346 CHAPTER 14. THE LEBESGUE INTEGRAL
Now these balls Bk are disjoint and contained in a bounded set so letting 1 > θ > θ 1, if nis large enough, and since the sum on the right is the tail of a convergent series,
∞
∑k=n
m(Bk)≤ (θ −θ 1) m̄(F)
Thus there exists n1 such that m̄(
F \∪n1j=1B j
)< θ m̄(F) . If m̄
(F \∪n1
j=1B j
)= 0, stop.
Otherwise, do for F \∪n1j=1B j exactly the same thing that was done for F. Since ∪n1
j=1B j isclosed, you can arrange to have the approximating open set be contained in the open set(∪n1
j=1B j
)C. It follows there exist disjoint closed balls from C called Bn1+1, · · · ,Bn2 such
thatm̄((
F \∪n1j=1B j
)\∪n2
j=n1+1B j
)< θ m̄
(F \∪n1
j=1B j
)< θ
2m̄(F)
continuing this way and noting that limn→∞ θn = 0 while m(F)< ∞, this shows the desired
result. Either the process stops because m̄(
F \∪nkj=1B j
)= 0 or else you obtain an infinite
sequence {Bk} and m̄(
F \∪∞j=1B j
)≤ m̄
(F \∪nk
j=1B j
)≤ θ
km̄(F) for each k, showing that
m̄(
F \∪∞j=1B j
)= 0.
14.13 Differentiation of Increasing FunctionsAs a spectacular application of the covering theorem, is the famous theorem that an increas-ing function has a derivative a.e. Here the a.e. refers to Lebesgue measure, the Stieltjesmeasure from the increasing function F (x) = x.
I will write yn ↑ x to mean limn→∞ yn = x and yn < x,yn ≤ yn+1. I will write yn ↓ x tomean limn→∞ yn = x and yn > x,yn ≥ yn+1.
Definition 14.13.1 The Dini derivates are as follows. In these formulas, f is areal valued function defined on R. yn ↓ x refers to a decreasing sequence as just described.
D+ f (x) ≡ sup{
lim supn→∞
f (yn)− f (x)yn− x
: yn ↓ x},
D+ f (x) ≡ inf{
lim infn→∞
f (yn)− f (x)yn− x
: yn ↓ x},
D− f (x) ≡ sup{
lim supn→∞
f (yn)− f (x)yn− x
: yn ↑ x},
D− f (x) ≡ inf{
lim infn→∞
f (yn)− f (x)yn− x
: yn ↑ x}
The notation means that the sup and inf refer to all sequences of the sort described in {} .
Lemma 14.13.2 The function f : R→ R has a derivative if and only if all the Diniderivates are equal for any sequence just described.
Proof: If D+ f (x) = D+ f (x) , then for any yn ↓ x, it must be the case that
lim supn→∞
f (yn)− f (x)yn− x
= lim infn→∞
f (yn)− f (x)yn− x
= limn→∞
f (yn)− f (x)yn− x
= D+ f (x) = D+ f (x)