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Contents 1. the Riemann and Lebesgue Integrals. 1 2. the Theory of the Lebesgue Integral

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Contents 1. the Riemann and Lebesgue Integrals. 1 2. the Theory of the Lebesgue Integral Contents 1. The Riemann and Lebesgue integrals. 1 2. The theory of the Lebesgue integral. 7 2.1. The Monotone Convergence Theorem. 7 2.2. Basic theory of Lebesgue integration. 9 2.3. Lebesgue measure. Sets of measure zero. 12 2.4. Nonmeasurable sets. 13 2.5. Lebesgue measurable sets and functions. 14 3. More on the Riemann integral. 18 3.1. The fundamental theorems of calculus. 21 3.2. Characterization of Riemann integrability. 22 1. The Riemann and Lebesgue integrals. Fix a positive integer n. Recall that Rn and Mn are the family of rectangles in Rn and the algebra of multirectangles in Rn, respec- tively. Definition 1.1. We let F + F B n ; n; n; be the set of [0; 1] valued functions on Rn; the vector space of real valued functions n on R ; the vector space of f 2 Fn such that ff =6 0g[rng f is bounded, respectively. We let S B \S M ⊂ F S+ S \F + ⊂ S+ M n = n ( n) n and we let n = n n ( n): n Thus s 2 Sn if and only if s : R ! R, rng s is finite, fs = yg is a multirectangle 2 R f 6 g 2 S+ 2 S ≥ for each y and s = 0 is bounded and s n if and only if s n and s 0. S+ S+ 2 S+ We let n;" be the set of nondecreasing sequences in n . For each s n;" we let 2 F + sup s n be such that n sup s(x) = supfsν (x): ν 2 Ng for x 2 R and we let n f + 2 N+g In;"(s) = sup In (sν ): ν : 2 S+ Remark 1.1. We shall prove below the nontrivial Theorem that if s; t n;" and n n sup s = sup t then In;"(s) = In;"(t). 2 1 2 S+ Proposition 1.1. Suppose c [0; ) and s; t n;". Then 2 S+ n n (i) cs n;" and In;"(cs) = cIn;"(s); 2 n n n n (ii) s + t In;" and In;"(s + t) = In;"(s) + In;"(t); ≤ n ≤ n (iii) if s t then In;"(s) In;"(t). Proof. Straightforward exercise for the reader. □ 1 2 2 F + Definition 1.2. For each f n we let f + 2 S+ ≤ g r(f) = inf In (s): s n and f s and we let f n 2 S+ ≤ g l(f) = inf In;"(s): s n;" and f sup s : Proposition 1.2. We have + 2 S+ r(s) = In (s) whenever s n . Proof. This should be obvious. □ Corollary 1.1. We have jIn(s)j ≤ r(jsj) whenever s 2 Sn. 2 S j j ≤ + j j □ Proof. Indeed, for any s n we have In(s) In ( s ). Remark 1.2. We also have + 2 S+ l(s) = In (s) whenever s n . We shall prove this nontrivial fact shortly. 2 F + 1 2 B Proposition 1.3. Suppose f n and r(f) < . Then f n. 2 S+ ≤ [ f g ⊂ Proof. There is s n such that f s and this implies rng f f > 0 rng s [ fs > 0g. □ 2 Rn 1 2 F + Remark 1.3. On the other hand, if a and f = 1fag n and f = supν ν1fag so l(f) = 0. Proposition 1.4. We have l ≤ r: 2 F + 2 S+ ≤ S+ Proof. Suppose f n , s n and f s. Let t be the sequence in n whose range equals s; that is, tν = s for all ν 2 N. Then sup t = s so ≤ n + l(f) In;"(t) = In (s) + 2 S+ which is to say l(f) is a lower bound for the set of In (u) corresponding to t n with f ≤ t. □ Proposition 1.5. Fn 3 f 7! r(jfj) and Fn 3 f 7! l(jfj) are extended seminorms on Fn. Proof. Straightforward exercise for the reader. □ Example 1.1. Let Q = (0; 1) \ Q: We will show that r(1Q) = 1 and that l(1Q) = 0: ≤ 2 S+ Since 1Q 1(0;1) n we find that ≤ + jj jj r(1Q) In (1(0;1)) = (0; 1) = 1: ≤ 2 + 2 1 Suppose 1Q s In . Let y [0; ]. Obviously, s(x) = s(q) ≥ 1 whenever x 2 s−1[fyg] and q 2 (0; 1) \ Q \ s−1[fyg]. 3 It follows that y ≥ 1 whenever (0; 1) \ int s−1[fyg] =6 ; since, in this case, Q \ s−1[fyg] =6 ;. Therefore, X + jj −1 f g jj In (s) = y s [ y ] 2 y Xrng s = yjjint s−1[fyg]jj 2 y Xrng s ≥ yjj(0; 1) \ int s−1[fyg]jj 2 y Xrng s ≥ jj(0; 1) \ int s−1[fyg]jj 2 y Xrng s = jj(0; 1) \ s−1[fyg]jj y2rng s = 1: Thus r(1Q) ≥ 1: Let q : N ! Q \ [0; 1] be univalent with range Q \ [0; 1]. For each ν 2 N let Xν 2 S+ sν = 1fqµg n : µ=0 2 S+ S+ Note that s n;" is a nondecreasing sequence in n , that Xν + jjf gjj I1 (sν ) = qµ = 0 µ=0 and that 1Q = sup s ≤ sup s: Thus ≤ n l(1Q) In;"(s) = 0: Theorem 1.1. Suppose A 2 Mn, B is a nondecreasing sequence in Mn and ⊂ [1 A ν=0Bν . Then jjAjj ≤ sup jjBν jj: ν + Proof. We define the sequence C in Mn by letting C0 = B0 and for each ν 2 N ∼ [ν 2 N letting Cν = Bν Bν−1. Then C is disjointed and Bν = µ=0Cµ for each ν . Suppose 1 < λ < 1. Choose a compact multirectangle K such that K ⊂ A and jjAjj ≤ λjjKjj. For each ν 2 N choose an open multirectangle Uν such that Cν ⊂ Uν jj jj ≤ jj jj ⊂ [1 2 N ⊂ [N and Uν λ Cν . Then K ν=0Uν so there is N such that K µ=0Uµ. Thus XN XN −1jj jj ≤ jj jj ≤ jj [N jj ≤ jj jj ≤ jj jj jj jj λ A K µ=0 Uµ Uµ λ Cµ = λ BN : µ=0 µ=0 Owing to the arbitrariness of λ the Lemma is proved. □ 4 Corollary 1.2. Suppose A 2 Mn, B is a countable subfamily of Mn and A ⊂ [B. Then X jjAjj ≤ jjBjj: B2B Proof. In case B is finite this follows from earlier work. So suppose B is infinite, B 2 N [ν let B be an enumeration of and, for each ν , let Cν = µ=0Bµ. Then C is a nondecreasing sequence in Mn whose union contains A so that, by the preceding Theorem, Xν X1 jjAjj ≤ sup jjCν jj ≤ sup jjBµjj = jjBν jj: ν ν µ=0 ν=0 □ Theorem 1.2. We have + 2 S+ l(s) = In (s) for any s n . 2 S+ ≤ + Proof. Suppose s n . Obviously, l(s) In (s) S+ ≤ ∼ Suppose t is a nondecreasing sequence in n and s supν tν . Let Y = (rng s) n f0g and for each y 2 Y let Ay = fx 2 R : s(x) = yg. Suppose 0 < λ < 1 and for 2 2 N f 2 g ⊂ [1 each y Y and ν let By,ν = x Ay : λy < tν (y) . Then Ay ν=0By,ν jj jj ≤ jj jj 2 2 N so Ay supν By,ν by the preceding Theorem. For each y Y and ν we have λy1B ≤ 1A tν and this implies that y,ν y ( ) jj jj ≤ jj jj jj jj ≤ λy Ay λy sup B ν = sup λy B ν In 1Ay tν : ν ν Thus X λIn(s) = λyjjAyjj 2 yXY ≤ sup In(1Ay tν ) 2 ν y Y 0 1 X @ A = sup In 1Ay tν ν y2Y ≤ sup In(tν ) Letting λ " 1 we find that In(s) ≤ sup In(tν ). Thus In(s) ≤ l(s). □ Corollary 1.3. We have jIn(s)j ≤ l(jsj) for any s 2 Sn. j j ≤ + j j Proof. This follows from the preceding Theorem since In(s) In ( s ) for any s 2 Sn. □ Definition 1.3. We let Riemn be the set of f 2 Fn such that for each ϵ > 0 there is s 2 Sn such that r(jf −sj) < ϵ. Thus Riemn is the closure of Sn with respect to the extended seminormFn 3 f 7! r(jfj). We say f 2 Fn is Riemann integrable if f 2 Riemn. 5 We let Lebn be the set of f 2 Fn such that for each ϵ > 0 there is s 2 Sn such that l(jf −sj) < ϵ. Thus Lebn is the closure of Sn with respect to the extended seminorm Fn 3 f 7! l(jfj) of Sn. We say f 2 Fn is Lebesgue integrable if f 2 Lebn. Proposition 1.6. Suppose f 2 Fn. Then f 2 Riemn if and only if for each ϵ > 0 there is g 2 Riemn such that r(jf − gj) < ϵ and f 2 Lebn if and only if for each ϵ > 0 there is g 2 Lebn such that r(jf − gj) < ϵ. Proof. This should be obvious. It boils down to the fact the the closure of the closure equals the closure. □ Theorem 1.3. Riemn is a linear subspace of Fn and there is one and only one linear function R : Riemn ! R such that (i) R(s) = In(s) whenever s 2 Sn; (ii) jR(f)j ≤ r(jfj) whenever f 2 Riemn. Lebn is a linear subspace of Fn and there is one and only one linear function L : Lebn ! R such that (i) L(s) = In(s) whenever s 2 Sn; (ii) jL(f)j ≤ l(jfj) whenever f 2 Lebn.
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