AN ELEMENTARY PROOF OF LEBESGUE S DIFFERENTIATION THEOREM

Transcription

1 AN ELEMENTARY PROOF OF LEBESGUE S DIFFERENTIATION THEOREM Elif Az Spring Abstract The fact that a continuous monotone function is differentiable almost everywhere was established by Lebesgue in Riesz gave a completely elementary proof of this theorem in 1932 by using his Rising Sun Lemma. Other than Riesz s, all the proofs of this theorem utilize measure theory. Also a geometric proof which involves measure theory and sets of measure zero was given by D. Austin. The purpose of this work is to give an easier alternative proof of this theorem that can be presented in an elementary analysis course. This proof uses the sets of measure zero and upper and lower derivatives. We will use Heine-Borel Theorem and an elementary covering lemma to show that a nondecreasing function f defined on [a, b] has derivative almost everywhere in this interval. Özet Sürekli ve monoton bir fonksiyonun hemen her yerde türevlenebilir olması durumu 1904 te Lebesgue tarafından tasdik edilmiştir. Reisz 1932 de Rising Sun önsavını kullanarak bu teoremin basit bir ispatını yapmıştır. Reisz ınki hariç ispatlarn hepsinde ölçü teorisi kullanılmıştır. Ayrıca D. Austin tarafından da ölçü teorisini ve ölçümü sıfır olan kümeleri içeren geometrik bir ispat verilmiştir. Bu çalışmanın amacı teoremin elementer analiz derslerinde sunulabilecek şekilde daha kolay alternatif bir ispatını vermektir. Bu ispatta ölçümü sıfır olan kümeleri ve alt ve üst türevler kullanılmaktadır. [a, b] de tanımlı azalmayan bir fonksiyon olan f nin bu aralıkta hemen her yerde türevlenebilir olduğunu göstermek için Heine-Borel Teoremin i ve aynı zamanda basit bir örtü kümesi önsavını kullanacağız. 1

2 1 INTRODUCTION The fact that a continuous monotone function is differentiable almost everywhere was established by Lebesgue in Many proofs of this theorem were given in many ways. In this work we will give an easier alternative proof of this theorem by using sets of measure zero. To achieve the result we have worked on the paper by Michael W. Botsko (Nov Sets of measure zero are just the sets that are negligible in theory of Lebesgue integration. If something happens except on a set of measure zero, it is said to happen almost everywhere [2, p. 63]. To show that a nondecreasing function f defined on [a, b] has derivative almost everywhere in this interval, first we prove a lemma by using Heine- Borel Theorem. In the proof of this lemma we will make some calculations on coverings and in the main part we will use this lemma in a technical part of our proof. 2 PRELIMINARIES In our proof we will use an elementary covering lemma which is stated below as Lemma 1. To prove Lemma 1 we need nothing more than the fact that an open set on the real line can be expressed as the countable disjoint union of open intervals and also the Heine-Borel Theorem which states that the compact sets of real numbers are exactly the sets that are both closed and bounded. Lemma 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 intervals that covers E, where J n denotes the lenght of J n. If I is any collection of open subintervals of [a, b] that covers E, then there exits a finite disjoint 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 I I = where {(a n, b n } is a disjoint sequence of open intervals. (a n, b n 2

3 Since E (a n, b n (b n a n ε. Now 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. Let n be a fixed positive integer. For each x in [a n, b n] there exists J x in I 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 finite subcovering {J 1, J 2,... J p } such that [a n, b n] p J k. Furthermore by discarding some of the intervals, we may assume that no interval in {J k } p is a subset of the union of the remaining intervals in {J k } p. 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 subcollections of I. It is obvious that either ( p J 2k 1 J k /2 or k ( p J 2k J k /2. k Depending on which one of the inequalities holds, we have found a finite disjoint subcollection I n of I such that ( p I J k /2 (b n a n/2. I I n Hence for each positive integer n there exist a finite disjoint subcollection I n of 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. I I n 3

4 Therefore I n = {I 1, I 2, I 3,...} is a countable disjoint subcollection of I for which I n ε/2c. Finally choose N so large that I k > ε/dc, where d R, d > 2 and dc R, dc > 2. Without loss of generality we can choose dc = 3 and I k > ε/3 to complete the proof. In this lemma we have shown that for a subset E of (a, b that is not of measure zero and I be an open covering for E, there exists a finite disjoint subcollection of open intervals of I whose total lenght is greater than ε/3. Here ε/3 is the maximum value that we can find such a collection. Remark 1: Suppose E and ε are exactly as in Lemma 1. Let P be a given finite subset of [a, b] and let I be a collection of open subintervals of [a, b] that covers E P. Then by adding to I sufficiently small open intervals centered at each point of P we can find a collection J such that J = I {K 1, K 2,... K i } that covers E. Now E J k and by Lemma 1 d R. Since I J, we have N 1 J k > ε/cd,where cd > 2, c and I k > ε/r, where r > cd > 2, r R. Hence there exists a finite disjoint subcollection {I 1, I 2,... I N } of I such that I k > ε/4. Clearly Remark 1 is an easy variation of Lemma 1. Lemma 2: Let f : [a, b] R, let P = {x 0, x 1, x 2,..., x n } be a partition of [a, b], let S be a nonempty subset of {1, 2, 3,..., n}, and let A > 0. If f(a f(b and < A x k x k 1 for each k in S, then > f(b f(a + AL, 4

5 where L = k S (x k x k 1. Proof : Since f(a f(b, it follows that, Therefore f(b f(a = f(b f(a = ( ( ( = + k S k / S < A (x k x k 1 + ( k S k / S AL + > f(b f(a + AL, which completes the proof. Remark 2: By putting g = f we can easily show that the conclusion of Lemma 2 holds if f(a f(b and for each k in S. x k x k 1 > A In Lemma 2 we have found a result by using the fact that the derivative of f defined on [a, b] with the properties as above, in a partition of [a, b] has a bound. We have shown these lemmas and their variations because we will use both Remark 1 and Lemma 2 in technical parts of our main proof. To prove our main result we will work on sets of measure zero and their open coverings.we will make some calculations on this coverings. Also we will use partitions on [a, b] with function defined on this intervals. 5

6 3 MAIN RESULT Now we are going to establish the main result. We will achieve the result by using the lemmas in preliminary part. We need them in some technical part of our proof. Our proof will depend on no measure theory beyond sets of measure zero. Also upper and lower derivatives are used. Theorem: If f is nondecreasing on [a, b], then f (x exists almost everywhere on [a, b]. Proof: We know that the only discontinuities of monotonic functions are jumps [2, p. 129]. Since f is continuous except at a countable number of points, it is sufficient to show that F = {x : x (a, b, f is continuous at x and Df(x > Df(x} has measure zero. Here Df(x = lim sup y x f(y f(x f(y f(x, Df(x = lim inf y x y x y x are the upper and lower derivatives of f at x, respectively. Clearly F is the countable union of the sets E r,s = {x : x (a, b, f is continuous at x and Df(x > r > s > Df(x} for rational numbers r and s with r > s. Thus we need to show that each E r,s has measure zero. Assume for a contradiction that for some choice of r and s the set E = E r,s does not have measure zero, and let ε be the positive number in Lemma 1. Let A = (r s/2, B = (r + s/2, and g = f Bx. and B are positive and Clearly both A E = {x : x (a, b, g is continuous at x, Dg(x > A and Dg(x < A}. Now { P g(x k g(x k 1 : P is partition of [a, b]} is bounded above. Namely, g(x k g(x k 1 = B(x k x k 1 P P + B (x k x k 1 P P = f(b f(a + B(b a. 6

7 Let T be the least upper bound for this set. Since both A and ε are positive, there exists a partition P = {x 0, x 1, x 2,... x n } of [a, b] for which g(x k g(x k 1 > T Aε 4. (1 Now let x belongs to E P, which means that x is in E (x k 1, x k for some k. Since Dg(x > A and Dg(x < A and g is continuous at x, we can choose a x and b x such that a x < x < b x, (a x, b x (x k 1, x k, and either g(x k 1 g(x k and g(x k g(x k 1 > A or g(x k 1 g(x k and x k x k 1 g(x k g(x k 1 < A holds. Thus I = {(a x, b x : x E P } is a collection x k x k 1 of open subintervals of [a, b] that covers E P. By Remark 1, there exists a finite disjoint subcollection {I 1, I 2,..., I N } of I such that I k > ε 4. (2 Now let Q = {y 0, y 1, y 2,..., y q } be the partition of [a, b] determined by the points of P and the endpoints of the intervals I 1, I 2,..., I N. For each [x k 1, x k ] containing at least one of the intervals {I 1, I 2,..., I N }, we get by Lemma 2 that g(y i g(y i 1 > g(x k g(x k 1 + AL k, (3 [y i 1,y i ] [x k 1,x k ] where the summation is taken over the closed intervals determined by Q that are contained in [x k 1, x k ] and L k is the sum of the lengths of those intervals I 1, I 2,..., I N that contained in [x k 1, x k ]. Summing inequality (3 over k and using (1 and (2, we get q g(y k g(y k 1 > g(x k g(x k 1 + A which contradicts the definition of T. We have shown that the set > T Aε ( ε 4 + A 4 = T, I k F = {x : x (a, b, f is continuous at x and Df(x > Df(x} 7

8 has measure zero. Hence upper and lower derivatives are equal but they can be both equal to infinity. For existence of the derivative this value must be finite. Now in this part of our proof we will show that f has a finite derivative almost everywhere on [a, b]. Since Df(x 0 for all x in (a, b, it is sufficient to prove that Df(x < almost everywhere on (a, b. To accomplish this we will show that has measure zero. E = {x : x (a, b and Df(x = } Suppose that E does not have measure zero. Let M be an arbitrarily large positive number, and let ε be as in Lemma 1. If x lies in E, then Df(x > M and there exist a x and b x such that a x < x < b x, (a x, b x (a, b, and f(b x f(a x b x a x > M. Thus I = {(a x, b x : x E} covers E and by Lemma 1 contains a finite disjoint subcollection {I 1, I 2,... I N } such that N I k > ε/3. For each k let I k = (a k, b k. Since f is increasing, clearly f(b f(a ( f(b k f(a k > M(b k a k > M ε 3. Therefore f(b f(a > Mε/3. Because ε > 0 and M is arbitrarily large, this contradicts the finiteness of f(b f(a. Hence we have shown that a continuous nondecreasing real-valued function f defined on [a, b] has a finite derivative almost everywhere in this interval. Similarly we can show that it is true for a nonincreasing function f. 4 CONCLUSION It is clear that if a fuction f is differentiable at a point x then it is continuous at x. But the inverse is not always true. In Lebesgue Differentiation Theorem it is stated that a continuous monotone function is differentiable almost everywhere. Actually this result is excellent in its own and made an extensive contribution to mathematics. It means nearly that a continuous function is differentiable. 8

9 In our work we proved this result with using elementary consepts. We used sets of measure zero, Heine-Borel Theorem, upper and lower derivatives which can be presented in an elementary analysis course. Mathematicians worked to prove this result many times in many ways. For instance a geometric proof was given by D. Austin. But the arguments in this work are more complicated to understand. It can be another work to study solely. References [1] M. W. Botsko. An Elementary Proof of Lebesgue s Differentiation Theorem, The American Mathematical Monthly, 110 (9, (Nov 2003: [2] R. P. Boas., A Primer of Real Funtions,(Rahway, New Jersey: Mathematical Association of America,