1. Outer Measures It Is Not Very Easy to Construct Non-Trivial Examples of Measures

Home , Outer measure

1. Outer Measures It is not very easy to construct non-trivial examples of measures. In this lecture we will give a general strategy to construct many examples of measures, and give a road map on how to construct the Lebesgue measure in Rn. Let X be a non-empty set, and let P(X) denote the family of all subsets of X. Definition 1.1. A function µ∗ : P(X) −→ [0, ∞] is an outer measure if i) µ∗(∅) = 0. ∈ P ⊂ ∗ ≤ ∗ ii) If A, B (X) and A B then∪ µ (A) µ∑(B). ∈ P ∈ N ∗ ∞ ≤ ∞ ∗ iii) If Aj (X), j , then µ ( j=1 Aj) j=1 µ (Aj). In what follows we have the following example in mind: Let ∪∞ F = {∅, [a, b); a, b ∈ Q}, then R = Xj,Xj ∈ F. j=1 Let V : F 7−→ [0, ∞) V (∅) = 0,V ([a, b)) = b − a In general, Proposition 1.1. Suppose there exists a family ∪∞ F = {∅,X1,X2,...,Xj ⊂ X} such that X = Xj. j=1 Let V : F 7−→ [0, ∞] be such that V (∅) = 0. For a subset E ⊂ X, define ∑∞ ∪∞ ∗ µ (E) = inf{ V (Cj); Cj ∈ F and E ⊂ Cj}. j=1 j=1 Then µ∗ is an outer measure on X. ∪ ∪ ∗ ∅ ∅ ⊂ ⊂ ∞ ∈ F ⊂ ∞ Proof. Obviously µ ( ) = V ( ) = 0. If A B, and if B j=1 Cj,Cj , then A j=1 Cj, ∗ ≤ ∗ and hence µ (A) µ (B). ∪ ∑ ⊂ ∈ N ∞ ∗ ≤ ∞ ∗ Finally, let Aj X,∑ j , and A = j=1 Aj. We want to show that µ (A) j=1 µ (Aj). ∞ ∗ ∞ We will assume that j=1 µ (Aj) < , otherwise the statement is obvious. In particular ∗ µ(Aj) < ∞, j ∈ N. From the definition of µ , given any ϵ > 0, there exist Cj,k ∈ F such that ∪∞ Aj ⊂ Cj,k and k=1 ∞ ∑ ϵ V (C ) < µ∗(A ) + . j,k j 2j k=1 1 2

So we conclude that ∞ ∞ ∞ ∪ ∪ ∑ ∑ ϵ ∑ A ⊂ C and V (C < (µ∗(A ) + ) < µ∗(A ) + ϵ. j j,k j,k j 2j j j=1 j,k=1 j,k j=1 Since ϵ > 0 is arbitrary, we obtain the result. Next we show a canonical way of constructing a measure from an outer measure. Deﬁnition 1.2. Let µ∗ be an outer measure on X. We say that A ⊂ X is µ∗ measurable if for every subset E ⊂ X (1.1) µ∗(A ∩ E) + µ∗(Ac ∩ E) = µ∗(E), where Ac = X \ A. ∗ Let Σ∗ denote the collection of all µ measurable subsets of X.

Theorem 1.1. Σ∗ is a σ-algebra and

µ :Σ∗ −→ [0, ∞] A 7−→ µ∗(A) ∗ is a measure on (X, Σ∗). Moreover, If A ⊂ X and µ (A) = 0, then A ∈ Σ∗ and µ(A) = 0. Proof. If A = ∅, then A ∩ E = ∅ and Ac ∩ E = E, so ∅ satisﬁes (1.1). Also, equation (1.1) is c ∈ c ∈ ∗ symmetric with respect to A and A , so if A Σ∗, then∪A Σ . ∈ ∈ N ∞ ∈ Next we want to show that if Aj Σ∗, j , then j=1 Aj Σ∗. We will prove this in four steps: ∪ ∈ N ∈ Step 1: If Aj Σ∗, j = 1,...,N, then j=1 Aj Σ∗. This is obvious for N = 1. Let A1,A2 ∈ Σ∗. We want to show that for any E ⊂ X, ∗ ∗ c ∗ µ ((A1 ∪ A2) ∩ E) + µ ((A1 ∪ A2) ∩ E) = µ (E). Since µ∗ is an outer measure and c E = (A1 ∪ A2) ∩ E) ∪ (A1 ∪ A2) ∩ E), it follows that ∗ ∗ ∗ c (1.2) µ (E) ≤ µ ((A1 ∪ A2) ∩ E) + µ ((A1 ∪ A2) ∩ E). We write ∪ ∩ ∪ ∩ c ∪ c ∩ A1 A2 = (A1 A2) (A1 A2) (A1 A2), and we recall that ∪ c c ∩ c (A1 A2) = a1 A2. Therefore, from (1.2), ∗ ≤ ∗ ∩ ∩ ∗ ∩ ∩ c ∗ ∩ c ∩ ∗ ∩ c ∩ c (1.3) µ (E) µ (E A1 A2) + µ (E A1 A2) + µ (E A1 A2) + µ (E A1 A2).

Since A1 ∈ Σ∗, ∗ ∩ ∩ ∗ ∩ c ∩ ∗ ∩ µ (E A1 A2) + µ (E A1 A2) = µ (A2 E) ∗ ∩ ∩ c ∗ ∩ c ∩ c ∗ c ∩ µ (E A1 A2) + µ (E A1 A2) = µ (A2 E). 3

But since A2 ∈ Σ∗, ∗ ∩ ∗ c ∩ ∗ (1.4) µ (A2 E) + µ (A2 E) = µ (E). Therefore ∗ ∗ ∗ c ∗ µ (E) ≤ µ ((A1 ∪ A2) ∩ E) + µ ((A1 ∪ A2) ∩ E) ≤ µ (E). ∪ N−1 This shows the result for N = 2, and by induction, if A ∈ Σ∗, the same argument shows ∪ j=1 j N ∈ that j=1 Aj Σ∗. Step 2: Let Aj ∈ Σ∗, j ∈ N, and suppose that Aj ∩ Ak = ∅, for j ≠ k. Then ∪∞ ∑∞ ∗ ∗ µ (E ∩ Aj) = µ (E ∩ Aj), for all E ⊂ X. j=1 j=1 ∈ ⊂ c Since A1 Σ∗, and A2 A1, we have ∗ ∩ ∪ ∗ ∩ ∪ ∩ ∗ ∩ ∪ ∩ c ∗ ∩ ∗ ∩ µ (E (A1 A2)) = µ (E (A1 A2) A1) + µ (E (A1 A2) A1) = µ (E A1) + µ (E A2). So, by induction, ∪N ∑N ∗ ∗ µ (E ∩ Aj) = µ (E ∩ Aj). j=1 j=1 But then, ∑N ∪N ∪∞ ∑∞ ∗ ∗ ∗ ∗ µ (E ∩ Aj) = µ (E ∩ Aj) ≤ µ (E ∩ Aj) ≤ µ (E ∩ Aj). j=1 j=1 j=1 j=1 Hence, by taking the limit as N → ∞ we prove the claim. ∪∞ Step 3: Let A ∈ Σ∗, j ∈ N, and suppose that A ∩ A = ∅, for j ≠ k. Then A ∈ Σ∗. ∪ j j k j=1 j N ∈ Since j=1 Aj Σ∗, ∪N ∪N ∪N ∪∞ ∗ ∗ ∗ c ∗ ∗ c µ (E) = µ (E ∩ Aj) + µ (E ∩ ( Aj) ) ≥ µ (E ∩ Aj) + µ (E ∩ ( Aj) ) = j=1 j=1 j=1 j=1 ∪∞ ∑N ∗ c ∗ µ (E ∩ ( Aj) ) + µ (E ∩ Aj). j=1 j=1 Taking the limit as N → ∞ and using step 2, we obtain ∪∞ ∪∞ ∗ ∗ ∗ c ∗ µ (E) ≥ µ (E ∩ Aj) + µ (E ∩ ( Aj) ) ≥ µ (E) j=1 j=1 The last inequality is due to the fact that µ∗ is an outer measure and that ∪∞ ∪∞ c E = (E ∩ Aj) ∪ (E ∩ ( Aj) ). j=1 j=1 ∪ ∈ ∈ N ∈ Step 4: If Aj Σ∗, j , then Aj Σ∗. ∪ \ n−1 ∩ ∅ ̸ ∈ To see that we let∪B1 = A∪1, and Bn = An j=1 Aj. So Bj Bk = if j = k, and Bj Σ∗. Hence, by Step 3, Aj = Bj ∈ Σ∗. 4

Step 5: µ is a measure on Σ∗.

∗ µ (∅) = 0 by deﬁnition. We need to show that if Aj ∈ Σ∗, j ∈ N, and Aj ∩ Ak = ∅ for j ≠ k, then ∪∞ ∑∞ ∗ ∗ µ ( Aj) = µ (Aj), j=1 j=1 but this follows from Step 2 by taking E = X..

∗ ∗ Step 6: If µ (B) = 0, then B ∈ Σ∗ and µ(B) = 0. Moreover, if C ⊂ B and µ (B) = 0, then C ∈ Σ∗. Since E = (E ∩ B) ∪ (E ∩ Bc), then µ∗(E ∩ Bc) ⊂ µ∗(E), and since E ∩ B ⊂ B, µ∗(E ∩ B) ≤ µ∗(B) = 0. Hence µ∗(E) ≤ µ∗(B ∩ E) + µ∗(E ∩ Bc) ≤ µ∗(E). ∗ ∗ Also, if C ⊂ B and µ (B) = 0, then µ (C) = 0, and hence C ∈ Σ∗.

2. The Lebesgue Measure in Rn A closed interval in Rn is a set of the form

(2.1) I = [a1, b1] × ... × [an, bn], −∞ < aj < bj < ∞, 1 ≤ j ≤ n. Similarly an open interval is a set of the form

I = (a1, b1) × ... × (an, bn), −∞ < aj < bj < ∞, 1 ≤ j ≤ n. The volume of either the open or closed interval is ∏n v(I) = (bj − aj). j=1 Let n F(R ) = {[a1, b1] × ... × [an, bn], aj, bj ∈ Q, aj < bj, 1 ≤ j ≤ n}. Given a set A ⊂ Rn we deﬁne its outer measure based in the family F(Rn) by ∑∞ ∪∞ ∗ n µ (A) = inf{ v(Ik): A ⊂ Ik,Ik ∈ F(R )}. k=1 k=1 As we proved above, µ∗ satisﬁes (OM1) If A ⊂ B ⊂ Rn, then µ∗(A) ≤ µ∗(B). n (OM2) If Ek ⊂ R , k = 1, 2, .... then ∪∞ ∑∞ ∗ ∗ µ ( Ek) ≤ µ (Ek) k=1 k=1 ∑ ∪ ⊂ Rn ∗ { ∞ ⊂ ∞ } Exercise 2.1. Show that if A , µ (A) = inf k=1 v(Ik): A k=1 Ik . In other words, one does not need Ik to have rational vertices. 5

2−k

x1 2−k m2−k

Figure 1. The lattice with end-points m2−k

n As above, we deﬁne Σ∗ as the family of subsets A ⊂ R that satisfy (1.1), and we know from Theorem 1.1 that µ :Σ∗ −→ [0, ∞] (2.2) A 7−→ µ∗(A)

is a measure. The σ-algebra Σ∗ is called the Lebesgue measurable sets, and µ is the Lebesgue measure. Our next step is to show that the Borel sets are contained in Σ∗. We begin by proving the following covering lemma

n Lemma 2.1. For any O ⊂ R nonempty open subset, there exist closed intervals Ij, j ∈ N, ◦ ◦ ∪ ∩ ∅ ̸ O ∞ such that Ij Ik= , j = k, and = j=1 Ij. n Proof. For k ∈ N, let Pk be the set of points of R whose coordinates are integral multiples of −k 2 . For each p = (p1, p2, ..., pn) ∈ Pk let −k n −k I(p, 2 ) = {x ∈ R : pj ≤ xj < pj + 2 , 1 ≤ j ≤ n}. −k Let Ωk = {I(p, 2 ): p ∈ Pk}. See ﬁg. 1 These sets satisfy the following properties: n −k P1) Fixed k, then for each point x ∈ R there exists a unique interval I(p, 2 ) in Ωk such that x ∈ I(p, 2−k). −k −k Just notice that any q ∈ R can be written as q = m2 + e1 where m ∈ N and |e1| < 2 .

−m −k −m P2) Let m > k, Z ∈ Pk and W ∈ Pm. Then either I(W, 2 ) ⊂ I(Z, 2 ) or I(W, 2 ) ∩ I(Z, 2−k) = ∅.

−m m−k To see this, we write W = (w1, . . . , wn) with wj = qj2 , and qj = pj2 + ej, with ej ∈ Z, ≤ m−k − qj pj ej −m ≤ pj 1 and ej 2 1. Hence we have wj = 2m = 2k + rm , and so wj + 2 2k + 2k . This means ∈ −k p1 pn −m ⊂ −k that W I(p, 2 ), with p = ( k ,..., k and I(W, 2 ) I(p, 2 ). But in view of P1, p is 2 2 ′ uniquely deﬁned, and so we either have p = Z, in which case I(W, 2−k ) ⊂ I(Z, 2−k), or p ≠ Z, ′ and in this case I(W, 2−k ) ∩ I(Z, 2−k) = ∅. 6

Now we can prove the lemma. Since O is open, then for every x ∈ O there exists I(p, 2−k) −k −k −k such that x ∈ I(p, 2 ) ⊂ O. Thus if F = {I(p, 2 ) ∈ Ωk : I(p, 2 ) ⊂ O}, ∪ O = I(p, 2−k). I(p,2−k)∈F

Now we just ﬁlter the collection, i.e. take those intervals in Ω1 and throw out all the intervals in Ωj, j > 1, which are contained in one of the intervals in Ω1 which are contained in O. From the remaining collection select the intervals in Ω2 which are contained in O and remove the ones in Ωj, j > 2, which are contained in any of these. Repeat∪ the process for n = 3, 4, ... The { ∈ N} O ∞ remaining intervals Ij, j are disjoint and satisfy = j=1 Ij. To ﬁnish the proof of the lemma just take the closures of these intervals.

We will then need to prove that intervals of the form (2.1) are Lebesgue sets. In view of Lemma 2.1 we have that open sets are Lebesgue measurable, and hence that the lebesgue sets contain the Borel sets. Before we prove that intervals are Lebesgue measurable, we need to establish that this outer measure has the following additional properties.

Theorem 2.1. Let µ∗ as deﬁned above, then ◦ ◦ (OM3) If Ij, j = 1, 2, ..., N be either closed intervals with Ik ∩ Ij = ∅ if k ≠ j, or pairwise disjoint open intervals, then ∪N ∑N ∗ µ ( Ij) = v(Ij). j=1 j=1 (OM4) For any E ⊂ Rn,

µ∗(E) = inf{µ∗(O): E ⊂ O and O is open }.

n ∗ ∗ (OM5) For any E ⊂ R there exists a Gδ set H such that E ⊂ H and µ (E) = µ (H).

(OM6) If r ∈ Rn,A ⊂ Rn, µ∗(A + r) = µ∗(A).

Proof. We first prove (OM3) for closed intervals,∪ and we begin by observing that the intervals ≤ ≤ N Ik, 1 k N form a cover of the set A = j=1 Ij. Therefore, by definition ∑N ∗ µ (A) ≤ v(Ij). j=1 ∑ ∗ ≥ N Now we want to prove that µ (A) j=1 v(Ij). Given ϵ > 0, it follows from the definition of infimum that there exist closed intervals Jm, m = 1, 2, ..., covering A, such that ∑∞ ∗ ∗ (2.3) µ (A) ≤ v(Jm) < µ (A)(1 + ϵ). m=1 ∏ m m × × m m n m − m Let us denote Jm = [a1 , b1 ] .... [an , bn ]. Then v(Jm) = i=1(bi ai ). Let δm > 0 ′ ′ m − m × × m − m and let Jm be the open interval Jm = (a1 δm, b1 + δm) .... (an δm, bn + δm). Then 7 ∏ ′ n m − m −∞ m m ∞ v(Jm) = i=1(bi ai + 2δm). Since < ai < bi < , i = 1, 2, ..., n, one can pick δm such that ′ (2.4) v(Jm) < (1 + ϵ)v(Jm). ∪ ⊂ ∞ ′ ′ Since A m=1 Jm,Jm is open, for all m, and A is compact, there exist a finite collection of ′ ≤ ≤ the Jm, 1 m M, covering A. ∪M ⊂ ′ A Jm. m=1 ′ ′ This implies that if χA and χJm are the characteristic functions of A and Jm respectively, ∑M ≤ ′ (2.5) χA χJm . m=1 Now we appeal to the fact that the volume of an interval is the Riemann integral of its characteristic function. Since A is the union of closed intervals Ik, 1 ≤ k ≤ N, with non- intersecting interiors, this implies that ∑m ∑M ≤ ′ (2.6) v(Ik) v(Jm). j=1 m=1 Putting together equations (2.3), (2.4), (2.5) and (2.6) we get ∑m ∑M ∑M ∑∞ ≤ ′ ≤ ≤ ≤ 2 ∗ v(Ik) v(Jm) (1 + ϵ)v(Jm) (1 + ϵ) v(Jm) (1 + ϵ) µ (A). j=1 m=1 m=1 m=1 ∑ ∗ ≥ N Since this holds for every ϵ, it shows that µ (A) j=1 v(Ij). This proves (OM3) when the intervals are closed. To prove (OM3) for pairwise disjoint open intervals, we first observe that ∪N ∪N Ij ⊂ Ij. j=1 j=1 By (OM1) and the first part of (OM3) ∪N ∪N ∑N ∗ ∗ (2.7) µ ( Ij) ≤ µ ( Ij) ≤ v(Ij). j=1 j=1 j=1 j j × × j j j − j On the other hand, if Ij = (a1, b1) ... (an, bn), then for 0 < δ < bk ak, k = 1, ..., n, ∗ j j − × × j j − ⊂ j = 1, ..., N, et Ij = [a1 + δ, b1 δ] ... [an + δ, bn δ] Ij, and since the Ij are paiwise ∗ disjoint, the Ij are disjoint. Therefore by P.1, the first case of P.2, and (2.7) ∑N ∪N ∪N ∑N ∗ ∗ ∗ ≤ ∗ ≤ v(Ij ) = µ ( Ij ) µ ( Ij) v(Ij). j=1 j=1 j=1 j=1 Since this holds for every δ, (OM3) also holds for open intervals. Since µ∗(E) ≤ µ∗(O) for every E ⊂ O, (OM4) is obvious if µ∗(E) = ∞. So we may assume that µ∗(E) < ∞. 8

For ϵ > 0, let Ik, k = 1, 2, ... be closed intervals such that ∪∞ ∑∞ ∗ E ⊂ Ik, and v(Ik) ≤ µ (E) + ϵ/2. k=1 k=1 ′ For each k, pick an open interval Ik such that ⊂ ′ ′ ≤ k+1 (2.8) Ik Ik and v(Ik) v(Ik) + ϵ/2 . ∪ O ∞ ′ O ⊂ O ∗ ≤ ∗ O Let = k=1 Ik. Then is open and E . By property (OM1), µ (E) µ ( ). By property (OM2) property (OM3) applied to a single interval, and (2.8) ∑∞ ∑ ∑ ∑ ∗ O ≤ ∗ ′ ′ ≤ k+1 ≤ ≤ ∗ µ ( ) µ (Ik) = v(Ik) (v(Ik) + ϵ/2 ) ϵ/2 + v(Ik) µ (E) + ϵ. k=1 k k k Thus for every ϵ > 0 there exists an open subset O such that E ⊂ O and that µ∗(E) ≤ µ∗(O) ≤ µ∗(E) + ϵ. This proves (OM4). Now we prove (OM5). If µ∗(E) = ∞, just take H = Rn. So we may assume that µ∗(E) < ∞. By property (OM4) there exists a sequence of open subsets Ok, k = 1, 2, ..., such that E ⊂ Ok and

∗ ∗ µ (Ok) ≤ µ (E) + 1/k. ∩ ∞ O ⊂ ⊂ O Let H k=1 k. Then H is a Gδ,E H k, k = 1, 2, ... Therefore, by property P.1 ∗ ∗ ∗ µ (E) ≤ µ (H) ≤ |Ok| ≤ µ (E) + 1/k, k = 1, 2, ...

Thus µ∗(H) = µ∗(E). This ends the proof of the theorem.

The following is an important generalization of Property (OM3):

n n Proposition 2.1. Let E1 ⊂ R and E2 ⊂ R be such that d(E1,E2) > 0, where

′ ′ d(E1,E2) = inf{|x − x | : x ∈ E1, x ∈ E2}.

∗ ∗ ∗ Then µ (E1 ∪ E2) = µ (E1) + µ (E2).

∗ ∗ Proof. If either µ (E1) = ∞ or µ (E2) = ∞, the statement is true. So we may assume that ∗ ∗ µ (E1) < ∞ and µ (E2) < ∞. From property (OM2) we know that

∗ ∗ ∗ µ (E1 ∪ E2) ≤ µ (E1) + µ (E2)

∗ So we only need to prove the opposite inequality. Since µ (E1 ∪ E2) < ∞, for any ϵ > 0 there exists a cover of E1 ∪ E2 by closed intervals {Ik, k ∈ N} such that ∑∞ ∗ v(Ik) ≤ µ (E1 ∪ E2) + ϵ. k=1 9

We assume that for every interval Ik in this cover Ik ∩ (E1 ∪ E2) ≠ ∅, otherwise we can just remove it from the collection. We then divide this family into three parts:

{Ik, k ∈ N} = F1 ∪ F2 ∪ F3, where

Ik ∈ F1 if and only if Ik ∩ E1 ≠ ∅,Ik ∩ E2 = ∅,

Ik ∈ F2 if and only if Ik ∩ E1 = ∅,Ik ∩ E2 ≠ ∅,

Ik ∈ F3 if and only if Ik ∩ E1 ≠ ∅,Ik ∩ E2 ≠ ∅.

Each interval in Ik ∈ F3 can be divided into a ﬁnite number of intervals Ij,k with diameter less than d(E1,E2). And we have ∑ (2.9) v(Ik) = v(Ij,k).

But each interval Ij,k falls into the family F1 or F2, or it does not intersect E1 ∪E2. We throw out the intervals that do not intersect E1 ∪ E2. Then we replace each of the intervals Ik ∈ F3 by the { } F F { ′ } F ∪F intervals in Ij,k which are in 1 or 2. We have now a family of closed intervals Ik = 1 2 which cover E1 ∪ E2, and consists of the intervals {Ik} which are in either F1 or F2, and the intervals in {Ijk} which are also in F1 ∪ F2. In view of (2.9) ∑∞ ′ ≤ | ∪ | (2.10) v(Ik) E1 E2 e + ϵ. k=1 Therefore, by deﬁnition of outer measure ∑ ∑ ∑ ∗ ∗ ≤ ′ ′ ′ ≤ ∗ ∪ µ (E1) + µ (E2) v(Ik) + v(Ik) = v(Ik) µ (E1 E2) + ϵ. ′ ∈F ′ ∈F Ik 1 Ik 2 k Since ϵ > 0 is arbitrary, this implies that ∗ ∗ ∗ µ (E1) + µ (E2) ≤ µ (E1cupE2).

Now we are ready to show that Intervals of the form (2.1) are Lebesgue sets.

Theorem 2.2. Let I = [a1, b1] × [a2, b2] × ... × [an, bn], aj, bj ∈ R, aj < bj, be a cosed interval. If E ⊂ Rn, then µ∗(E) = µ∗(E ∩ I) + µ∗(E ∩ Ic).

Proof. We know that in general µ∗(E) ≤ µ∗(E ∩ I) + µ∗(E ∩ Ic), and so need to show that µ∗(E) ≥ µ∗(E ∩ I) + µ∗(E ∩ Ic). We will assume µ∗(E) < ∞, otherwise the statement is obvious. The ﬁrst step is to prove this result when E is an interval with sides parallel to the axes, i.e.

E = [α1, β1] × [α2, β2] × ... [αn, βn]. 10

The key point is that one can write E = (E ∩ I) ∪ (E ∩ Ic) and E ∩ I is an interval and (2.11) ∩ c ∪M E I = m=1Im, where Im are non-overlapping intervals, see Fig. 2. In view of property OM3, ∑M ∗ ∗ ∗ ∗ ∗ c (2.12) µ (E) = µ (E ∩ I) + µ (Im) = µ (E ∩ I) + µ (E ∩ I ). m=1 The next step is to prove the result when E is an open subset. In this case ∪∞ E = j=1Ij, with Ij non-overlapping closed intervals,

and by applying (2.11) to each interval Ij,, we can write ∩ ∪ ∪N Ij = (I Ij) ( k=1Ljk), with Ljk intervals with disjoint interiors, see Fig. 2. c Notice that the intervals Ljk cover E ∩ I . Hence if we renumber {Ljk} = {Lj}, then for any N ∈ N, ∩ ∪∞ ∩ ∩ c ∪∞ E I = j=1(Ij I), and E I = j=1Lj, c where I ∩ Ij and Lj are non overlapping intervals. Since E = (E ∩ I) ∪ (E ∩ I ), we deduce that for any N ∈ N, ( ) ( ) ⊃ ∪N ∩ ∪ ∪N E j=1(I Ij) j=1Lj, and in view of (OM3) we have ∑N ∑N ∗ ∗ ∗ µ (E) ≥ µ (I ∩ Ij) + µ (Lj),N ∈ N. j=1 j=1 Therefore, ∑∞ ∑∞ ∗ ∗ ∗ ∗ ∗ c µ (E) ≥ µ (I ∩ Ij) + µ (Lj) ≥ µ (E ∩ I) + µ (E ∩ I ). j=1 j=1 This proves the property for the case where E is open. For a general set E, we use property (OM4), which says that for any ϵ > 0 there exists an open subset O ⊂ Rn such that E ⊂ O, and µ∗(O) < µ∗(E) + ϵ. But we have just shown that, since O is open, µ∗(O) = µ∗(O ∩ I) + µ∗(O ∩ Ic), and since O ⊃ E, µ∗(O ∩ I) ≥ µ∗(E ∩ I) and µ∗(O ∩ Ic) ≥ µ∗(E ∩ Ic), and therefore µ∗(E) + ϵ ≥ µ∗(E ∩ I) + µ∗(E ∩ Ic), ϵ > 0. Since this is true for every ϵ > 0, the result follows. Finally we will show that the Lebesgue integral is in fact a generalization of the Riemann integral. This is expressed in 11

Lj1 Lj2

Lj4 I

Lj3

Figure 2. The intersection of two intervals

Theorem 2.3. Let f :[a, b] −→ R be Riemann integrable. Then f is Lebesgue measurable, and ∫ ∫ b f(x) dx = f dµ. a [a,b] Proof. First of all, by working with the positive and the negative parts of f, we may assume that f ≥ 0. Let Pn, n ∈ N, be a family of partitions of [a, b] with Pn ⊂ Pn+1, and assume |Pn| → 0. Let’s say

Pn = {a = x1n < x2n < . . . x(k−1)n < xkn = b},Ijn = [x(j−1)n, xjn],

Let mjn = infIjn f, Mjn = supIjn f, and let ∑k ∑k

Ln(x) = mjnχIjn ,Un(x) = MjnχIjn . j=1 j=1

We know that Ln and Un are Lebesgue measurable, and that

Ln(x) ≤ f(x) ≤ Un(x) (2.13) Ln(x) ≤ Ln+1(x) and Un+1(x) ≤ Un(x). Since f is bounded, these sequences converge, so we deﬁne

L(x) = lim Ln(x), and U(x) = lim Un(x), n→∞ n→∞ and in view of (2.13) (2.14) L(x) ≤ f(x) ≤ U(x). We also know, by deﬁnition of the integral that ∫ ∑k ∫ ∑k Ln dµ = mjn|Ijn| = L(P, f), and Un dµ = Mjn|Ijn| = U(P, f). j=1 j=1 Again, since f is bounded, the dominated convergence theorem guarantees that ∫ ∫ ∫ ∫

L dµ = lim Ln dµ and U dµ = lim Un dµ. n→∞ n→∞ Since f is Riemann integrable, ∫ ∫ ∫ ∫

L dµ = lim Ln dµ = lim Un dµ = U dµ. n→∞ n→∞ 12 ∫ Since U ≥ L, and (U − L) dµ = 0, we deduce that U − L except on a set of measure zero. But in view of (2.14), we conclude that U = f = L a.e., and therefore, if [α, β) ⊂ R, then U −1([α, β)) \ f −1([α, β) and f −1([α, β)) \ U −1([α, β) are sets of Lebesgue measure equal to zero and hence both U −1([α, β)) \ f −1([α, β)) and f −1([α, β)) \ U −1([α, β)) are measurable. But in general if A and B are sets, and if A, A\B and B \A are measurable sets, then B is measurable. Indeed, since A and A \ B are measurable, we write A = (A ∩ B) ∪ (A \ B), and deduce that A ∩ B is measurable. Since B \ A is measurable, and B = (B ∩ A) ∪ (B \ A), then B is measurable. If we apply this for A = U −1([α, β)) and B = f −1([α, β)) we deduce that f −1([α, β)) is measurable, and since f is bounded, f is measurable and ∫ ∫ ∫ b f dµ = U dµ = f(x) dx. b