Divulgaciones Matem´aticasVol. 8 No. 1 (2000), pp. 75{85 The Fundamental Theorem of Calculus for Lebesgue Integral El Teorema Fundamental del C´alculo para la Integral de Lebesgue Di´omedesB´arcenas([email protected]) Departamento de Matem´aticas.Facultad de Ciencias. Universidad de los Andes. M´erida.Venezuela. Abstract In this paper we prove the Theorem announced in the title with- out using Vitali's Covering Lemma and have as a consequence of this approach the equivalence of this theorem with that which states that absolutely continuous functions with zero derivative almost everywhere are constant. We also prove that the decomposition of a bounded vari- ation function is unique up to a constant. Key words and phrases: Radon-Nikodym Theorem, Fundamental Theorem of Calculus, Vitali's covering Lemma. Resumen En este art´ıculose demuestra el Teorema Fundamental del C´alculo para la integral de Lebesgue sin usar el Lema del cubrimiento de Vi- tali, obteni´endosecomo consecuencia que dicho teorema es equivalente al que afirma que toda funci´onabsolutamente continua con derivada igual a cero en casi todo punto es constante. Tambi´ense prueba que la descomposici´onde una funci´onde variaci´onacotada es ´unicaa menos de una constante. Palabras y frases clave: Teorema de Radon-Nikodym, Teorema Fun- damental del C´alculo,Lema del cubrimiento de Vitali. Received: 1999/08/18. Revised: 2000/02/24. Accepted: 2000/03/01. MSC (1991): 26A24, 28A15. Supported by C.D.C.H.T-U.L.A under project C-840-97. 76 Di´omedesB´arcenas 1 Introduction The Fundamental Theorem of Calculus for Lebesgue Integral states that: A function f :[a; b] R is absolutely continuous if and only if it is ! 1 differentiable almost everywhere, its derivative f 0 L [a; b] and, for each t [a; b], 2 2 t f(t) = f(a) + f 0(s)ds: Za This theorem is extremely important in Lebesgue integration Theory and several ways of proving it are found in classical Real Analysis. One of the better known proofs relies on the non trivial Vitali Covering Lemma, perhaps influenced by Saks monography [9]; we recommend Gordon's book [5] as a recent reference. There are other approaches to the subject avoiding Vitali's Covering Lemma, such as using the Riesz Lemma: Riez-Nagy [7] is the clas- sical reference. Another approach can be seen in Rudin ([8], chapter VIII) which treats the subject by differentiating measures and of course makes use of Lebesgue Decomposition and the Radon-Nikodym Theorem. The usual form of proving this fundamental result runs more or less as follows: First of all, Lebesgue Differentiation Theorem is established: Every bounded variation function f :[a; b] R is differentiable almost everywhere with derivative belonging to L1[a; b!]: If the function f is non- decreasing, then b f 0(s)ds f(b) f(a): ≤ − Za In the classical proof of the above theorem Vitali's Covering Lemma (Riesz Lemma either) is used, as well as in that of the following Lemma, previous to the proof of the theorem we are interested in. If f :[a; b] R is absolutely continuous with f 0 = 0 almost everywhere then f is constant! . An elementary and elegant proof of this Lemma using tagged partitions has recently been achieved by Gordon [6]. In this paper we present another approach to the subject; indeed, we start with Lebesgue Decomposition and Radon Nikodym Theorem and soon after, we derive directly the Fundamental Theorem of Calculus and get relatively simple proofs of well known results. The Fundamental Theorem of Calculus for Lebesgue Integral 77 We start outlining the proof of the Radon Nikodym Theorem given by Bradley [4] in a slightly different way; indeed, instead of giving the proof of Radon Nikodym Theorem as in [4], we make a direct (shorter) proof of the Lebesgue Decomposition Theorem which has as a corollary the Radon Nikodym Theorem. Readers familiarized with probability theory will soon recognize the presence of martingale theory in this approach. The needed preliminaries for this paper are all studied in a regular graduate course in Real Analysis, especially we need the Monotone and Dominated Convergence theorems, the Cauchy-Schwartz inequality (an elementary proof is provided by Apostol [1]), the fact that if (Ω; Σ; µ) is a measure space and 1 g L (µ), then ν :Σ R defined for ν(E) = E g dµ is a real measure and f 2 L1(ν) if and only if!fg L1(µ) and 2 2 R f dν = fg dµ, E Σ: 8 2 ZE ZE We also need the definitions of mutually orthogonal measures (µ λ) and measure λ absolutely continuous with respect to µ (λ µ). While? these preliminary facts, previous to Lebesgue Decomposition Theorem and Radon Nikodym Theorem, are nicely treated in Rudin [8], a good account of dis- tribution functions (bounded variation functions) can be found in Burrill [3]. Particularly important are the following results: Every bounded variation function f :[a; b] R determines a unique Lebesgue-Stieljes measure µ. The function f is absolutely! continuous if and only if its corresponding Lebesgue-Stieljes measure µ is absolutely continuous with respect to Lebesgue measure. It is also important for our purposes the following fact: If f :[a; b] R is a bounded variation function with associated Lebesgue- Stieljes measure!µ, then the following statements are equivalent: a) f is differentiable at x and f 0(x) = A. b) For each > 0 there is a δ > 0 such that µ(I) A < , whenever I is j m(I) − j an open interval with Lebesgue measure m(I) < δ and x I, 2 Both references Rudin [8] and Burrill [3] are worth to look for the proof of this equivalence. 78 Di´omedesB´arcenas 2 Lebesgue Decomposition and Radon-Nikodym Theorem The Radon Nikodym Theorem plays a key role in our proof of the Fundamen- tal Theorem of Calculus, particularly the proof given by Bradley [4], so we will outline this proof but deriving it from Lebesgue Decomposition Theorem (Theorem 1 below). Theorem 1. (Lebesgue Decomposition Theorem). Let (Ω; Σ; µ) be a finite, positive measure space and λ :Ω R a bounded ! variation measure. Then there is a unique pair of measures λa and λs so that λ = λa + λs with λa µ and λs µ. ? 1 1 2 2 Proof. We first prove the uniqueness: if λa + λs = λa + λs with λi µ; λi µ qquadi = 1; 2; a s? then λ1 λ2 = λ2 λ1 a − a s − s with both sides of this equation simultaneously absolutely continuous and singular with respect to µ. This quickly yields 1 2 1 2 λa = λa and λs = λs: Existence: This portion of the proof is modelled on Bradley's idea [4]. Let us denote by λ the variation of λ and put σ = µ + λ . Then µ and λ are absolutely continuousj j with respect to the finite positivej j measure σ. Now for any measurable finite partition P = A1;A2;:::;An of Ω we define f g n hP = ciχAi ; i=1 X where 0 if σ(A ) = 0; c = i i λ(Ai) otherwise: ( σ(Ai) (Our proof omits some details which can be found in [4]). Notice that the following three statements hold for each finite measurable partition P : The Fundamental Theorem of Calculus for Lebesgue Integral 79 a) 0 hP (x) 1. ≤ j j ≤ b) λ(Ω) = hP dσ: ZΩ c) If P and Π are finite measurable partitions of Ω with Π a refinement of P , then 2 2 2 h dσ = h dσ + (hΠ hP ) dσ Π P − ZΩ ZΩ ZΩ h2 dσ: ≥ P ZΩ If we put 2 k = sup hP dσ; ZΩ where the supremum is taken over all finite measurable partitions of Ω, then, since for any finite measurable partition P of Ω we have that n λ(A )2 h2 dσ = i dσ P σ(A )2 Ω Ω i=1 i Z Z X n 2 λ(Ai) = j j σ(A ) i=1 i Xn λ(Ai) ≤ j j i=1 X λ (Ω) < ; ≤ j j 1 we conclude that 0 k < : ≤ 1 For each n = 1; 2;::: take as Pn a finite measurable partition of Ω such that 1 k h2 dσ − 4n ≤ Pn ZΩ and let Πn be the least common refinement of P1;P2; :::; Pn: By (c), 1 k h2 dσ h2 dσ k: − 4n ≤ Pn ≤ Πn ≤ ZΩ ZΩ 80 Di´omedesB´arcenas So for each n > 1, we have 2 2 2 1 (hΠ hΠ ) dσ = h dσ h dσ n+1 − n Πn+1 − Πn ≤ 4n ZΩ ZΩ ZΩ and by the Cauchy-Schwartz inequality we get 1 1 hΠ hΠ dσ (σ(Ω)) 2 < j n+1 − n j ≤ 2n 1 ZΩ and so n 1 hΠ hΠ dσ (σ(Ω)) 2 < j n+1 − n j ≤ 1 Ω i=1 Z X which implies that n hΠ = hΠ + hΠ hΠ n+1 1 i+1 − i i=1 X converges σ-almost everywhere. Put limn hΠn (x) if the limits exists; h(x) = !1 (0 otherwise: If A Σ and n N take Rn as the smallest common refinement of Πn and A; Ω2\A to get2 f g 2 1 hRn hΠ dσ : j − n j ≤ 4n ZΩ By (c) and the Cauchy-Schwartz inequality, for n 1, ≥ 2 2 1 (hRn hΠ ) dσ hRn hΠ dσ < ; j − n j ≤ j − n j 2n ZΩ ZΩ which implies that lim (hRn hΠ ) dσ = 0: n − n !1 ZA Since by (b), λ(A) = hΠ dσ + (hRn hΠ ) dσ; n − n ZA ZA The Fundamental Theorem of Calculus for Lebesgue Integral 81 we conclude that λ(A) = lim hΠ dσ n n !1 ZA and applying the Dominated Convergence Theorem, we get λ(A) = h dσ A Σ: 8 2 ZA Summarizing we have gotten a σ-integrable function hλ, depending on λ, such that λ(A) = hλ dσ; A Σ: 8 2 ZA Similarly we can find a σ-integrable function hµ (depending on µ) such that µ(A) = hµ dσ A Σ: 8 2 ZA Put Ω1 = x : hµ(x) 0 ; f ≤ g Ω2 = x : hµ(x) > 0 : f g Plainly µ(Ω1) = 0 and defining, for A Σ, λs(A) = λ(A Ω1) and λa(A) = 2 \ λ(A Ω2), we see that λs µ and λa + λs = λ.