1. No category

Vitali coverings and differentiating increasing functions
These notes follow the treatment in Royden, Real Analysis.
1. Vitali coverings
The following lemma is a version of the Vitali covering lemma adapted to the case of
possibly non-measurable subsets of Rd . The sets to which we will apply this construction—
sub-level sets of the derivates, or Dini number functions, of an increasing function of
bounded variation—are in fact measurable, but the proof that they are measurable is
not at all short, and so we will proceed to prove the Vitali covering lemma without the
assumption of measurability.
We begin by recalling that for every E ⊂ Rd with exterior measure m∗ (E) < ∞, there
is an open O ⊃ E so that m(O) < m∗ (E)+. Moreover, exterior measure m∗ is countably
sub-additive. These are the only properties we will need. (Recall that if in addition E is
measurable, then m(O − E) < , and we have countable additivity for disjoint measurable
sets.)
Let B be a collections of balls of finite positive radius in Rd . It will be useful for us to
assume the balls in B are closed. Given a subset E ⊂ Rd , we say B is a Vitali covering
of E if for all x ∈ E and > 0, there is a ball B ∈ B which contains x and has radius
r(B) < .
Lemma 1 (Vitali covering). Let E ⊂ Rd have finite exterior measure, and let B be a
Vitali covering of E. Let > 0. Then there is a finite disjoint collection {B1 , . . . , BN } of
balls in B so that
!
N
G
Bn < .
m∗ E −
n=1
Proof. Since E has finite exterior measure, we may choose O to be an open set of finite
measure containing E. The set of ball in B which are contained in O is still a Vitali
covering, and so we assume that every ball in B is contained in O.
Now we choose by induction a disjoint sequence of balls Bn in B. In general, we expect
Bn to be an infinite sequence, although it may be finite. Assume B1 , . . . , Bn are disjoint
and have been chosen already. Then define
B0 = B,
Bn = {B ∈ B : B ∩ Bi = ∅ for i = 1, . . . , n} for n ≥ 1,
kn = sup r(B).
B∈Bn
Note kn < ∞, since all B ⊂ O and m(O) < ∞. Consider x ∈ E − tni=1 Bi . The Vitali
covering condition, together with the fact that tni=1 Bi is compact, show there is a ball
B ∈ Bn which contains x. Now there are two possibilities. If kn = 0, then there is no
x ∈ E − tni=1 Bi , and so E ⊂ tni=1 Bi and the proof is complete. If kn > 0, then choose
Bn+1 ∈ Bn so that r(Bn+1 ) > 21 kn .
1
2
Now if the process terminates at any step, we are finished. So we may assume we have
produced a disjoint sequence {Bn }∞
n=1 . Note that
∞
X
m(Bn ) = m(tBn ) ≤ m(O) < ∞,
n=1
which implies that m(Bn ) → 0, and thus the radii r(Bn ) → 0. Moreover, we may choose
N so that
∞
X
m(Bn ) <
n=N +1
.
5d
Now consider the set R = E − tN
i=1 Bi . We will show m∗ (R) < . To show this, consider
x ∈ R, consider as above a ball B ∈ Bn containing x, for n ≥ N (note we are guaranteed
that there is such a B ∈ BN , but we consider the possibility of B ∈ Bn for larger n as
well). Then by the choice of Bn+1 ,
r(B) ≤ kn < 2r(Bn+1 ).
(1)
We also claim that B ∩ Bn 6= ∅ for some n > N . To prove the claim, note that
r(Bn ) → 0 and the bound (1) imply that for n large enough, B ∈
/ Bn , or equivalently,
there is at least one n so that B ∩ Bn 6= ∅. This proves the claim, and we may choose n
to be the smallest integer so that B ∩ Bn 6= ∅. By our choice of B, n > N .
If c(B) denotes the center of the ball, we have by (1) that the distance
kx − c(Bn )k ≤ kx − c(B)k + kc(B) − c(Bn )k ≤ r(B) + [r(B) + r(Bn )] ≤ 5r(Bn ).
˜n to be the closed ball with the same center as Bn and 5 times the radius,
So if we define B
˜n for some n > N . Thus
we find x ∈ B
R⊂
∞
[
˜i ,
B
i=N +1
and so
m∗ (R) ≤ m
∞
G
i=N +1
by our choice of N .
˜i
B
!
≤
∞
X
i=N +1
˜i ) = 5d
m(B
∞
X
m(Bi ) < i=N +1
3
2. Differentiating functions of bounded variation
In order to apply this theory, we revert to the case d = 1. Define the derivates, or Dini
numbers, of a function f at x by
F (x + h) − F (x)
,
h
D+ (F )(x) = lim sup ∆h (F )(x),
∆h (F )(x) =
h→0+
D+ (F )(x) = lim inf
∆h (F )(x),
+
h→0
−
D (F )(x) = lim sup ∆h (F )(x),
h→0−
D− (F )(x) = lim inf
∆h (F )(x),
−
h→0
Note that F is differentiable at x if and only if the four Dini numbers at x are finite
and equal.
Theorem 1. Let F be an increasing real-valued function on the interval [a, b]. Then F
is differentiable almost everywhere, the derivative F 0 is measurable, and
Z b
F 0 (x) dx ≤ F (b) − F (a).
a
Proof. We first show that F is differentiable almost everywhere. For simplicity, we only
consider the case of E = {x : D+ (F )(x) > D− (F )(x)}
S to show m∗ (E) = 0. There are
other cases which are handled similarly. Write E = u,v∈Q Eu,v , where
Eu,v = {x : D+ (F )(x) > u > v > D− (F )(x)}.
Since F is increasing, each of the Dini numbers is nonnegative, and thus u > v > 0. Since
E is a countable union, it suffices to prove m∗ (Eu,v ) = 0. Note we have not proved that
D+ (F ) and D− (F ) are measurable. Let s = m∗ (Eu,v ).
Let > 0. Consider an open set O ⊃ Eu,v so that m(O) < m∗ (Eu,v ) + . Then by
the definition of D− (F )(x) and D− (F )(x) < v, there are arbitrarily small intervals of the
form [x − h, x] for h > 0 which are contained in O so that f (x) − f (x − h) < vh. The set
of all such intervals thus forms a Vitali covering of Eu,v .
Lemma 1 then applies to find a finite disjoint collection I1 , . . . , IN of these intervals so
N
◦
that m∗ [Eu,v − (tN
n=1 In )] < . This implies that A = Eu,v ∩ (tn=1 In ) has exterior measure
greater than s − (recall that countable sub-additivity applies for m∗ ). Here In◦ is the
interior of In . We discard the set of endpoints of In , which has measure zero.
Let In = [xn − hn , xn ] and compute
N
X
n=1
[F (xn ) − F (xn − hn )] < v
N
X
hn < v m(O) < v(s + ).
n=1
Now if y ∈ A, we apply the definition of D+ (F )(x) and the inequality D+ (F )(x) > u
to find a Vitali covering of A of the form [y, y + k] for arbitrarily small k > 0 so that each
such interval is contained in one of the In◦ and F (y + k) − F (y) > uk. Lemma 1 applies to
4
show there is a finite disjoint collection J1 , . . . , JM of these intervals so that the exterior
measure of A ∩ (tM
m=1 Jm ) is at least s − 2. Summing over these intervals gives
M
X
M
X
F (ym + km ) − F (ym ) > u
m=1
km > u(s − 2).
m=1
Now each interval Jm = [ym , ym + km ] is contained in some interval In = [xn − hn , xn ],
and since F is increasing, we have for each n
X
F (ym + km ) − F (ym ) ≤ F (xn ) − F (xn − hn ).
m:Jm ⊂In
This implies
N
X
F (xn ) − F (xn − hn ) ≥
n=1
M
X
F (ym + km ) − F (ym ),
m=1
which in turn implies v(s + ) > u(s − 2) for all > 0. Letting → 0+ , we see vs ≥ us.
Since u > v, this implies m∗ (Eu,v ) = s = 0. Thus m∗ (E) = 0 and F is differentiable
almost everywhere.
Therefore,
F (x + h) − F (x)
h→0
h
is defined almost everywhere, and F is differentiable wherever f is finite. For x > b, define
F (x) = F (b), and define
fn (x) = n[F (x + n1 ) − F (x)].
Then fn (x) → f (x) for almost every x and so f is measurable. Since F is increasing, each
fn ≥ 0. Thus we may apply Fatou’s Lemma to find
Z b
Z b
f (x) dx ≤ lim inf
fn (x) dx
n→∞
a
a
Z b
= lim inf n
[F (x + n1 ) − F (x)] dx
n→∞
" aZ 1
#
1
Z
f (x) = lim
b+ n
= lim inf n
n→∞
b
"
= lim inf F (b) − n
n→∞
a+ n
F (x) dx − n
F (x) dx
a
Z
1
a+ n
F (x) dx
#
a
≤ F (b) − F (a).
The last inequality is because F is increasing. Therefore, f is integrable, and so is
finite almost everywhere. So F is differentiable almost everywhere and F 0 = f almost
everywhere.
Corollary 2. If F is a function of bounded variation, then F is differentiable almost
everywhere.
Proof. Apply the previous theorem and the fact that F is the difference of two increasing
functions.