14.13. DIFFERENTIATION OF INCREASING FUNCTIONS 347

whenever yn ↓ x. Therefore, it would follow that the limit of these difference quotientswould exist and f would have a right derivative at x. Therefore, D+ f (x)> D+ f (x) if andonly if there is no right derivative. Similarly D− f (x) > D− f (x) if and only if there is noderivative from the left at x. Also, there is a derivative if and only if there is a derivativefrom the left, right and the two are equal. Thus this happens if and only if all Dini derivatesare equal.

The Lebesgue measure of single points is 0 and so we do not need to worry aboutwhether the intervals are closed in using Corollary 14.12.9.

Let ∆ f (I) = f (b)− f (a) if I is an interval having end points a < b. Now suppose{

J j}

are disjoint intervals contained in I. Then, since f is increasing, ∆ f (I)≥∑ j ∆ f (J j). In thisnotation, the above lemma implies that if D− f (x)> b or D+ f (x)> b, then for each ε > 0there is an interval J of length less than ε which is centered at x and ∆ f (J)

m(J) > b where m(J)is the Lebesgue measure of J which is the length of J. If either D− f (x) or D+ f (x) < a,the above lemma implies that for each ε > 0 there exists I centered at x with m(I)< ε and∆ f (I)m(I) < a. For example, if D− f (x)< a, there exists a sequence yn ↑ x with

f (yn)− f (x)yn− x

=f (x)− f (yn)

x− yn< a

so let In be the interval (yn,x) . For large n it is smaller than ε .

Theorem 14.13.3 Let f :R→R be increasing. Then f ′ (x) exists for all x off a setof measure zero.

Proof: Let Nab for 0 < a < b denote either{x : D+ f (x)> b > a > D+ f (x)

},{

x : D− f (x)> b > a > D− f (x)},

or {x : D− f (x)> b > a > D+ f (x)

},{

x : D+ f (x)> b > a > D− f (x)}.

The function f is increasing and so it is a Borel measurable function. Indeed, f−1 (a,∞)is either open or closed. Therefore, all these derivates are also Borel measurable, henceLebesgue measurable. Assume that Nab is bounded and let V be open with

V ⊇ Nab, m(Nab)+ ε > m(V )

By Corollary 14.12.9 and the above discussion, there are open, disjoint intervals {In} , eachcentered at a point of Nab such that

∆ f (In)

m(In)< a, m(Nab) = m(Nab∩∪iIi) = ∑

im(Nab∩ Ii)

Now do for Nab∩ Ii what was just done for Nab and get disjoint intervals J ji contained in Ii

with∆ f(

J ji

)m(

J ji

) > b, m(Nab∩ Ii) = ∑j

m(

Nab∩ Ii∩ J ji

)