity, it follows that n−1 X µ(En r An) ≤ µ(An r Ak), k=1 which forces µ(En r An) = 0. This gives (11) µ(En) = µ(An) + µ(En r An) = µ(An), ∀ n ≥ 1. S∞ S∞ Since n=1 En = n=1 An, and we have the inclusions E1 ⊂ E2 ⊂ ... , by Lemma 4.1, combined with (11), we get ∞ ∞ [ [ µ An = µ En = lim µ(En) = lim µ(An). n→∞ n→∞ n=1 n=1 Part (ii) is proven exactly as part (ii) from Lemma 4.1. Exercise 4. Let µ be a measure on a ring R. Prove that, for A, B ∈ R, one has the implication A ⊂ B ⇒ µ(A) ≤ µ(B). µ Example 4.3. Fix some integer n ≥ 1. Consider the semiring of “half-open boxes” in Rn n Y n Jn = {∅} ∪ [aj, bj): a1 < b1, . . . , an < bn ⊂ P(R ). j=1

For a non-empty box A = [a1, b1) × · · · × [an, bn) ∈ Jn, we deﬁne n Y voln(A) = (bk − ak). k=1 We also deﬁne voln(∅) = 0.

Theorem 4.2. With the above notations, the map voln : J → [0, ∞] is a measure on Jn. Proof. First we prove additivity. Using Proposition 4.3, it suﬃces to analyze only the case n = 1, i.e. the case of half-open intervals in R. We need to show the implication p [a, b) = S [a , b )  p k=1 p p  X (12) =⇒ b − a = (bk − ak). p  k=1 [ak, bk) k=1 pair-wise disjoint We can prove this using induction on p. The case p = 1 is trivial. Assuming that the above fact holds for p = N, let us prove it for p = N + 1. Pick k1 ∈ {1,...,N + 1} such that ak1 = a. Then we clearly have [ [ak, bk) = [bk1 , b), 1≤k≤N+1 k6=k1 212 CHAPTER III: MEASURE THEORY so by the inductive hypothesis we get X b − bk1 = (bk − ak), 1≤k≤N+1 k6=k1 so we get N+1 X (bk − ak) = (bk1 − ak1 ) + (b − bk1 ) = b − ak1 = b − a, k=1 and we are done. We now prove that voln is σ-sub-additive. Suppose we have A ∈ Jn and a ∞ S∞ sequence (Ak)k=1 ⊂ Jn, such that A ⊂ k=1 Ak, and let us prove the inequality ∞ X (13) voln(A) ≤ voln(Ak). k=1 It will be helpfull to introduce the following notations. For every half-open box

B = [x1, y1) × · · · × [xn, yn), and every δ > 0, we deﬁne the boxes δ B = [x1 − δ, y1) × · · · × [xn − δ, yn),

Bδ = [x1, y1 − δ) × · · · × [xn, yn − δ).

It is clear that, for any box B ∈ Jn we have δ (14) Bδ ⊂ B ⊂ Int(B ), δ (15) voln(B) = lim voln(B ) = lim voln(Bδ). δ→0+ δ→0+ ∞ To prove (13), we ﬁx some ε > 0, and we choose positive numbers δ and (δk)k=1, such that ε (16) vol (A ) > vol (A) − ε, and vol (A )δk < + vol (A ), ∀ k ∈ . n δ n n k 2k n k N Notice now that, using (14), we have the inclusions ∞ ∞ [ [ δk Aδ ⊂ A ⊂ Ak ⊂ Int (Ak) , k=1 k=1 and using the compactness of Aδ, there exists some N ≥ 1, such that N [ δk Aδ ⊂ Int (Ak) . k=1 This immediately gives the inclusion N [ δk Aδ ⊂ (Ak) . k=1 Using sub-additivity (see Corollary 4.1) we now get

N X δk voln(Aδ) ≤ voln (Ak) , k=1 §4. The concept of measure 213 and using (16) we have N N ∞ X ε X X vol (A) − ε ≤ + vol (A ) ≤ ε + vol (A ) ≤ ε + vol (A ). n 2k n k n k n k k=1 k=1 k=1 This gives ∞ X voln(A) − 2ε ≤ voln(Ak). k=1 Since this inequality holds for all ε > 0, the inequality (13) immediately follows.