9.13. DIFFERENTIATION OF INCREASING FUNCTIONS 267

Proof: The set of jumps J is countable so it is a Borel set of measure zero, an Fσ set.Since f is increasing, its only points of discontinuity are points where it has a jump. Henceit is continuous off this set J. f−1 ([c,∞)) is an interval of the form [d,∞) or (d,∞). Thusf is Borel measurable. Consider D+ f for x /∈ J. Let gr (x)≡ sup0<u≤r

f (x+u)− f (x)u . Thus it

is the supremum of functions continuous on JC. Hence if x ∈ g−1r (c,∞) ,c≥ 0, and x /∈ JC,

f (x+u)− f (x)u > c for some 0 < u≤ r. It follows that, since f is continuous at x, if x̂ is close

enough to x, it is also true that f (x̂+u)− f (x̂)u > c. Thus if x∈ g−1

r (c,∞)∩JC, then for some δ x

small enough, (x−δ x,x+δ x)⊆ g−1r (c,∞). Hence, g−1

r (c,∞)∩JC is the intersection of anopen set, the union of the intervals (x−δ x,x+δ x) for x ∈ g−1

r (c,∞)∩ JC, with JCa Borelset. It follows that gr is decreasing in r and is measurable because, since J is countable,[gr > c] is the union of a countable set with a Borel set. Thus for all x,

D+ f (x) = limr→0+

(sup

0<u≤r

f (x+u)− f (x)u

)= lim

r→0+(gr (x)) = lim

rn→0grn (x)

where rn is a decreasing sequence converging to 0. It follows that x→ D+ f (x) is Borelmeasurable as claimed because it is the limit of Borel measurable functions. Similar rea-soning shows that the other derivates are measurable also. ■

Theorem 9.13.4 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)}

From the above lemma, Nab is measurable. Assume that Nab is bounded and let V be openwith V ⊇ Nab, m(Nab)+ ε > m(V ) . By Corollary 9.12.3 and the above discussion, thereare open, disjoint intervals {In} , each centered at a point of Nab such that

2∆ 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

with2∆ f

(J j

i

)m(

J ji

) > b, m(Nab∩ Ii) = ∑j

m(

Nab∩ Ii∩ J ji

)Then

a(m(Nab)+ ε)> am(V )≥ a∑i

m(Ii)> ∑i

2∆ f (Ii)≥∑i

∑j

2∆ f(

J ji

)≥ b∑

i∑

jm(

J ji

)≥ b∑

i∑

jm(

J ji ∩Nab

)= b∑

im(Nab∩ Ii) = bm(Nab)

Since ε is arbitrary and a < b, this shows m(Nab) = 0. If Nab is not bounded, apply theabove to Nab∩ (−r,r) and conclude this has measure 0. Hence so does Nab.