AN ELEMENTARY PROOF OF LEBESGUE’S DIFFERENTIATION THEOREM Elif Az 2003102278 2008 Spring Abstract The fact that a continuous monotone function is differentiable almost everywhere was established by Lebesgue in 1904. 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. ¨ Ozet S¨ urekli ve monoton bir fonksiyonun hemen her yerde t¨ urevlenebilir olması durumu 1904’te Lebesgue tarafından tasdik edilmi¸stir. Reisz 1932’de ’Rising Sun’ o¨nsavını kullanarak bu teoremin basit bir ispatını yapmı¸stır. Reisz’ınki hari¸c ispatlarn hepsinde o¨l¸cu ¨ teorisi kullanılmı¸stır. Ayrıca D. Austin tarafından da ¨ol¸cu ¨ teorisini ve o¨l¸cu ¨m¨ u sıfır olan k¨ umeleri i¸ceren geometrik bir ispat verilmi¸stir. Bu ¸calı¸smanın amacı teoremin elementer analiz derslerinde sunulabilecek ¸sekilde daha kolay alternatif bir ispatını vermektir. Bu ispatta ¨ol¸cu ¨m¨ u sıfır olan k¨ umeleri ve alt ve u ¨st t¨ urevler kullanılmaktadır. [a, b] de tanımlı azalmayan bir fonksiyon olan f nin bu aralıkta hemen her yerde t¨ urevlenebilir oldu˘gunu g¨ostermek i¸cin Heine-Borel Teoremin’i ve aynı zamanda basit bir o¨rt¨ u k¨ umesi o¨nsavını kullanaca˘gız.

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1

INTRODUCTION

The fact that a continuous monotone function is differentiable almost everywhere was established by Lebesgue in 1904. 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 2003). 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 HeineBorel 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 ∞ X |Jn | ≥ ε for any sequence {Jn } of open interthere exist ε > 0 such that n=1

vals that covers E, where |Jn | denotes the lenght of Jn .) If = is any collection of open subintervals of [a, b] that covers E, then there exits a finite disjoint N X subcollection {I1 , I2 , . . . , IN } of = such that |Ik | > ε/3. k=1

Proof: Since

S

I∈=

I is an open set, we can state that

[ I∈=

where {(an , bn )} is a disjoint sequence of open intervals.

2

I =

∞ [

(an , bn )

n=1

∞ [

Since E ⊆

(an , bn )

n=1

∞ X (bn − an ) ≥ ε. n=1 0

0

Now for all (an , bn ) choose a closed subinterval [an , bn ] such that c > 1, c ∈ R, ∞ X 0 0 0 0 bn − an = (bn − an )/c. Thus (bn − an ) ≥ ε/c. n=1 0

0

Let n be a fixed positive integer. For each x in [an , bn ] there exists Jx in 0 0 = such that x ∈ Jx ⊆ (an , bn ). So {Jx : x ∈ [an , bn ]} is an open cover of 0 0 0 0 [an , bn ]. Since [an , bn ] closed and bounded by Heine-Borel theorem it is com0 0 0 0 bn ] can be reduced to a finite pact. So an open cover {Jx : x ∈ [an , bn ]} of [an ,S 0 0 subcovering {J1 , J2 , . . . Jp } such that [an , bn ] ⊆ pk=1 Jk . Furthermore by discarding some of the intervals, we may assume that no interval in {Jk }pk=1 is a subset of the union of the remaining intervals in S {Jk }pk=1 . Thus each Ji contains a point xi which does not belong to k6=i Jk , and renumbering Jk s, if necessary, we can assume that x1 < x2 < . . . < xp . Therefore both {J1 , J3 , J5 , . . .} and {J2 , J4 , J6 , . . .} are finite disjoint subcollections of =. It is obvious that either p X  X |J2k−1 | ≥ |Jk | /2 or k

X k

k=1

|J2k | ≥

p X

 |Jk | /2.

k=1

Depending on which one of the inequalities holds, we have found a finite disjoint subcollection =n of = such that p X  X 0 0 |I| ≥ |Jk | /2 ≥ (bn − an )/2. I∈=n

k=1

Hence for each positive integer n there exist a finite disjoint subcollection =n of I each of whose open intervals is a subset of (an , bn ) such that X 0 0 |I| ≥ (bn − an )/2. I∈=n

Summing both sides of the inequality, we have ∞ X ∞ X  X 0 0 |I| ≥ (bn − an ) /2 ≥ ε/2c. n=1 I∈=n

n=1

3

Therefore

S∞

= for which

n=1 =n ∞ X

= {I1 , I2 , I3 , . . .} is a countable disjoint subcollection of N X |In | ≥ ε/2c. Finally choose N so large that |Ik | > ε/dc,

n=1

k=1

where d ∈ R, d > 2 and dc ∈ R, dc > 2. Without loss of generality we can N X choose dc = 3 and |Ik | > ε/3 to complete the proof. k=1

In this lemma we have shown that for a subset E of (a, b) that is not of measure zero and = be an open covering for E, there exists a finite disjoint subcollection of open intervals of = 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 = be a collection of open subintervals of [a, b] that covers E − P . Then by adding to = sufficiently small open intervals centered at each point of P we can find a collection J such that J = = ∪ {K1 , K2 , . . . Ki } that covers E. Now E ⊆

S∞

k=1

Jk and by Lemma 1

∞ X

|Jk | > ε/cd,where cd > 2, c and

k=1

d ∈ R. Since = ⊆ J, we have

N1 X

|Ik | > ε/r, where r > cd > 2, r ∈ R.

k=1

Hence there exists a finite disjoint subcollection {I1 , I2 , . . . IN } of = such N X that |Ik | > ε/4. k=1

Clearly Remark 1 is an easy variation of Lemma 1. Lemma 2: Let f : [a, b] → R, let P = {x0 , x1 , x2 , . . . , xn } 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 f (xk ) − f (xk−1 ) < −A xk − xk−1 for each k in S, then n X

|f (xk ) − f (xk−1 )| > |f (b) − f (a)| + AL,

k=1

4

where L =

P

k∈S (xk

− xk−1 ).

Proof : Since f (a) ≤ f (b), it follows that, |f (b) − f (a)| = f (b) − f (a) =

n  X

 f (xk ) − f (xk−1 )

k=1  X  X = f (xk ) − f (xk−1 ) + f (xk ) − f (xk−1 ) k∈S /

k∈S

 X X < −A (xk − xk−1 ) + f (xk ) − f (xk−1 ) k∈S /

k∈S

≤ −AL +

n X

|f (xk ) − f (xk−1 )|

k=1

Therefore

n X

|f (xk ) − f (xk−1 )| > |f (b) − f (a)| + AL,

k=1

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 f (xk ) − f (xk−1 ) >A xk − xk−1 for each k in S. 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.

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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. 0

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) , y−x

Df (x) = lim inf y→x

f (y) − f (x) y−x

are the upper and lower derivatives of f at x, respectively. Clearly F is the countable union of the sets Er,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 Er,s has measure zero. Assume for a contradiction that for some choice of r and s the set E = Er,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. Clearly both A and B are positive and E = {x : x ∈ (a, b), g is continuous at x, Dg(x) > A and Dg(x) < −A}. P Now { P |g(xk )−g(xk−1 )|: P is partition of [a, b]} is bounded above. Namely, X X |g(xk ) − g(xk−1 )| = |f (xk ) − f (xk−1 ) − B(xk − xk−1 )| P

P

≤

X

|f (xk ) − f (xk−1 )| +

P

X P

= f (b) − f (a) + B(b − a). 6

B|(xk − xk−1 )|

Let T be the least upper bound for this set. Since both A and ε are positive, there exists a partition P = {x0 , x1 , x2 , . . . xn } of [a, b] for which n X

|g(xk ) − g(xk−1 )| > T −

k=1

Aε . 4

(1)

Now let x belongs to E − P , which means that x is in E ∩ (xk−1 , xk ) for some k. Since Dg(x) > A and Dg(x) < −A and g is continuous at x, we can choose ax and bx such that ax < x < bx , (ax , bx ) ⊆ (xk−1 , xk ), and g(xk ) − g(xk−1 ) either g(xk−1 ) ≤ g(xk ) and > A or g(xk−1 ) ≥ g(xk ) and xk − xk−1 g(xk ) − g(xk−1 ) < −A holds. Thus = = {(ax , bx ) : x ∈ E − P } is a collection xk − xk−1 of open subintervals of [a, b] that covers E − P . By Remark 1, there exists a finite disjoint subcollection {I1 , I2 , . . . , IN } of = such that N X ε |Ik | > . (2) 4 k=1 Now let Q = {y0 , y1 , y2 , . . . , yq } be the partition of [a, b] determined by the points of P and the endpoints of the intervals I1 , I2 , . . . , IN . For each [xk−1 , xk ] containing at least one of the intervals {I1 , I2 , . . . , IN }, we get by Lemma 2 that X |g(yi ) − g(yi−1 )| > |g(xk ) − g(xk−1 )| + ALk , (3) [yi−1 ,yi ]⊆[xk−1 ,xk ]

where the summation is taken over the closed intervals determined by Q that are contained in [xk−1 , xk ] and Lk is the sum of the lengths of those intervals I1 , I2 , . . . , IN that contained in [xk−1 , xk ]. Summing inequality (3) over k and using (1) and (2), we get q X

|g(yk ) − g(yk−1 )| >

k=1

n X

|g(xk ) − g(xk−1 )| + A

k=1

>T−

N X

|Ik |

k=1

ε Aε = T, +A 4 4

which contradicts the definition of T . We have shown that the set F = {x : x ∈ (a, b), f is continuous at x and Df (x) > Df (x)} 7

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 E = {x : x ∈ (a, b) and Df (x) = ∞} has measure zero. 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 ax and bx such that ax < x < bx , (ax , bx ) ⊆ (a, b), and f (bx ) − f (ax ) > M. b x − ax Thus = = {(ax , bx ) : x ∈ E} covers E and byPLemma 1 contains a finite disjoint subcollection {I1 , I2 , . . . IN } such that N k=1 |Ik | > ε/3. For each k let Ik = (ak , bk ) . Since f is increasing, clearly f (b) − f (a) ≥

N  X



f (bk ) − f (ak ) >

k=1

N X

ε M (bk − ak ) > M . 3 k=1

Therefore f (b) − f (a) > M ε/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

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): 834-838. [2] R. P. Boas., A Primer of Real Funtions,(Rahway, New Jersey: Mathematical Association of America, 1960).

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