Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 4: The Lebesgue Differentiation Theorem
In this lecture, we will see how the covering theorems proven in Lecture 3 give rise to
a fundamental differentiability result.
1. Lebesgue’s differentiation theorem for monotone functions
The following theorem is at the heart of modern analysis. It is deeper than one might
expect.
Theorem 1.1. Let f : R → R be a monotone function. Then f is differentiable almost
everywhere with respect to Lebesgue measure m.
Proof. It suffices to show that given a bounded open interval (a, b) ⊆ R, the function
f is differentiable at m-almost every point of (a, b); we may also assume that f is nondecreasing.
Let x ∈ (a, b) and consider the functions
f (x + t) − f (x)
,
h→0 0<|t|<h
t
Df (x) = lim sup
f (x + t) − f (x)
.
h→0 0<|t|<h
t
Df (x) lim inf
For each α > 0 and any interval (a0 , b0 ) ⊆ (a, b)
Eα = {x ∈ (a0 , b0 ) : Df (x) ≥ α}.
We first claim that for each α > 0,
f (b0 ) − f (a0 )
.
α
To this end, fix any 0 < α0 < α and let F be the collection of all closed intervals
[c, d] ⊆ (a0 , b0 ) such that
f (d) − f (c)
≥ α0 .
d−c
By definition, this forms a fine cover of Eα . Hence, we may apply the Vitali Covering
Theorem to F and find a countable, disjointed subcollection {[ci , di ]}i∈N such that
!
[
m Eα \ [ci , di ] = 0.
(1.1)
m (Eα ) ≤
i∈N
Since f is increasing and the subcollection is disjoint,
X
X f (di ) − f (ci )
f (b0 ) − f (a0 )
≤
.
m(Eα ) ≤
(di − ci ) ≤
α0
α0
i∈N
i∈N
Letting α0 tend to α proves claim 1.1. Note that E∞ ⊆ Eα for every α > 0, and so
m(E∞ ) = 0.
Now, fix numbers 0 < α < β < ∞, and define the set
E := Eα,β := {x ∈ (a, b) : Df (x) > α > β > Df (x)}.
1
For > 0, we may find an open set U ⊆ (a, b) such that E ⊆ U and m(U ) < m(E) + .
Let F now denote the collection of intervals [c, d] ⊆ U such that
f (d) − f (c)
< β.
d−c
Since Df > β on E, the collection F is a fine covering of E. Hence we may find a
countable, disjointed subcollection {[ci , di ]}i∈N such that that
!
[
m E\ [ci , di ] = 0.
i∈N
0
Thus applying the claim (1.1) with (a , b0 ) = (ci , di ) for each i ∈ N,
X
X f (di ) − f (ci )
m(E) ≤
m(E ∩ (ci , di )) ≤
α
i∈N
i∈N
Xβ
≤
(di − ci )
α
i∈N
β
m(U )
α
β
≤ (m(E) + ).
α
Since β < α, this is a contradiction when is sufficiently small unless m(E) = 0.
Now, note that the set of points at which f fails to be differentiable is contained in the
union of E∞ and countably many sets of the form Eα,β .
≤
2. Lebesgue’s differentiation theorem for metric measure spaces
A very similar argument gives the following theorem in a very general setting.
Theorem 2.1. Let (X, d, µ) be a Vitali space, and let f be a locally integrable function
on X. Then for µ-almost every point x ∈ X, it holds that
Z
lim −
f dµ = f (x).
r→0 B(x,r)
It is worth remarking how this is a differentiation result. Roughly speaking, one should
think of the integral as the analogue of the monotone function, and the function f itself
as the derivative of the monotone function.
Proof. Let E be the set of points in X such that
Z
lim −
f dµ = f (x)
r→0 B(x,r)
fails to hold. It suffices to consider the case that E is contiained in some ball B on which
f is integrable.
We first claim that for each α > 0, if A ⊆ B is a set such that for all a ∈ A,
Z
lim sup −
f dµ ≥ α,
r→0
B(a,r)
then
R
(2.1)
µ(A) ≤
A
f dµ
α
Similarly, if A ⊆ B is a set such that for all a ∈ A
Z
f dµ ≤ β,
lim inf −
r→0
B(a,r)
then
R
A
µ(A) ≥
(2.2)
f dµ
.
β
Suppose that these claims hold. By (2.1),
Z
lim sup −
f dµ < ∞
r→0
B(a,r)
at µ-almost every point of B.
For 0 < α < β < ∞, set
Z
Eα,β := x ∈ (a, b) : lim sup −
r→0
Then
Z
f dµ ≥ α > β ≥ lim inf −
r→0
B(x,r)
Z
αµ(Eα,β ) ≤ −
f dµ .
B(x,r)
f dµ ≤ βµ(Eα,β ),
Eα,β
and so µ(Eα,β ) = 0. This implies that
Z
g(x) := lim −
r→0 B(x,r)
f dµ
exists and is finite µ-almost everywhere in B. To see that g and f agree µ-almost everywhere, let F ⊆ B, > 0, and n ∈ N be arbitrary, and define
An = {x ∈ F : (1 + )n ≤ g(x) < (1 + )n+1 }.
Then
Z
XZ
g dµ =
F
n∈N
g dµ ≤
An
X
n+1
(1 + )
µ(An ) ≤ (1 + )
n∈N
XZ
n∈N
Z
f dµ ≤ (1 + )
An
f dµ.
F
Since and F are arbitrary, this implies that f = g almost everywhere.
It now suffices to prove the claim (2.1); the claim (2.2) is proven analogously. Since µ
is Borel regular, there is an open set U ⊆ A such that
µ(U ) ≤ µ(A) + .
By the Vitali covering theorem, we may find a countable, disjointed collection of closed
balls {Bi }i∈N such that
!
[
µ A\
Bi = 0
i∈N
and
Z
− f dµ ≤ α + .
Bi
Then
Z
f dµ ≤
A
XZ
i∈N
Bi
≤
X
(α + )µ(Bi ) ≤ (1 + )(µ(A) + ).
i∈N
Letting tend to 0 yields the desired result.
We can derive the following interesting corollary:
Corollary 2.2. Let (X, d, µ) be a doubling metric measure space, and let E ⊆ X be a
measurable set. Then, for almost every x ∈ X,
µ(E ∩ B(x, r))
lim
= χE (x).
r→0
µ(B(x, r)
This result says that from the perspective of measure, when zooming in on a point in
a set E, we eventually see only E, almost everywhere.
Definition 2.3. A point x for which the conclusion of Corollary 2.2 holds is called a
Lebesgue point or a density point for E.