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 ex- ists 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.