6. the Lebesgue Measure

6. The Lebesgue measure

In this section we apply various results from the previous sections to a very basic example: the Lebesgue measure on Rn. Notations. We ﬁx an integer n ≥ 1. In Section 21 we introduced 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 defined its n-dimesnional volume by n Y voln(A) = (bk − ak). k=1 We also defined voln(∅) = 0. By Theorem 4.2, we know that voln is a finite measure on Jn.

Definitions. The maximal outer extension of voln is called the n-dimensional ∗ outer Lebesgue measure, and is denoted by λn. ∗ n The λn-measurable sets in R will be called n-Lebesgue measurable. The σ- n n ∗ algebra mλ∗ ( ) will be denoted simply by m( ). The measure λn n is n R R m(R ) simply denoted by λn, and is called the n-dimensional Lebesgue measure. Although ∗ this notation may appear to be confusing, it turns out (see Proposition 5.3) that λn is indeed the maximal outer extension of λn. In the case when n = 1, the subscript will be ommitted. We know (see Section 21) that n S(Jn) = Σ(Jn) = Bor(R ). n Using the fact that the semiring Jn is σ-total in R , by the deﬁnition of the outer Lebesgue measure, we have ∞ ∞ ∗ X ∞ [ n (1) λn(A) = inf voln(Bk):(Bk)k=1 ⊂ Jn, Bk ⊃ A , ∀ A ⊂ R k=1 k=1 Using Corollary 5.2, we have the equality n m(R ) = Bor(Rn), n n where Bor( ) is the completion of Bor( ) with respect to the measure λn n . R R Bor(R ) This means that a subset A ⊂ Rn is Lebesgue measurable, if and only if there exists a Borel set B and a neglijeable set N such that A = B ∪ N. (The fact N is ∗ neglijeable means that λn(N) = 0, and is equivalent to the existence of a Borel set C ⊃ N with λn(C) = 0.) n Exercise 1. Let A = [a1, b1) × · · · × [an, bn) be a half-open box in R . Assume A 6= ∅ (which means that a1 < b1, . . . , an < bn). Consider the open box Int(A) and the closed box A, which are given by

Int(A) = (a1, b1) × · · · × (an, bn) and A = [a1, b1] × · · · × [an, bn].

200 §6. The Lebesgue measure 201

Prove the equalities λn Int(A) = λn A = voln(A).

n Remarks 6.1. If D ⊂ R is a non-empty open set, then λn(D) > 0. This is a consequence of the above exercise, combined with the fact that D contains at least one non-empty open box. The Lebesgue measure of a countable subset C ⊂ Rn is zero. Using σ-additivity, it suﬃces to prove this only in the case of singletons C = {x}. If we write x in coordinates x = (x1, . . . , xn), and if we consider half-open boxes of the form

Jε = [x1, x1 + ε) × · · · × [xn, xn + ε), then the obvious inclusion {x} ⊂ Jε will force n 0 ≤ λn {x} ≤ λn(Jε) = ε , so taking the limit as ε → 0, we indeed get λn {x} = 0. The (outer) Lebesgue measure is completely determined by its values on open sets. More explicitly, one has the following result.

Proposition 6.1. Let n ≥ 1 be an integer. For every subset A ⊂ Rn one has: ∗ n (2) λn(A) = inf{λn(D): D open subset of R , with D ⊃ A}. Proof. Throughout the proof the set A will be ﬁxed. Let us denote, for simplicity, the right hand side of (2) by ν(A). First of all, since every open set is ∗ Lebesgue measurable (being Borel), we have λn(D) = λn(D), for all open sets D, ∗ so by the monotonicity of λn, we get the inequality ∗ λn(A) ≤ ν(A). ∗ We now prove the inequality λn(A) ≥ ν(A). Fix for the moment some ε > 0, and ∞ S∞ use (1). to get the existence of a sequence (Bk)k=1 ⊂ Jn, such that k=1 Bk ⊃ A, and ∞ X ∗ voln(Bk) < λn(A) + ε. k=1 For every k ≥ 1, we write

(k) (k) (k) (k) Bk = [a1 , b1 ) × · · · × [an , bn ), Qn (k) (k) so that voln(Bk) = j=1(b1 − aj ). Using the obvious continuity of the map n Y (k) (k) R 3 t 7−→ (b1 − aj − t) ∈ R, j=1

(k) (k) (k) (k) we can ﬁnd, for each k ≥ 1 some numbers c1 < a1 , ... , cn < an , with n n Y (k) (k) ε Y (k) (k) (3) (b − c ) < + (b − a ). 1 j 2k 1 j j=1 j=1 Notice that, if we deﬁne the half-open boxes

(k) (k) (k) (k) Ek = [c1 , b1 ) × · · · × [cn , bn ), 202 CHAPTER III: MEASURE THEORY then for every k ≥ 1, we clearly have Bk ⊂ Int(Ek), and by Exercise 1, combined with (3), we also have the inequality ε λ Int(E ) = vol (E ) < + vol (B ). n k n k 2k n k Summing up we then get ∞ ∞ ∞ X X ε X (4) λ Int(E ) < + vol (B ) = ε + vol (B ) < 2ε + λ∗ (A). n k 2k n k n k n k=1 k=1 k=1 Now we observe that by σ-sub-additivity we have ∞ ∞ [ X λn Int(Ek) ≤ λn Int(Ek) , k=1 k=1 S∞ so if we define the open set D = k=1 Int(Ek), then using (4) we get ∗ (5) λn(D) < 2ε + λn(A). It is clear that we have the inclusions ∞ ∞ [ [ A ⊂ Bk ⊂ Int(Ek) = D, k=1 k=1 so by the definition of ν(A), combined with (5), we finally get ∗ ν(A) ≤ λn(D) < 2ε + λn(A). ∗ Up to this moment ε > 0 was fixed. Since the inequality ν(A) < 2ε + λn(A) holds ∗ for any ε > 0 however, we finally get the desired inequality ν(A) ≤ λn(A). The Lebesgue measure can also be recovered from its values on compact sets. Proposition 6.2. Let n ≥ 1 be an integer. For every Lebesgue measurable subset A ⊂ Rn one has: n (6) λn(A) = sup{λn(K): K compact subset of R , with K ⊂ A}. Proof. Let us denote, for simplicity, the right hand side of (6) by µ(A). First of all, by the mononoticity we clearly have the inequality

λn(A) ≥ µ(A).

To prove the inequality λn(A) ≤ µ(A), we shall ﬁrst use a reduction to the bounded case. For each integer k ≥ 1, we deﬁne the compact box

Bk = [−k, k] × · · · × [−k, k]. S∞ n Notice that we have B1 ⊂ B2 ⊂ ... , with k=1 Bk = R . We then have

B1 ∩ A ⊂ B2 ∩ A ⊂ ..., S∞ with k=1(Bk ∩ A) = A, so using the Continuity Lemma 4.1, we have (7) λn(A) = lim λn(Bk ∩ A) = sup λn(Bk ∩ A): k ≥ 1 . k→∞ Fix for the moment some ε > 0, and use the (7) to ﬁnd some k ≥ 1, such that λn(A) ≤ λn(Bk ∩ A) + ε. Apply Proposition 6.1 to the set Bk r A, to ﬁnd an open set D, with D ⊃ Bk r A, and λn(Bk r A) ≥ λn(D) − ε. On the one hand, we have λn(Bk) = λn(Bk ∩ A) + λn(Bk r A) ≥ λn(Bk ∩ A) + λn(D) − ε ≥ (8) ≥ λn(Bk ∩ A) + λn(Bk ∩ D) − ε. §6. The Lebesgue measure 203

On the other hand, we have

λn(Bk) = λn(Bk r D) + λn(Bk ∩ D), so using (8) we get the inequality

λn(Bk r D) + λn(Bk ∩ D) ≥ λn(Bk ∩ A) + λn(Bk ∩ D) − ε, and since all numbers involved in the above inequality are ﬁnite, we conclude that

λn(Bk r D) ≥ λn(Bk ∩ A) − ε ≥ λn(A) − 2ε. Obviously the set K = Bk r D is compact, with K ⊂ Bk ∩ A ⊂ A, so we have µ(A) ≥ λn(K), hence we get the inequality

µ(A) ≥ λn(A) − 2ε.

Since this is true for all ε > 0, the desired inequality µ(A) ≥ λn(A) follows. Corollary 6.1. For a set A ⊂ Rn, the following are equivalent: (i) A is Lebesgue measurable; ∞ (ii) there exists a neglijeable set N and a sequence of (Kj)j=1 of compact subsets of Rn, such that ∞ [ A = N ∪ Kj. j=1 Proof. (i) ⇒ (ii). Start by using the boxes

Bk = [−k, k] × · · · × [−k, k] S∞ n S∞ which have the property that k=1 Bj = R , so we get A = k=1(Bk ∩ A). Fix k ∞ for the moment k. Apply Proposition 6.2. to find a sequence (Cr )r=1 of compact k subsets of Bk ∩ A, such that limr→∞ λn(Cr ) = λn(Bk ∩ A). Consider the countable k ∞ ∞ family (Cr )k,r=1 of compact sets, and enumerate it as a sequence (Kj)j=1, so that we have ∞ ∞ ∞ [ [ [ k Kj = Cr . j=1 k=1 r=1 S∞ k If we define, for each k ≥ 1, the sets Ek = r=1 Cr ⊂ Bk∩A and Nk = (Bk∩A)rEk, k then, because of the inclusion Cr ⊂ Ek ⊂ Bk ∩ A, we have the inequalities k (9) 0 ≤ λn(Nk) = λn(Bk ∩ A) − λn(Ek) ≤ λn(Bk ∩ A) − λn(Cr ), ∀ r ≥ 1. Using the fact that k lim λn(Cr ) = λn(Bk ∩ A) ≤ λn(Bk) < ∞, r→∞ the inequalities (9) force λn(Nk) = 0, ∀ k ≥ 1. Now if we define the set N = S∞ A r j=1 Kj , we have ∞ ∞ ∞ ∞ [ [ [ [ N = (Bk ∩ A) r Kj = (Bk ∩ A) r Ep ⊂ k=1 j=1 k=1 p=1 ∞ ∞ [ [ ⊂ (Bk ∩ A) r Ek = Nk, k=1 k=1 which proves that λn(N) = 0. The implication (ii) ⇒ (i) is trivial. 204 CHAPTER III: MEASURE THEORY

Proposition 6.2 does not hold if A ⊂ Rn is non-measurable. In fact the equality ∗ (6), with λn replaced by λn, essentially forces A to be measurable, as shown by the following. n ∗ Exercise 2. Let A ⊂ R be am arbitrary subset, with λn(A) < ∞. Prove that the following are equivalent: (i) A is Lebesgue measurable; ∗ n (ii) λn(A) = sup{λn(K): K compact subset of R , with K ⊂ A}. Propositions 6.1 and 6.2 are regularity properties. The following terminology is useful: Definitions. Suppose A is a σ-algebra on X, and µ is a measure on A. Sup- pose we have a sub-collection F ⊂ A. (i) We say that µ is regular from below, with respect to F, if µ(A) = sup µ(F ): F ⊂ A, F ∈ F . (ii) We say that µ is regular from above, with respect to F, if µ(A) = inf µ(F ): F ⊃ A, F ∈ F . With this terminology, Proposition 6.1 gives the fact that the Lebesgue measure is regular from above with respect to open sets, while Proposition 6.2 gives the fact that the Lebesgue measure is regular from below with respect to compact sets.

Exercise 3. For a subset A ⊂ Rn, prove that the following are equivalent: (i) A is Lebesgue measurable; ∞ (ii) There exist a sequence of compact sets (Kj)j=1, and a dequence of open ∞ S∞ T∞ sets (Dj)j=1, such that j=1 Kj ⊂ A ⊂ j=1 Dj, and the difference T∞ S∞ j=1 Dj r j=1 Kj is neglijeable. Hint: For the implication (i) ⇒ (ii) analyze first the case when λ∗(A) < ∞. Then write A as a countable union of sets of finite outer measure. In the one-dimensional case n = 1, the Lebesgue measure of open sets can be computed with the aid of the following result.

Proposition 6.3. For every open set D ⊂ R, there exists a countable (or S ﬁnite) pair-wise disjoint collection {Ji}i∈I of open intervals with D = i∈I Ji. Proof. For every point x ∈ D, we deﬁne

ax = inf{a < x :(a, x) ⊂ D} and bx = sup{b > x :(x, b) ⊂ D}. (The fact that D is open guarantees the fact that both sets above are non-empty.) It is clear that, for every x ∈ D, the open interval Jx = (ax, bx) is contained in D, so S we have the equality D = x∈D Jx. The problem at this point is the fact that the collection {Jx}x∈D is not pair-wise disjoint. What we need to ﬁnd is a countable (or ﬁnite) subset X ⊂ D, such that the sub-collection {Jx}x∈X is pair-wise disjoint, S and we still have D = x∈X Jx. One way to do this is based on the following Claim: For two points x, y ∈ D, the following are equivalent: (i) x ∈ Jy; (ii) Jx ⊃ Jy; (ii) Jx ∩ Jy 6= ∅; (iii) Jx = Jy. §6. The Lebesgue measure 205

To prove the implication (i) ⇒ (ii) we observe that if x ∈ Jy, then ay < x < by, so we have (ay, x) ⊂ D and (x, by) ⊂ D, which means that ax ≤ ay and bx ≥ by, therefore we have the inclusion Jx = (ax, bx) ⊃ (ay, by) = Jy. The implication (ii) ⇒ (iii) is trivial. To prove (iii) ⇒ (iv), assume Jx ∩ Jy 6= ∅, and pick a point z ∈ Jx ∩ Jy. Using the implication (i) ⇒ (ii) we have the inclusions Jz ⊃ Jx and Jz ⊃ Jy. In particular we have x ∈ Jz, so again using the inplication (i) ⇒ (ii) we get Jx ⊃ Jz, which means that we have in fact the equality Jx = Jz. Likewise we have the equality Jy = Jz, so (iv) follows. The implication (iv) ⇒ (i) is trivial. Going back to the proof of the Proposition, we now see that, using the fact that any open interval contains a rational number, if we put X0 = D ∩ Q, then for any y ∈ D, there exists x ∈ X0, such that Jx = Jy. This gives the equality D = S J , this time with the indexing set X countable. Finally, if we equip x∈X0 x 0 the set X0 with the equivalence relation

x ∼ y ⇐⇒ Jx = Jy, and we choose X ⊂ X0 to the a list of all equivalence classes. This means that, for every y ∈ X0, there exists a unique x ∈ X with Jx = Jy. It is clear now that we S 0 0 0 still have D = x∈X Jx, but now if x, x ∈ X are such that x 6= x , then x 6∼ x , so we have Jx 6= Jx0 , which by the Claim gives Jx ∩ Jx0 = ∅. Comments. When we want to compute the Lebesgue measure of an open set S D ⊂ R, we should ﬁrst try to write D = i∈I Ji with (Ji)i∈I a countable (or ﬁnite) pair-wise collection of open intervals. If we succeed, then we would have X λ(D) = λ(Ji). i∈I For intervals (open or not) the Lebesgue measure is the same as the length. There are instances when we can manage only to write a given open set D as a S∞ union D = k=1 Jk, with the J’s not necessarily disjoint. In that case we can only get the estimate ∞ X λ(D) ≤ λ(Jk). k=1

Example 6.1. Consider the ternary Cantor set K3 ⊂ [0, 1], discussed in III.3. ∞ We know (see Remarks 3.5) that one can ﬁnd a pair-wise sequence (Dn)n=0 of open S∞ subsets of (0, 1) such that K3 = [0, 1] r n=0 Dn, and such that, for each n ≥ 0, n n+1 the open set Dn is a disjoint union of 2 intervals of length 1/3 . In particular, n n+1 this means that λ(Dn) = 2 /3 , so ∞ ∞ ∞ [ X X 2n λ(K ) = λ[0, 1] − λ D = 1 − λ(D ) = 1 − = 0. 3 n n 3n+1 n=0 n=0 n=0

What is interesting here (see Remarks 3.5) is the fact that card K3 = c. Remark 6.2. An interesting consequence of the above computation is the fact that all subsets of K3 are Lebesgue measurable, i.e. one has the inclusion P(K3) ⊂ m(R). This gives the inequality

card K3 c card m(R) ≥ card P(K3) = 2 = 2 . Since we also have m(R) ⊂ P(R), we get card c card m(R) ≤ card P(R) = 2 R = 2 , 206 CHAPTER III: MEASURE THEORY so using the Cantor-Bernstein Theorem we get the equality c card m(R) = 2 . We also know (see Corollary 2.5) that card Bor(R) = c. As a consequence of this diﬀerence in cardinalities, one gets the fact that we have a strict inclusion (10) Bor(R) ( m(R). Later on we shall construct (more or less) explicitly a Lebesgue measurable set which is not Borel.

Exercise 4. The strict inclusion (10) holds also if R is replaced with Rn, with n ≥ 2. In this case, instead of using Cantor sets, one can proceed as fol- n−1 lows. Consider the set S = R × {0}. Prove that λn(S) = 0. Conclude that c card m(Rn) = 2 . One key feature of the Lebesgue (outer) measure is the translation invariance property, described in the following result. To formulate it we introduce the following notation. For an integer n ≥ 1, a point x ∈ Rn, and a subset A ⊂ Rn, we deﬁne the set A + x = {a + x : a ∈ A}. n n Remark that the map Θx : R 3 a 7−→ a + x ∈ R is a homeomorphism. In −1 particular, both Θx and Θx = Θ−x are Borel measurable, which means that, for a set A ⊂ Rn, one has the equivalence n n A ∈ Bor(R ) ⇐⇒ A + x ∈ Bor(R ). Proposition 6.4. Let n ≥ 1 be an integer. For any set A ⊂ Rn one has the equality ∗ ∗ λn(A + x) = λn(A).

Proof. Fix A and x. First remark that, for every half-open box B ∈ Jn, its translation B + x is again a half-open box, and we have the equality

voln(B + x) = voln(B). ∞ Fix for the moment ε > 0, and choose a sequence (Bk)k=1 ⊂ Jn, such that A ⊂ S∞ k=1 Bk, and ∞ X ∗ voln(Bk) ≤ λn(A) + ε. k=1 S∞ Then, using the obvious inclusion A + x ⊂ k=1(Bk + x), by the remark made at the begining of the proof, combined with the monotonicity of the outer Lebesgue measure, we have ∞ ∞ ∗ ∗ [ X ∗ λn(A + x) ≤ λn (Bk + x) ≤ λn(Bk + x) = k=1 k=1 ∞ ∞ X X ∗ = voln(Bk + x) = voln(Bk) ≤ λn(A) + ε. k=1 k=1 ∗ ∗ Since the inequality λn(A + x) ≤ λn(A) + ε holds for all ε > 0, we get ∗ ∗ λn(A + x) ≤ λn(A). §6. The Lebesgue measure 207

The other inequality follows from the above one applied to the set A + x and the translation by −x.

Corollary 6.2. For a subset A ⊂ Rn, one has the equivalence n n A ∈ m(R ) ⇐⇒ A + x ∈ m(R ). Proof. Write A = B ∪ N, with B Borel, and N neglijeable. Then we have A + x = (B + x) ∪ (N + x). The set B + x is Borel. By the above result we have ∗ ∗ λn(N + x) = λn(N) = 0, i.e. N + x is neglijeable. Therefore A + x is Lebesgue measurable. As we have seen, the fact that there exist Lebesgue measurable sets that are c not Borel is explained by the diﬀerence in cardinalities. Since card m(Rn) = 2 = card P(Rn), it is legitimate to ask whether the inclusion m(Rn) ⊂ P(Rn) is strict. In other words, do there exist sets that are not Lebesgue measurable? The answer is aﬃrmative, as discussed in the following.

Example 6.2. Equipp R with the equivalence relation x ∼ y ⇐⇒ x − y ∈ Q. Denote by R/Q the quotient space (this is in fact the quotient group of (R, +) with respect to the subgroup Q), and denote by π : R → R/Q the quotient map. Since for every x ∈ R, one can ﬁnd some y ∼ x, with y ∈ [0, 1), it follows that the map

π [0,1) : [0, 1) → R/Q is surjective. Choose then a map φ : R/Q → [0, 1), such that φ ◦ π = Id, and put E = φ(R/Q). The set E is a complete set of representatives for the equivalence relation ∼. In other words, E ⊂ [0, 1) has the property that, for every x ∈ R, there exists exactly one element y ∈ E, with x ∼ y. In particular, the S collection of sets (E + q)q∈ is pair-wise disjoint, and satisﬁes (E + q) = . Q q∈Q R Using σ-sub-additivity, we get

X ∗ ∞ = λ(R) ≤ λ (E + q). q∈Q Since (by Proposition 6.5) we have λ∗(E + q) = λ∗(E), the above inequality forces λ∗(E) > 0. Claim: The set E is not Lebesgue measurable Assume E is Lebesgue measurable. If we deﬁne the set X = Q ∩ [0, 1), then the sets E + q, q ∈ X are pair-wirse disjoint. On the one hand, the measurability of E, combined with the Corollary 6.2 would imply the measurability of the set S S = q∈X (E + q). On the other hand, the equalities λ(E + q) = λ(E) > 0 will force λ(S) = ∞. But this is impossible, since we obviously have S ⊂ [0, 2), which forces λ(S) ≤ 2.

Exercise 5. Let E ∈ m(Rn). Prove that the map n R 3 x 7−→ λ E ∪ (E + x) ∈ [0, ∞] is continuous. Hint: Analyze ﬁrst the case when E is compact. In this particular case, show that for every n x0 ∈ R and every open set D ⊃ E ∪ (E + x0), there exists some neighborhood V of x0, such that D ⊃ E ∪ (E + x), ∀ x ∈ V. 208 CHAPTER III: MEASURE THEORY

Use then regularity from above, combined with the inequality1 n |λ(A) − λ(B)| ≤ λ(A4B), for all A, B ∈ m(R ), with λ(A), λ(B) < ∞. In the general case, use regularity from below. (The case λ(E) = ∞ is trivial.)

n Exercise 6. Let E ∈ m(R ), be such that λn(E) > 0. Prove that the set E − E = {x − y : x, y ∈ E} is a neighborhood of 0. ∞ n Hint: Assume the contrary, which means that there exists a sequence (xp)p=1 ⊂ R r (E − E), with limp→∞ xp = 0. This will force E ∩ (E + xp) = ∅, ∀ p ≥ 1. Use the preceding Exercise to get a contradiction. We are now in position to construct a Lebesgue measurable set which is not Borel.

Example 6.3. In Section 3 we discussed the compact space T = {0, 1}ℵ0 and the maps ∞ X αn φ : T 3 (α )∞ 7−→ (r − 1) ∈ [0, 1]. r n n=1 rn n=1 For each r ≥ 2 the map φr : T → [0, 1] is continuous so the set Kr = φr(T ) is compact. We have K2 = [0, 1], and K3 is the ternary Cantor set. We also know (see Theorem 3.5) that, for a set A ⊂ T , one has the equivalence

(11) A ∈ Bor(T ) ⇐⇒ φr(A) ∈ Bor(Kr). Choose now a set E ⊂ [0, 1] which is not Lebesgue measurable. In particular, E is not Borel, so E 6∈ Bor([0, 1]). Since φ2 : T → [0, 1] is surjective, by (11) it follows −1 that the set A = φ2 (E) is not in Bor(T ). Again, by (11) it follows that the set S = φ3(A) is not in Bor(K3). Since

Bor(K3) = Bor( ) R K3 this gives S 6∈ Bor(R). Notice however that since S ⊂ K3, it follows that S is Lebesgue measurable.

Comment. When one wants to prove that a Lebesgue measurable set M ⊂ R has positive measure, a suﬃcient condition for this property is that Int(M) 6= ∅ (see Remark 6.1). It turns out however that this condition is not always necessary, as seen from the following: Exercise 7. Start with an arbitrary inerval [0, 1], and list all rational numbers ∞ in [0, 1] as a sequence Q ∩ [0, 1] = {xn}n=1. Fix some ε > 0, and consider the open set ∞ [ ε ε D = x − , x + . n 2n+1 n 2n+1 n=1 Consider the compact set K = [0, 1] r D. (i) Prove that λ(D) ≤ ε. (ii) Prove that λ(K) ≥ 1 − ε. (iii) Prove that Int(K) = ∅.

Hint: For (iii) use the fact that K ∩ Q = ∅.

1 This inequality holds for any additive map deﬁned on a ring. §6. The Lebesgue measure 209

Exercise 8*. Prove that, for every non-empty open set D ⊂ R, and any two positive numbers α, β with α + β < λ(D), there exist compact sets A, B ⊂ D, with λ(A) > α, λ(B) > β, such that A ∩ B = ∅ and (A ∪ B) ∩ Q = ∅. ∞ Hint: Write D as a union of a pair-wise disjoint sequence (Jn)n=1 of open intervals, so that P∞ ∞ ∞ λ(D) = n=1 λ(Jn). Find then two sequences (αn)n=1 and (βn)n=1 of positive numbers, such P∞ P∞ that n=1 αn > α, n=1 βn > β, and αn + βn < λ(Jn), for all n ≥ 1. This reduces essentially the problem to the case when D is an open interval, for which one can use the construction outlined in Exercise 7.

Exercise 9*. Construct o Borel set A ⊂ R, such that, for every open interval I ⊂ R one has λ(I ∩ A) > 0 and λ(I r A) > 0. ∞ Hints: List all open intervals with rational endpoints as a sequence (In)n=1. Start (use exercise 8) oﬀ by choosing two compact sets A1,B1 ⊂ I1, with A1 ∩ B1 = ∅,(A1 ∪ B1) ∩ Q = ∅, ∞ ∞ and λ(A1), λ(B1) > 0. Use Exercise 5 to construct two sequences (An)n=1 and (Bn)n=1 of compact sets, such that, for all n ≥ 1 we have: (i) An ∩ Bn = ∅; (ii) (An ∪ Bn) ∩ Q = ∅; Sn S∞ (iii) λ(An), λ(Bn) > 0; (iv) An+1 ∪ Bn+1 ⊂ In+1 r k=1(Ak ∪ Bk) . Put A = n=1 An and S∞ B = n=1 Bn. Notice that A ∩ B = ∅, λ(A), λ(B) > 0, and λ(A ∩ In), λ(B ∩ In) > 0, ∀ n ≥ 1. In the remainder of this section we discuss some applications of the Lebesgue measure to the theory of Riemann integration. The following techincal result will be very useful.

Lemma 6.1. Let f :[a, b] → R be a non-negative Riemann integrable function, let A, B ⊂ [a, b] be two disjoint sets, with A∪B = [a, b]. Then one has the estimates Z b λ∗(A) · inf f(z) ≤ f(t) dt ≤ (b − a) · sup f(x) + λ∗(B) · sup f(y). z∈A a x∈A y∈B Proof. Deﬁne the numbers α = sup f(x), β = sup f(y), and γ = inf f(z). x∈A y∈B z∈A

Recall ﬁrst that, if for each partition ∆ = (a = x0 < x1 < ··· < xn = b) of [a, b], we deﬁne the lower and the upper Darboux sums of f with respect to ∆: n X L(∆, f) = (xk − xk−1) · inf f(t), t∈[x ,x ] k=1 k−1 k n X U(∆, f) = (xk − xk−1) · sup f(t), k=1 t∈[xk−1,xk] then one has the equalities Z b f(t) dt = sup L(∆, f) : ∆ partition of [a, b] = (12) a = inf U(∆, f) : ∆ partition of [a, b] .

Fix now a partition ∆ = (a = x0 < x1 < ··· < xn = b) of [a, b], and deﬁne the set S = k ∈ {1, . . . , n} :[xk−1, xk] ∩ A 6= ∅ . 210 CHAPTER III: MEASURE THEORY

It is clear that inf f(x) ≤ α, sup f(x) ≥ γ, ∀ k ∈ S, x∈[x ,x ] k−1 k x∈[xk−1,xk] inf f(y) ≤ β, sup f(y) ≥ 0, ∀ k ∈ {1, . . . , n} r S, y∈[x ,x ] k−1 k y∈[xk−1,xk] so we get X X (13) L(∆, f) ≤ (xk − xk−1) · α + (xk − xk−1) · β k∈S k6∈S X (14) U(∆, f) ≥ (xk − xk−1) · γ k∈S Consider now the sets [ [ M = [xk−1, xk] and N = [xk−1, xk]. k∈S k6∈S Since the intervals involded in both M and N have at most singleton overlaps, it follows that we have the equalities X X (xk − xk−1) = λ(M) and (xk − xk−1) = λ(N), k∈S k6∈S so the estimates (13) and (14) read (15) L(∆, f) ≤ λ(M) · α + λ(N) · β (16) U(∆, f) ≥ λ(M) · γ Since we clearly have A ⊂ M ⊂ [a, b] and N ⊂ B, we have the inequalities λ∗(A) ≤ λ(M) ≤ b − a and λ(N) ≤ λ∗(B), so the inequalities (15) and (16) give L(∆, f) ≤ (b − a) · α + λ∗(B) · β and U(∆, f) ≥ λ∗(A) · γ.

Since ∆ is arbitrary, the desired inequality then follows from (12). One application of the above result is the following.

Proposition 6.5. If f :[a, b] → R is Riemann integrable, and the set N = {x ∈ [a, b]: f(x) 6= 0} is neglijeable, then Z b (17) f(x) dx = 0. a Proof. Since f is bounded, there exists some constant C > 0, such that the Riemann integrable functions C+f and C−f are both non-negative. Apply Lemma

6.1 to these two functions with A = [a, b] r N and B = N. Since f [a,b] N = 0, we r get (C ± f) = C, so we get [a,b]rN Z b [C ± f(x)] dx ≤ (b − a) · C, a §6. The Lebesgue measure 211 which yields Z b Z b Z b ± f(x) dx = [C ± f(x)] − C dx = [C ± f(x)] dx − (b − a) · C ≤ 0, a a a from which (17) immediately follows. In order to make the exposition a bit easier to follow, it will be helpful to introduce the following

Convention. Given two functions f1, f2 :[a, b] → R, and a relation r on R (in our case r will be either “=,” or “≥,” or “≤”), we write

f1 r f2, a.e. if the set A = x ∈ [a, b]: f1(x) r f2(x) ∗ has neglijeable complement in [a, b], i.e. λ [a, b] r A = 0. The abreviation “a.e.” stands for “almost everywhere.” For example, using this convention, Proposition 6.6 reads: if f :[a, b] → R is R b Riemann integrable, and f = 0, a.e., then a f(x) dx = 0. Exercise 10. A. Prove that “= a.e” is an equivalence relation, and “≥ a.e” and “≤ a.e” are transitive relations on the collection of all function [a, b] → R. B. Prove that f1 ≥ f2, a.e. and f1 ≤ f2, a.e. imply f1 = f2, a.e. C. Prove that these relations are compatible with the arithmetic operations, in the exact way as their “honest” versions. For example, if r is one of “=,” or “≥,” or “≤”, and if f1 r f2, a.e. and g1 r g2, a.e., then (f1 + g1) r (f2 + g2), a.e. Exercise 11. Let f, g :[a, b] → R be continuous functions, such that f ≥ g, a.e. Prove that f ≥ g.

Exercise 12. Let f :[a, b] → R be a non-negative Riemann integrable function, R b with a f(x) dx = 0. Prove that f = 0, a.e. Comment. Riemann integrability is quite a rigid condition. For example the characteristic function of the set of rational numbers in [a, b] is not Riemann κ Q∩[a,b] integrable. By the above result however, we can introduce a slightly weaker notion, which will make such functions integrable, in a weaker sense. This will be a ﬁrst “improvement” of the Riemann integration theory. Eventually (see Chapter IV), a more soﬁsticated theory - the Lebesgue integral - will emerge.

Definition. We say that a function f :[a, b] → R is almost Riemann integrable, if there exists a Riemann integrable function g :[a, b] → R, with f = g, a.e. Of course, such a g is not unique. Notice however that, if h :[a, b] → R is another Riemann integrable function, with f = h, a.e., then g = h, a.e., so by Proposition 6.6, we immediately get the equality Z b Z b g(x) dx = h(x) dx. a a This observation shows that we can unambiguously deﬁne Z b Z b ≈ f(x) dx = g(x) dx. a a 212 CHAPTER III: MEASURE THEORY

Example 6.4. Consider the function f = . Since ∩[a, b] is neglijeable, κ Q∩[a,b] Q we have f = 0, a.e. So f is almost Riemann integrable (althought it is not Riemann integrable), and we have Z b ≈ f(x) dx = 0. a We now focus our attention to (honest) Riemann integrability, with an eye on the role played by continuity. For a function f :[a, b] → R we deﬁne the set Df = x ∈ [a, b]: f not continuous at x . It is well-known that continuous functions are Riemann integrable. There are dis- continuous functions which are still Riemann integrable, for instance we know that

(18) Df finite =⇒ f Riemann integrable. Notations. Let f :[a, b] → R be a bounded function. Suppose ∆ = (a = x0 < x1 < ··· < xn = b) is a partition. For each k ∈ {1, . . . , n} we consider the numbers Mk = sup f(t) and mk = inf f(t), t∈[x ,x ] t∈[xk−1,xk] k−1 k and we define the functions f = m · + m · + ··· + m · , ∆ 1 κ [x0,x1] 2 κ (x1,x2] n κ (xn−1,xn] f ∆ = M · + M · + ··· + M · . 1 κ [x0,x1] 2 κ (x1,x2] n κ (xn−1,xn] ∆ Clearly the functions f∆ and f have only finitely many points of discontinuity, so they are Riemann integrable. With these notations we have the following

Proposition 6.6. For a bounded function f :[a, b] → R, the following are equivalent: (i) f is Riemann integrable; R b ∆ (ii) inf a [f (x) − f∆(x)] dx : ∆ partition of [a, b] = 0; ∞ (iii) there exists a sequence (∆p)p=1 of partitions of [a, b], with ∆1 ⊂ ∆2 ⊂ ... , b R ∆p and limp→∞ a f (x) − f∆p (x) dx = 0. Proof. From the deﬁnition of Riemann integrability, we know that (i) is equivalent to any of the following two conditions (ii’) inf U(∆, f) − L(∆, f) : ∆ partition of [a, b] = 0; ∞ (iii’) there exists a sequence (∆p)p=1 of partitions of [a, b], with ∆1 ⊂ ∆2 ⊂ ... , and limp→∞ U(∆p, f) − L(∆p, f) = 0. Then the Proposition follows immediately from the fact that, for every partition ∆ one has the equalities Z b Z b ∆ f∆(x) dx = L(∆, f) and f (x) dx = U(∆, f). a a The following result gives a complete description of the relationship between Riemann integrability and continuity. Theorem 6.1 (Lebesgue’s criterion for Riemann integrability). Let f :[a, b] → R be a bounded function. The following are equivalent: §6. The Lebesgue measure 213

(i) f is Riemann integrable; (ii) the discontinuity set Df is neglijeable. Proof. (i) ⇒ (ii). Assume f is Riemann integrable. Using Proposition 6.7, ∞ there exists a sequence (∆p)p=1 of partitions of [a, b], such that ∆1 ⊂ ∆2 ⊂ ... and Z b ∆p lim f (x) − f∆p (x) dx = 0. p→∞ a Notice that

∆1 ∆2 ∆3 (19) f ≥ f ≥ f ≥ · · · ≥ f ≥ · · · ≥ f∆3 ≥ f∆2 ≥ f∆1 .

∆p Deﬁne the Riemann integrable functions hp = f − f∆p , p ∈ N. We then clearly have

(α) hp ≥ hp+1 ≥ 0, ∀ p ∈ N; R b (β) limp→∞ a hp(x) dx = 0. Using (α) we can deﬁne the function h :[a, b] → R by

h(x) = lim hp(x), ∀ x ∈ [a, b]. p→∞ Claim 1: The set N = {x ∈ [a, b]: h(x) 6= 0} is neglijeable.

First of all, the functions hp are all Lebesgue measurable. Secondly, since h is a point-wise limit of a sequence of Lebesgue measurable functions, it follows (see Theorem 3.2) that h itself is Lebesgue measurable. In particular N is Lebesgue measurable. For every integer j ≥ 1, define 1 N = x ∈ [a, b]: h(x) > , j j S∞ so that the sets Nj, j ≥ 1 are again Lebesgue measurable, and N = j=1 Nj. In order to prove that N is neglijeable, it then suffices to prove that λ(Nj) = 0, for all j ≥ 1. Fix for the moment j ≥ 1. Since hp ≥ h ≥ 0, it follows that 1 inf hp(x) ≥ , ∀ p ≥ 1, x∈Nj j so by Lemma 6.1 we get the inequality Z b λ(Nj) ≤ hp(x) dx, ∀ p ≥ 1, j a so by (β) we indeed get λ(Nj) = 0. S∞ Define the set S = p=1 ∆p. Claim 2: If y ∈ [a, b] r (N ∪ S), then f is continuous at y. Fix y ∈ [a, b] r (N ∪ S). In order to prove that f is continuous at y, we must find, for every ε > 0, some open interval Jε 3 y, such that

(20) |f(z) − f(y)| < ε, ∀ z ∈ Jε ∩ [a, b].

Since y 6∈ N, we have limp→∞ hp(y) = 0. Fix ε and choose p ≥ 1, such that 0 ≤ hp(y) < ε. Write the partition ∆p as

∆p = (a = x0 < x1 < ··· < xn = b). 214 CHAPTER III: MEASURE THEORY

Using the fact that y 6∈ ∆p, if we deﬁne k = min j ∈ {1, . . . , n} : y < xj , we have y ∈ (xk−1, xk). In particular, we get

∆p f (y) = sup f(t) and f∆p (y) = inf f(s), s∈[x ,x ] t∈[xk−1,xk] k−1 k so the inequality 0 ≤ hp(y) < ε gives sup f(t) − inf f(s) < ε, s∈[x ,x ] t∈[xk−1,xk] k−1 k so if we choose Jε = (xk−1, xk), we clearly have (20). Now we are done, because using the fact that S is countable, it follows that S is neglijeable, so N ∪ S is also neglijeable. Since by Claim 2, we have Df ⊂ N ∪ S, it follows that Df itself is neglijeable. (ii) ⇒ (i). Assume now the discontinuity set Df is neglijeable, and let us prove ∞ that f is Riemann integrable. Fix a sequence (∆p)p=1 of partitions of [a, b], with 2 S∞ ∆1 ⊂ ∆2 ⊂ ... , and limp→∞ |∆p| = 0. As before, we deﬁne the set S = p=1 ∆p. S Claim 3: For any point y ∈ [a, b] r (Df S), one has the equalities

∆p lim f (y) = lim f∆ (y) = f(y). p→∞ p→∞ p

Fix for the moment ε > 0. Since f is continuous at y, there exists some δε > 0, such that

(21) |f(z) − f(y)| < ε, ∀ z ∈ (y − δε, y + δε) ∩ [a, b].

Choose now q ≥ 1, such that |∆q| < δε. Write ∆q = (a = x0 < x1 < ··· < xn = b). Using the fact that y 6∈ ∆q, we can ﬁnd k ∈ {1, . . . , n} such that y ∈ (xk−1, xk). Since xk − xk−1 < δε, we have the inclusion [xk−1, xk] ⊂ (y − δε, y + δε), so by (21) we immediately get f(y) ≤ f ∆q (y) = sup f(z) ≤ f(y) + ε; z∈[xk−1,xk]

f(y) ≥ f∆q (y) = inf f(z) ≥ f(y) − ε. z∈[xk−1,xk] ∞ ∞ ∆p Since the sequence f (y) p=1 is non-increasing, and the sequence f∆p (y) p=1 is non-decreasing, the above inequalities give

∆p |f (y) − f(y)| ≤ ε and |f∆p (y) − f(y)| ≤ ε, for all p ≥ q, and the Claim follows. Going back to the proof of the Theorem, we will now prove that f satsifies condition (iii) in Proposition 6.6. Fix ε > 0. Since Df ∪ S is also neglijeable, using regularity from above with respect to open sets, we can find an open set E ⊂ R such that E ⊃ Df ∪ S, and λ(E) < ε. Define the compact set A = [a, b] r E, and put B = [a, b] ∩ E. We clearly have (22) λ(B) ≤ λ(E) < ε.

∞ ∆p Deﬁne the sequence (hp)p=1 by hp = f − f∆p . Since A ∩ ∆p = ∅, it follows that hp A is continuous, for each p ≥ 1. Since A ∩ (Df ∪ S) = ∅, by Claim 3, we know

2 Recall that, for a partition ∆ = (a = x0 < ··· < xn = b), the number |∆| is deﬁned as |∆| = max xk − xk−1 : 1 ≤ k ≤ n . §6. The Lebesgue measure 215

∞ that limp→∞ hp(y) = 0, ∀ y ∈ A. Since (hp)p=1 is monotone, by Dini’s Theorem (see ??) it follows that lim max hp(y) = 0. p→∞ y∈A

In particular, there exists pε ≥ 1, such that

(23) hpε (y) ≤ ε, ∀ y ∈ A. Let M = sup f(x) and m = inf f(x). x∈[a,b] x∈[a,b]

Using Lemma 6.1 for hpε and the sets A and B, combined with (22), we have Z b ∗ hpε (x) dx ≤ (b − a) · sup hpε (y) + λ (B) · sup hpε (z) ≤ a y∈A z∈B ≤ ε(b − a) + λ∗(B)(M − m) ≤ ε(b − a + M − m).

Since hpε ≥ hp ≥ 0, for all p ≥ pε, we get the inequalities Z b 0 ≤ hp(x) dx ≤ ε(b − a + M − m), ∀ p ≥ pε. a R b The above argument proves that limp→∞ a hp(x) dx = 0, i.e. Z b ∆p lim [f (x) − f∆p (x)] dx = 0. p→∞ a By Proposition 6.6, it follows that f is Riemann integrable. Exercise 13. Prove that a Riemann integrable function f :[a, b] → R is Lebesgue measurable. ∞ Hint: Use a sequence of partitions (∆p)p=1, with ∆1 ⊂ ∆2 ⊂ ... , and limp→∞ |∆p| = 0. Use the arguments given in the proof of the implication (ii) ⇒ (i), to ﬁnd a neglijeable set N ⊂ [a, b], such that lim f∆ (x) = f(x), ∀ x ∈ [a, b] N. p→∞ p r ∞ The sequence (f∆p )p=1 is non-decreasing, so it has a point-wise limit, say g, which is Lebesgue measurable. Use the fact that

f(x) = g(x) ∀ x ∈ [a, b] r N, to show that f itself is Lebesgue measurable.

Exercise 14. Let K ⊂ [0, 1] be a compact set with K ∩ Q = ∅, and λ(K) > 0 (see Exercise 7 for the existence of such sets). Prove that the characteristic function κ K : [0, 1] → R is not Riemann integrable. In fact, f cannot be almost Riemann integrable either.

Hint: Examine the discontinuity set Df , and prove that K ⊂ Df .

Exercise 15. Let fn :[a, b] → R, n ≥ 1 be a sequence of Riemann integrable Q∞ functions. Consider the product space P = n=1 Ran fn, equipped with the product topology (the sets Ran fn, n ≥ 1, are equipped with the topology induced from ∞ R), and the function F :[a, b] → P , deﬁned by F (x) = fn(x) n=1. Prove that, for every bounded continuous function g : P → R, the composition g ◦ F :[a, b] → R is Riemann integrable. In other words, the result of a bounded continuous operation, involving a sequence of Riemann integrable functions, is again a Riemann integrable function. 216 CHAPTER III: MEASURE THEORY

Hint: Examine the relationship between the discountinuity set Dg◦F and the dsicontinuity sets

Dfn , n ≥ 1.

Exercise 16. Let M be an arbitrary subset of [a, b], and let f :[a, b] → R be a Riemann integrable function, such that f ≤ κ M . Prove the inequality Z b f(x) dx ≤ λ∗(M). a

Hint: Consider the function g :[a, b] → R deﬁned by g(x) = max{f(x), 1}. Then f ≥ g ≥ κ M , and g is still Riemann integrable. Apply Lemma 6.1 (the ﬁrst inequality) to the function 1 − g.

Exercise 17*. Let f :[a, b] → R be a bounded function. Prove that the following are equivalent: (i) f is Riemann integrable; (ii) for every ε > 0, there exist continuous functions g, h :[a, b] → R with R b g ≥ f ≥ h, and a [g(x) − h(x)] dx < ε; (iii) for every ε > 0, there exist Riemann integrable functions g, h :[a, b] → R R b with g ≥ f ≥ h, and a [g(x) − h(x)] dx < ε.

Hints: For the implication (i) ⇒ (ii) analyze ﬁrst the particular case when f = κ J , with J ∆ a sub-interval of [a, b]. Then analyze the functions of the type f and f∆. For the implication (iii) ⇒ (i), analyze the relationship among lower/upper Darboux sums of f, g and h.

Comment. The statement of Theorem 6.1 shows that, appart from trivial cases, the problem of checking that a function f :[a, b] → R is Riemann integrable, is a rather diﬃcult one. The main diﬃculty arises from the fact that, if N ⊂ [a, b] is a neglijeable set, and f is continuous, then f need not be continuous [a,b]rN at all points in [a, b] r N. For instance, if we consider the characteristic function f = κ ∩[a,b] of the rationals in [a, b], and N = Q∩[a, b], then clearly N is neglijeable, Q f is continuous (because it is constant zero), but Df = [a, b]. [a,b]rN As earlier suggested, in the hope that such an anomaly can be eliminated, it is reasonable to consider the slightly weaker notion of almost Riemann integrabilty. In the remainder of this section, we take a closer look at this notion, and we will eventually show (see Theorem 6.2) that this indeed removes the above anomaly.

We begin with an “almost” version of Exercise 17.

Lemma 6.2. For a function f :[a, b] → R, the following are equivalent: (i) f is almost Riemann integrable; (ii) for every ε > 0, there exist continuous functions g, h :[a, b] → R with R b g ≥ f ≥ h a.e., and a [g(x) − h(x)] dx < ε; (iii) for every ε > 0, there exist Riemann integrable functions g, h :[a, b] → R R b with g ≥ f ≥ h a.e., and a [g(x) − h(x)] dx < ε. Proof. The implication (i) ⇒ (iii) is trivial. The implication (iii) ⇒ (ii) follows from Exercise 17. We now prove (ii) ⇒ (i). Assume f has property (ii). For each integer n ≥ 1, choose continuous functions gn, hn :[a, b] → R, such that gn ≥ f ≥ hn, a.e., and §6. The Lebesgue measure 217

R b a [gn(x) − hn(x)] dx ≤ 1/n. Deﬁne the functions Gn,Hn :[a, b] → R, n ≥ 1, by Gn(x) = min g1(x), . . . , gn(x) , Hn(x) = max g1(x), . . . , gn(x) . It is clear that

(α) Gm ≥ f ≥ Hn, a.e., ∀ m, n ≥ 1; (β) G1 ≥ G2 ≥ ... and H1 ≤ H2 ≤ ... ; R b R b (γ) a [Gn(x) − Hn(x)] dx ≤ a [gn(x) − hn(x)] dx ≤ 1/n, ∀ n ≥ 1. Notice that, since the Gm’s and the Hn’s are continuous, by Exercise ??, we also have 0 (α ) Gm ≥ Hn (everywhere!), ∀ m, n ≥ 1. Use (β) to deﬁne the functions G, H :[a, b] → R, by

G(x) = lim Gn(x) and H(x) = lim Hn(x), ∀ x ∈ [a, b], n→∞ n→∞ 0 so by (α ) we clearly have Gn ≥ G ≥ H ≥ Hn, ∀ n ≥ 1. Using then (γ), by Exercise 17 it follows that both G and H are Riemann integrable. Moreover, we have G − H ≥ 0 and Z b Z b 0 ≤ [G(x) − H(x)] dx ≤ [Gn(x) − Hn(x)] dx ≤ 1/n, ∀ n ≥ 1, a a R b Which forces a [G(x) − H(x)] dx = 0, so by Exercise ??, we get G = H, a.e. By (α) it follows that f = G, a.e., so f in indeed almost Riemann integrable. We are now in position to prove the “almost” version of Theorem 6.1.

Theorem 6.2. Let f :[a, b] → R be a bounded function. The following are equivalent: (i) f is almost Riemann integrable;

(ii) there exists a neglijeable set N ⊂ [a, b] such that f is continuous. [a,b]rN Proof. (i) ⇒ (ii). Assume f is almost Riemann integrable, so there exists a Riemann integrable function g :[a, b] → R, such that f = g, a.e. By Theorem 6.1, the discontinuity set Dg is neglijeable. Take M = {x ∈ [a, b]: f(x) 6= g(x)}.

Since f = g, a.e., the set M is neglijeable, and so is the set N = M ∪ Dg. On the one hand, since Dg ⊂ N, the restriction g [a,b] N , is continuous. On the other r hand, since M ⊂ N, we have f = g , so (ii) follows. [a,b]rN [a,b]rN (ii) ⇒ (i). We are going to imitate the proof of Theorem 6.1, with some minor modifications. Fix N ⊂ [a, b] neglijeable, such that f is continuous. Fix [a,b]rN ∞ also a sequence (∆p)p=1 of partitions, with ∆1 ⊂ ∆2 ⊂ ... , and limp→∞ |∆p| = 0. S∞ Put S = p=1 ∆p. Since S is countable, the set N ∪ S is still neglijeable. We put ∆p T = [a, b] r (N ∪ S), and we define the analogues of the functions f and f∆p as follows. Write each partition as ∆ = (a = xp < xp < ··· < xp = b), and define, p 0 1 np for each k ∈ {1, . . . , np}, the numbers p p p p p p Mk = sup f(t): t ∈ [xk−1, xk] ∩ T and mk = inf f(t): t ∈ [xk−1, xk] ∩ T . 218 CHAPTER III: MEASURE THEORY

We then deﬁne, for each p ≥ 1, the functions p p p gp = m · p p + m · p p + ··· + m · p p , 1 κ [x0 ,x1 ] 2 κ (x1 ,x2 ] n κ (xn−1,xn] p p p p g = M · p p + M · p p + ··· + M · p p . 1 κ [x0 ,x1 ] 2 κ (x1 ,x2 ] n κ (xn−1,xn] p Note that we have the inequalities g (x) ≥ f(x) ≥ gp(x), ∀ x ∈ T , which give p (24) g ≥ f ≥ gp, a.e., ∀ p ≥ 1. p It is obvious that g and gp, p ≥ 1, are all Riemann integrable. We are now going R b p p to estimate the integrals [g (x) − gp(x)] dx. Put hp = g − gp, p ≥ 1. First we a observe that, since f T is continuous, and T ∩ ∆p = ∅, ∀ p ≥ 1, we clearly have the p equalities limp→∞ g (x) = limp→∞ gp(x) = f(x), ∀ x ∈ T , which give

(25) lim hp(x) = 0, ∀ x ∈ T. p→∞ Fix some ε > 0, and use regularity from above, to ﬁnd an open set D with D ⊃ N ∪S and λ(D) < ε. Take the compact set A = [a, b] r D. Note that f A is continuous, p since A ⊂ [a, b] r N. Note also that, since A ⊂ [a, b] r S, the functions g A and p ∞ gp A are also continuous, and so will be hp A, for every p ≥ 1. Since g (x) p=1 ∞ is non-increasing, and gp(x) p=1 is non-decreasing, for all x, it follows that the ∞ sequence (hp)p=1 is monotone, so by Dini’s Theorem, (25) gives lim max hp(x) = 0. p→∞ x∈A

In particular, there exists some pε ≥ 1, such that

(26) hp(x) ≤ ε, ∀ p ≥ pε, x ∈ A.

Put B = [a, b] r A, and take M = supx∈[a,b] f(x) and m = infx∈[a,b] f(x). Using the inclusion B ⊂ D, we get λ∗(B) ≤ λ(D) ≤ ε, so by Lemma 6.1, (the functions hp, p ≥ 1, are clearly non-negative), combined with (26), we get Z b ∗ hp(x) dx ≤ (b − a) · sup hp(x) + λ (B) · sup hp(x) ≤ a x∈A x∈B ∗ ≤ (b − a)ε + λ (B)(M − m) ≤ ε(b − a + M − m), ∀ p ≥ pε. R b This estimate then proves that limp→∞ a hp(x) dx = 0, i.e. Z b p lim [g (x) − gp(x)] dx = 0. p→∞ a Combining this with (24), and applying Lemma 6.2, yields the fact that f is almost Riemann integrable. Comment. The hypothesis that f is bounded can be replaced with a slightly weaker one, which assumes that f is almost bounded, meaning that there exists a neglijeable set U ⊂ [a, b], such that f is bounded. [a,b]rU

Exercise 18. Let fn :[a, b] → R, n ≥ 1, be almost Riemann integrable functions, such that

(i) fn ≥ fn+1 ≥ 0, a.e., ∀ n ≥ 1; (ii) limn→∞ fn(x) = 0, for “almost all” x ∈ [a, b], i.e. there exists a neglijeable set N ⊂ [a, b], such that limn→∞ fn(x) = 0, ∀ x ∈ [a, b] r N. §6. The Lebesgue measure 219

Prove that Z b lim ≈ fn(x) dx = 0. n→∞ a