an elementary proof of lebesgue's differentiation theorem

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1 INTRODUCTIONThe fact that a continuous monotone function is differentiable almost everywherewas established by Lebesgue in 1904. Many proofs of this theoremwere given in many ways. In this work we will give an easier alternativeproof of this theorem by using sets of measure zero. To achieve the result wehave worked on the paper by Michael W. Botsko (Nov 2003).Sets of measure zero are just the sets that are negligible in theory of Lebesgueintegration. If something happens except on a set of measure zero, it is saidto happen almost everywhere [2, p. 63].To show that a nondecreasing function f defined on [a, b] has derivativealmost everywhere in this interval, first we prove a lemma by using Heine-Borel Theorem. In the proof of this lemma we will make some calculationson coverings and in the main part we will use this lemma in a technical partof our proof.2 PRELIMINARIESIn our proof we will use an elementary covering lemma which is stated belowas Lemma 1. To prove Lemma 1 we need nothing more than the fact thatan open set on the real line can be expressed as the countable disjoint unionof open intervals and also the Heine-Borel Theorem which states that thecompact sets of real numbers are exactly the sets that are both closed andbounded.n=1Lemma 1: Let E be a subset of (a, b) that is not of measure zero. (Thus∞∑there exist ε > 0 such that |J n | ≥ ε for any sequence {J n } of open intervalsthat covers E, where |J n | denotes the lenght of J n .) If I is any collectionof open subintervals of [a, b] that covers E, then there exits a finite disjointN∑subcollection {I 1 , I 2 , . . . , I N } of I such that |I k | > ε/3.Proof: Since ⋃ I∈I I is an open set, we can state that ⋃ I∈II =k=1where {(a n , b n )} is a disjoint sequence of open intervals.∞⋃(a n , b n )n=12
Since E ⊆∞⋃(a n , b n )n=1∞∑(b n − a n ) ≥ ε.n=1Now for all (a n , b n ) choose a closed subinterval [a ′ n, b ′ n] such that c > 1, c ∈ R,∞∑b ′ n − a ′ n = (b n − a n )/c. Thus (b ′ n − a ′ n) ≥ ε/c.n=1Let n be a fixed positive integer. For each x in [a ′ n, b ′ n] there exists J x inI such that x ∈ J x ⊆ (a n , b n ). So {J x : x ∈ [a ′ n, b ′ n]} is an open cover of[a ′ n, b ′ n]. Since [a ′ n, b ′ n] closed and bounded by Heine-Borel theorem it is compact.So an open cover {J x : x ∈ [a ′ n, b ′ n]} of [a ′ n, b ′ n] can be reduced to a finitesubcovering {J 1 , J 2 , . . . J p } such that [a ′ n, b ′ n] ⊆ ⋃ pk=1 J k.Furthermore by discarding some of the intervals, we may assume that nointerval in {J k } p k=1is a subset of the union of the remaining intervals in{J k } p k=1 . Thus each J i contains a point x i which does not belong to ⋃ k≠i J k,and renumbering J k s, if necessary, we can assume that x 1 < x 2 < . . . < x p .Therefore both {J 1 , J 3 , J 5 , . . .} and {J 2 , J 4 , J 6 , . . .} are finite disjoint subcollectionsof I.It is obvious that either∑ ( ∑p )|J 2k−1 | ≥ |J k | /2 orkk=1∑ ( ∑p )|J 2k | ≥ |J k | /2.kk=1Depending on which one of the inequalities holds, we have found a finitedisjoint subcollection I n of I such that∑ ( ∑p )|I| ≥ |J k | /2 ≥ (b ′ n − a ′ n)/2.I∈I nk=1Hence for each positive integer n there exist a finite disjoint subcollection I nof I each of whose open intervals is a subset of (a n , b n ) such that∑I∈I n|I| ≥ (b ′ n − a ′ n)/2.Summing both sides of the inequality, we have∞∑ ∑ ( ∑ ∞ )|I| ≥ (b ′ n − a ′ n) /2 ≥ ε/2c.n=1 I∈I n n=13
Page 1: AN ELEMENTARY PROOF OFLEBESGUE’S
Page 5 and 6: where L = ∑ k∈S (x k − x k−
Page 7: Let T be the least upper bound for

an elementary proof of lebesgue's differentiation theorem

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