Examensarbete
Lebesgue points, H¨
older continuity
and Sobolev functions
Lebesgue points, H¨
older continuity
and Sobolev functions
Applied Mathematics, Link¨opings Universitet John Karlsson
LiTH - MAT - EX - - 08 / 15 - - SE
Examensarbete: 30 hp Level: D
Supervisor: Jana Bj¨orn,
Applied Mathematics, Link¨opings Universitet Examiner: Jana Bj¨orn,
Matematiska Institutionen 581 83 LINK ¨OPING SWEDEN January 2008 x x http://www.ep.liu.se/exjobb/mai/2008/tm/015/ LiTH - MAT - EX - - 08 / 15 - - SE
Lebesgue points, H¨older continuity and Sobolev functions
John Karlsson
This paper deals with Lebesgue points and studies properties of the set of Lebesgue points for various classes of functions. We consider continuous functions, L1functions
and Sobolev functions. In the case of uniformly continuous functions and H¨older continuous functions we develop a characterization in terms of Lebesgue points. For Sobolev functions we study the dimension of the set of non-Lebesgue points.
Lebesgue point, Hausdorff dimension, Hausdorff measure, H¨older continuity, Maximal
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Abstract
This paper deals with Lebesgue points and studies properties of the set of Lebesgue points for various classes of functions. We consider continuous func-tions, L1 functions and Sobolev functions. In the case of uniformly
continu-ous functions and H¨older continucontinu-ous functions we develop a characterization in terms of Lebesgue points. For Sobolev functions we study the dimension of the set of non-Lebesgue points.
Keywords: Lebesgue point, Hausdorff dimension, Hausdorff measure, H¨older continuity, Maximal function, Poincar´e inequality, Sobolev space, Uniform continuity.
Acknowledgements
I would like to thank my supervisor, Jana Bj¨orn, for guidance and for showing seemingly endless patience.
Notation
The symbol Ω will generally denote an open set in Rn and S will denote any
subset of Rn. Points in Rn are denoted by x = (x
1, ..., xn), where xi ∈ R, 1 ≤
i ≤ n. In most cases f is used as the name for functions, however when studying functions in the Sobolev space W1,p(Ω), u is used instead. The gradient of a
real-valued function u is denoted by ∇u(x) = µ ∂u(x) ∂x1 , ..., ∂u(x) ∂xn ¶ . The open ball with center x and radius r is denoted by
B(x, r) = {y : |x − y| < r}. If S ⊂ Rn then |S| will denote the Lebesgue measure of S.
Unless otherwise stated, the letter C denotes various positive constants whose exact values are unimportant and may vary with each usage.
Contents
1 Introduction 1
1.1 Topics covered . . . 1
2 Lebesgue points 3
2.1 Definition and examples . . . 3 2.2 Uniform Continuity . . . 8 2.3 H¨older continuity . . . 10
3 Lebesgue points for L1-functions 13
3.1 Introducing L1 . . . . 13
3.2 The Hardy–Littlewood maximal function . . . 14 3.3 The set of Lebesgue points for L1-functions . . . . 15
4 Lebesgue points and Sobolev spaces 17
4.1 Introducing Sobolev spaces . . . 17 4.2 H¨older continuous Sobolev functions . . . 19 4.3 Lebesgue points and the Hausdorff dimension . . . 22
5 Summary and thoughts of further studies 27
5.1 Summary . . . 27 5.2 Further studies . . . 27
Chapter 1
Introduction
This text is written in LATEX as a master of science thesis at Link¨opings universitet
by John Karlsson with Jana Bj¨orn as supervisor and examiner, during 2006–2008. In this thesis a definition of Lebesgue points is given and some properties of the set of Lebesgue points and the set of non-Lebesgue points are studied. We show that continuous functions have Lebesgue points everywhere but that the converse implication is not true. In the uniformly continuous case we get an equivalence between uniformly continous functions and functions for which the decay rate of the corresponding Lebesgue point integral is uniform. A similar characterization is obtained for H¨older continuous functions. Later we consider locally integrable functions. In this case we will see that a funtion f has Lebesgue points almost everywhere. Lastly we study functions in the Sobolev space W1,p(Rn) where we
use the results from earlier chapters to show that for p > n these Sobolev functions are H¨older continuous. We also introduce Hausdorff dimension and a Poincar´e inequality. By using this inequality and other lemmas we are able to obtain an upper bound for the Hausdorff dimension of the set of non-Lebesgue points of Sobolev functions.
1.1
Topics covered
This thesis consists of this introduction and four other chapters. It is organized as follows:
Chapter 2: We define Lebesgue points and study the relationship between continuous functions and functions having Lebesgue points everywhere. We are also considering different types of continuity such as uniform con-tinuity or H¨older concon-tinuity. In the end of the chapter we prove that a function can have Lebesgue points everywhere but still be discontinuous at many points.
Chapter 3: In this chapter we define the L1-functions and we introduce the
Hardy–Littlewood maximal function that is used later to study the rela-tionship between L1-functions and Lebesgue points. We arrive at a
classi-cal result that every f ∈ L1(Rn) has Lebesgue points almost everywhere.
Chapter 4: We narrow the set of available functions and enter the Sobolev space W1,p(Rn). We then study the relationship between this class of
2 Chapter 1. Introduction
functions and Lebesgue points and as a result we get an upper bound for the dimension of the set of non-Lebesgue points. For p > n we also study the possibility of choosing another representant of f having Lebesgue points everywhere. This representant of f turns out to be H¨older con-tinuous.
Chapter 5: We summarize the results derived in the earlier chapters and give examples of problems that could be studied further in the subject of Lebesgue points.
Chapter 2
Lebesgue points
In this section we start by defining Lebesgue points and then study the relation between Lebesgue points and continuity properties of functions.
2.1
Definition and examples
Definition 2.1.1. A Lebesgue point of an integrable function f : Rn→ R is a
point x0∈ Rn satisfying lim r→0 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx = 0, (2.1)
i.e. the mean value oscillation of f tends to zero at x0.
The equation (2.1) is a form of continuity in integral average sense. Conti-nuity turns out to be a stronger property than all points being Lebesgue points as the following proposition shows.
Proposition 2.1.2. If f : Rn→ R is continuous then equation (2.1) holds for
all x0∈ Rn i.e. all x0 are Lebesgue points of f .
Proof. We want to show that for any ² > 0 and any x0 ∈ Rn there exists a
δ > 0 so that 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < ²
whenever r < δ. By the definition of continuity there exists a δ0 > 0 so that
|f (x) − f (x0)| < ² whenever |x − x0| < δ0. For all r < δ0, we then have
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx ≤ 1 |B(x0, r)| Z B(x0,r) ² dx = ². This concludes the proof.
There is not an equivalence between having Lebesgue points everywhere and being continuous i.e. the converse of Proposition 2.1.2 does not hold. The following examples will illustrate that.
4 Chapter 2. Lebesgue points
Example 2.1.3. If all x are Lebesgue points of f then it does not imply that f is continuous.
To show this we find a function f where equation (2.1) holds for every point but has at least one point where f is not continuous, in this example f : R → R. We shall prove that f (x) =P∞n=1un(x) where
un(x) = 2n3x − 2n2, if 1 n ≤ x ≤ n1 +2n13; −2n3x + 2n2+ 2, if 1 n +2n13 ≤ x ≤ n1 +n13; 0, otherwise;
has the desired qualities.
The function f is not continuous at x = 0 since f (0) = 0 and arbitrarily close to 0 there are points where f (x) = 1. To convince ourselves that f is continuous for x > 0 we observe that for every n ≥ 1, the functions un(x) and
un+1(x) have pairwise disjoint supports. Indeed we have
1 n + 1+ 1 (n + 1)3 − 1 n = − n2+ n + 1 (n + 1)3n < 0, i.e. 1 n + 1+ 1 (n + 1)3 < 1 n,
which proves the claim. Since f is continuous for x > 0 we know that every x > 0 is a Lebesgue point of f according to Theorem 2.1.2. Now we only have
2.1. Definition and examples 5
to prove that x = 0 is a Lebesgue point as well. For x = 0 we have 1 2r Z r −r |f (t) − f (0)| dt = 1 2r Z r 0 |f (t)| dt = 1 2r Z r 0 ∞ X n=1 un(t) dt = 1 2r ∞ X n=1 Z r 0 un(t) dt,
where the last equality follows from the Beppo Levi theorem since
∞ X n=1 Z R un(t) dt = ∞ X n=1 1 2n3 < ∞.
For 0 < r < 1, there exists an integer N > 0 so that 1
N > r ≥ N +11 , which then gives 1 2r ∞ X n=1 Z r 0 un(t) dt < N + 1 2 ∞ X n=N Z r 0 un(t) dt < N + 1 2 ∞ X n=N Z R un(t) dt = N + 1 2 ∞ X n=N 1 2n3.
By estimating this sum with an integral we get N + 1 2 ∞ X n=N 1 2n3 < N + 1 2 Z ∞ N −1 1 2x3dx = N + 1 2 1 4(N − 1)2 ≤ N + 1 8(N − 1)2.
This tends to 0 as N → ∞, or equivalently as r → 0. This shows that x = 0 is a Lebesgue point of f and concludes the proof.
The previous example showed that a function can be discontinuous at one point but still have Lebesgue points everywhere. This can be generalised so that a function is discontinuous at an infinite number of points but still has Lebesgue points everywhere.
Example 2.1.4. We can get a function which is discontinuous at infinitely many points by creating a similar function as in Example 2.1.3. This function g(x) has similar triangle functions accumulating at every 21n, n = 1, 2 . . .. We
get this by using the functions un(x) from Example 2.1.3 and constructing the
functions gn(x) = ∞ X i=2n ui(x − 2−(n−1)).
Here gn(x) is discontinuous at the point 2−(n−1) and thus the function g(x) =
P∞
n=1gn(x) has the desired qualities.
Example 2.1.5. We can also construct a function which is discontinuous at uncountably many points and still every point is a Lebesgue point of f . For this purpose we will study the Cantor set described e.g. in Abott[1] or Rudin [17].
The usual Cantor set, C, is constructed by starting with the interval [0, 1] and then removing the middle third, obtaining the set C . The set C is the
6 Chapter 2. Lebesgue points
two intervals gives C2. Recursively we define Cn as the set obtained when the
middle thirds of Cn−1’s 2n−1 constituent intervals are removed. We will call
the center point of these constituent intervals cni where 1 ≤ i ≤ 2n−1. For the
Cantor set the length of each removed interval from Cn−1to obtain Cnis 1/3n,
i.e. Cn has gaps of sizes 1/3i, i = 1, . . . , 2n−1 with centers cni.
Now construct the sequence of functions fn where fn(x) = 0 when x ∈ Cn
and fn(x) consists of spine-like functions with height 1, whenever x /∈ Cn. Let
gn(x) denote the spine function first introduced in generation n, i.e. the spine
function with base 1/3n. This spine function will be a translation of xn. The
expression for gn(x) with the spine centered at 0 will be
gn(x) = (2 · 3nx + 1)n, if − 1 2·3n ≤ x ≤ 0; (1 − 2 · 3nx)n, if 0 ≤ x ≤ 1 2·3n; 0, otherwise; We can write fn(x) as fn(x) = n X i=1 2i−1 X j=1 gi(x − cij).
2.1. Definition and examples 7
From this it follows that
f1(x) = g1(x − 1/2)
f2(x) = f1(x) + g2(x − 1/6) + g2(x − 5/6)
f3(x) = f2(x) + g3(x − 1/18) + g3(x − 5/18) + g3(x − 13/18) + g3(x − 17/18)
and so on. A result that we will need later is the integral value of gn
Z 1
0
gndx = 1
3n(n + 1).
Let f = limn→∞fn. It is clear that f is not continuous at any x ∈ C since
f (x) = 0 for all x ∈ C and we can find a spine arbitrarily close to x. If x /∈ C then x is a Lebesgue point of f since f is continuous in a neighborhood of x. We will prove that all x ∈ C are Lebesgue points of f as well. Let x0∈ C, then
we must show that
1 |B(x0, r)| Z B(x0,r) |f (x)−f (x0)| dx = 1 |B(x0, r)| Z B(x0,r) f (x) dx → 0, as r → 0.
Let an = 31n be the length of the smallest removed interval in Cn. Also let
8 Chapter 2. Lebesgue points
R
Inf (x) dx will be called An. We then have by translation of f that
1 |In| Z In f (x) dx ≤ 1 |In| ¡Z [0,an] f (x) dx + An ¢ = 1 |In| ∞ X i=n+1 1 3i(i + 1)2 i−n+ An |In| ≤ 3n ∞ X i=n+1 1 3in2 i−n+ An |In| = 2 n+ An |In|.
We must now show that An/|In| → 0 when n → ∞. We know that An is at
most an integral of length 2·31n over a spine of generation n or lower. From the
symmetry we then have An |In| ≤ 3n Z 1 2·3n 0 (2 · 3α)αxαdx = 3n(2 · 3α)α (α + 1)(2 · 3n)α+1 = 3α(α−n) 2(α + 1) =: h(α) for some (integer) α ≤ n. We will show that h(α) → 0 when 1 ≤ α ≤ n and n → ∞. First assume that 1 ≤ α ≤ n/2. Then we have
h(α) = 3α(α−n) 2(α + 1) ≤ 3−n/2 4 → 0 when n → ∞. If n/2 ≤ α ≤ n we get h(α) = 3α(α−n) 2(α + 1) ≤ 1 n → 0 when n → ∞. We have now shown that
An
|In| → 0, when n → ∞.
Let an+1< r < an. We then have
1 |B(x0, r)| Z B(x0,r) f (x) dx ≤ 1 |In+1| Z In f (x) dx = |In| |In+1| 1 |In| Z In f (x) dx, thus if |In|/In+1| is bounded then the integral tends to 0. We have
|In|
|In+1|
= 3n+1 3n = 3.
From this it follows that all x0 in C are Lebesgue points for f .
2.2
Uniform Continuity
We have seen how continuity is related to Lebesgue points. In this section we will study the relation between uniform continuity and Lebesgue points. In particular, we will get an equivalence between uniform continuity and a quantitative version of Lebesgue points.
2.2. Uniform Continuity 9
Definition 2.2.1. A function f : Rn → R is uniformly continuous in S ⊂ Rn
if for every ² > 0 there exists δ > 0 so that |f (x) − f (y)| < ² whenever x, y ∈ S and |x − y| < δ.
Uniformly continuous functions are continuous. Proposition 2.1.2 implies that they have Lebesgue points everywhere. The following proposition shows that more is true for uniformly continuous functions. It shows that the speed with which the Lebesgue point integrals converge to zero is uniform.
Proposition 2.2.2. If f : Rn → R is uniformly continuous on Rn then the
convergence in equation (2.1) is uniform as well, i.e. for every ² > 0 there exists a δ > 0 so that for all x0∈ Rn and all 0 < r < δ,
1 |B(x0, r)|
Z
B(x0,r)
|f (x) − f (x0)| dx < ².
Proof. It follows from the definition of uniform continuity that there exists a δ > 0 so that |f (x) − f (x0)| < ² whenever |x − x0| < δ. For all r < δ this gives
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < 1 |B(x0, r)| Z B(x0,r) ² dx = ². This concludes the proof.
The converse is also true.
Theorem 2.2.3. If the convergence in equation (2.1) is uniform then f is uni-formly continuous.
Proof. We will show that if f is not uniformly continuous then it follows that the integral in equation (2.1) does not tend to 0 uniformly, which is equivalent to the statement of the theorem. Assuming that f is not uniformly continuous it follows that there exists an ²0 > 0 so that for every δ > 0 there exist x0, y0
such that |f (x0) − f (y0)| > ²0 and |x0− y0| < δ. We shall prove that there
exists ² > 0 so that for every δ > 0 there exists an x0 and r < δ so that
1 |B(x0, r)|
Z
B(x0,r)
|f (x) − f (x0)| dx > ². (2.2)
Given any δ > 0 choose x0and y0so that |f (x0)−f (y0)| > ²0and |x0−y0| < δ/2.
Let r = 2|x0− y0| < δ and study the integrals
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx + 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx.
We can do a downwards estimate by using the facts that B(x0, r/2) ⊂ B(x0, r),
B(x0, r/2) ⊂ B(y0, r) and |B(x0, r)| = |B(y0, r)|. It then follows that
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx + 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx ≥ 1 Z ¡ |f (x) − f (x )| + |f (x) − f (y )|¢dx.
10 Chapter 2. Lebesgue points
The triangle equality yields
|f (x0) − f (y0)| = |f (x0) − f (x) − f (y0) + f (x)| ≤ |f (x0) − f (x)| + |f (y0) − f (x)|
and it follows that 1 |B(x0, r)| Z B(x0,r/2) (|f (x) − f (x0)| + |f (x) − f (y0)|) dx ≥ 1 |B(x0, r)| Z B(x0,r/2) |f (y0) − f (x0)| dx ≥ 1 |B(x0, r)| Z B(x0,r/2) ²0dx = 1 |B(x0, r)||B(x0, r/2)|²0= ²0 2n.
To sum up we have shown that for every δ > 0 there exist x0, y0 and r < δ so
that |x0− y0| < δ and 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx + 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx ≥ ²0 2n.
From this it follows that at least one of the following statements is true 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx ≥ ²0 2n+1, 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx ≥ ²0 2n+1.
That means that for every δ > 0 there exist x0and 0 < r < δ so that
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx ≥ ²0 2n+1 =: ².
The theorem then follows.
An important thing to note here is the difference between the continuous case and the uniformly continuous case. In the uniformly continuous case there is an equivalence between uniform continuity and uniform convergence in the Lebesgue point integral while in the continuous case there is no such equivalence. Remark 2.2.4. Proposition 2.2.2 and Theorem 2.2.3 show that a function is uniformly continuous if and only if the speed of the convergence in equation (2.1) is uniform.
2.3
H¨
older continuity
A special case of uniform continuity is the so called H¨older continuity. It tells us something about just how continuous a function is.
Definition 2.3.1. A function f : Rn → R is H¨older continuous with exponent
α, 0 < α ≤ 1, if there exists C > 0 so that for every x, y ∈ Rn,
2.3. H¨older continuity 11
Remark 2.3.2. If α = 1 in Definition 2.3.1 the function is said to be Lipschitz continuous.
It will be shown in the same way as in Proposition 2.2.2 that H¨older conti-nuity implies a similar H¨older type estimate for the Lebesgue points.
Theorem 2.3.3. If f : Rn → R is H¨older continuous with exponent α, then
for every x0∈ Rn, 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < Drα
for some D independent of x0.
Proof. It follows from the definition of H¨older continuity that 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < 1 |B(x0, r)| Z B(x0,r) C|x − x0|αdx = C |B(0, r)| Z B(0,r) |x|αdx ≤ C |B(0, r)| Z B(0,r) |r|αdx = Crα. The statement follows.
The converse of Theorem 2.3.3 also holds, as the following result shows. Theorem 2.3.4. Assume that there exist D > 0 and 0 < α ≤ 1 so that for every x0∈ Rn and r > 0, 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < Drα.
Then f is H¨older continuous with exponent α.
Proof. We will show that if f is not H¨older continuous then there can not exist any D > 0 so that 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx < Drα
holds for every x0 ∈ Rn and every r > 0. Assume therefore that for every
C0> 0 there exist x0, y0∈ Rn so that
|f (x0) − f (y0)| > C0|x0− y0|α.
Put r = 2|x0− y0|. It then follows as in the proof of Theorem 2.2.3 that
1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx + 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx ≥ 1 |B(x0, r)| Z B(x0,r/2) |f (y0) − f (x0)| dx ≥ 1 |B(x0, r)| Z B(x0,r/2) C0|y0− x0|αdx
12 Chapter 2. Lebesgue points
This implies that either 1 |B(x0, r)| Z B(x0,r) |f (x) − f (x0)| dx > C1 2 r α or 1 |B(y0, r)| Z B(y0,r) |f (x) − f (y0)| dx > C1 2 r α.
This concludes the proof.
Just as there was an equivalence in the case of uniform continuity versus uniform convergence in the Lebesgue point integral, we have now shown that there is a similar connection between having a H¨older continuous function and having a H¨older estimate for the convergence of the Lebesgue point integral.
Chapter 3
Lebesgue points for
L
1
-functions
If x is a Lebesgue point of f then f is clearly integrable on a neighborhood of x. In this section we will study the converse implication.
3.1
Introducing L
1We will start this section with defining the L1-functions. To do this we first
need to define the terms almost everywhere and equivalence class.
Definition 3.1.1. A property is said to hold almost everywhere if the set of points where it does not hold has Lebesgue measure 0.
Definition 3.1.2. An equivalence class [f ] of the function f is defined as [f ] = { ˜f : ˜f = f almost everywhere}.
With these terms we can now define L1.
Definition 3.1.3. L1(Rn) is the set of equivalence classes of functions with
finite integrals, i.e.
L1(Rn) = ½ [f ] : Z Rn f dx < ∞ ¾ . (3.1)
Note that an element in L1(Rn) is a function defined almost everywhere.
We can get a larger class of functions if we remove the restrictions for the integrability of f at infinity.
Definition 3.1.4. A function f : Rn → R defined almost everywhere is called
locally integrable if f g ∈ L1(Rn) for every non-negative continuous function g
with compact support in Rn. The set of locally integrable functions is denoted
L1
loc(Rn).
The set of continuous functions is a subset of L1
loc(Rn). If a function belongs
to L1(Rn) then all points are not necessarily Lebesgue points. However, the set
of points which are not Lebesgue points is small. This will be the main goal of this section.
14 Chapter 3. Lebesgue points for L1-functions
3.2
The Hardy–Littlewood maximal function
To give an upper bound of the size of the set of non-Lebesgue points we will need the Hardy–Littlewood maximal function, which is one of the fundamental functions in analysis and was first studied in 1930 by Hardy–Littlewood [10]. Definition 3.2.1. For f ∈ L1(Rn), the Hardy–Littlewood maximal function is
defined as ˜ f (x0) = sup B 1 |B| Z B |f (x)| dx, where the supremum is taken over all balls B containing x0.
The Hardy–Littlewood maximal function is used in the following Hardy-Littlewood maximal theorem which will be needed later. It provides a bound for the size of the set where the Hardy–Littlewood maximal function is greater than a particular value.
Theorem 3.2.2. (Hardy–Littlewood maximal theorem). If f ∈ L1(Rn), then
the following inequality holds for all s > 0, ¯ ¯{x ∈ Rn: ˜f (x) > s}¯¯ ≤ 5n s Z Rn |f (x)| dx.
To prove this we will need the following lemma. The proof is from Claesson and H¨ormander [6].
Lemma 3.2.3. If F is a family of open balls with uniformly bounded radii then there is a subfamily F0 of F consisting of disjoint balls so that
[ B∈F B ⊂ [ B∈F0 ˜ B
where ˜B is the ball with the same center as B and five times greater radius. Proof. Let B1 be a ball in F such that
rad(B1) ≥1
2B∈Fsup
rad(B),
where rad(B) is the radius of the ball B. Let F1be the family of balls that are
disjoint with B1. All balls that are not disjoint with B1 will be contained in ˜B1
since their radii are at most twice the radius of B1. We can now in a similar
manner choose a ball B2∈ F1 so that
rad(B2) ≥1
2B∈Fsup1
rad(B)
and let F2 be all balls that are disjoint with B1 and B2. It follows that
F0= {B1, B2, . . . }
has the desired quality, i.e. [ B∈F B ⊂ ∞ [ i=1 ˜ Bi= [ B∈F0 ˜ B.
3.3. The set of Lebesgue points for L1-functions 15
The following proof of Theorem 3.2.2 can also be found in Claesson and H¨ormander [6].
Proof of Theorem 3.2.2. Let Es = {x ∈ Rn : ˜f (x) > s}. For every x ∈ Es,
there exists a ball Bx3 x so that
1 |Bx|
Z
Bx
|f | dx > s. (3.2)
Note that Bx⊂ Es. Hence the set Es is the union of the family of open balls
where the average value of |f | is larger than s. We call this family F . Since f ∈ L1(Rn) we know that 1 s Z Rn |f | dx > |Bx|,
thus the balls have uniformly bounded radii as required in Lemma 3.2.3. By applying Lemma 3.2.3 we know that there is a disjoint subfamily F0 of F so
that Es= [ B∈F B ⊂ [ B∈F0 ˜ B. We now have |Es| ≤ X B∈F0 | ˜B| = 5n X B∈F0 |B| ≤±Using (3.2)±≤ 5n X B∈F0 Z B |f | s dx. The balls in F0 are disjoint and thus it follows that
5n s X B∈F0 Z B |f | dx ≤ 5 n s Z Rn |f (x)| dx. The theorem follows.
3.3
The set of Lebesgue points for L
1-functions
By using the Hardy-Littlewood maximal theorem we can prove the main theo-rem of this section. The proof can be found in Claesson and H¨ormander [6] or in Ziemer [18].
Theorem 3.3.1. If f ∈ L1
loc(Rn), then almost every x0 ∈ Rn is a Lebesgue
point of f . Proof. If f ∈ L1
loc(Rn), then we have f χB(0,r) ∈ L1(Rn) for all r > 0. Also,
every x ∈ B(0, r) is a Lebesgue point of f if and only if it is a Lebesgue point of f χB(0,r). Thus we can assume that f ∈ L1(Rn). Define
T f (x0) = lim sup |B|→0 1 |B| Z B |f (x) − f (x0)| dx,
where the lim sup is taken over all balls containing x0with radii tending to zero.
Since lim sup 1 Z |f (x) − f (x )| dx ≥ lim 1 Z |f (x) − f (x )| dx
16 Chapter 3. Lebesgue points for L1-functions
we get that T f (x0) = 0 implies that x0 is a Lebesgue point of f . Now let
E²= {x ∈ Rn: T f (x) > ²}.
For all x0∈ E², we have T f (x0) 6= 0, i.e. x0 is not a Lebesgue point of f . We
want to show that |E²| = 0 for every ² as it implies that
|{x ∈ Rn: T f (x) > 0}| = 0,
i.e. almost every x0 is a Lebesgue point of f . We start out by observing that
the triangle inequality yields 1 |B| Z B |f (x) − f (x0)| dx ≤ 1 |B| Z B (|f (x)| + |f (x0)|) dx = 1 |B| Z B |f (x)| dx + |f (x0)|.
It then follows that for every ball containing x0,
˜ f (x0) + |f (x0)| ≥ 1 |B| Z B |f (x) − f (x0)| dx.
Taking lim sup over shrinking balls containing x0 yields for all x0∈ E2²,
˜
f (x0) + |f (x0)| ≥ 2².
This implies that for every x0∈ E2² either ˜f (x0) > ² or |f (x0)| > ². From this
it follows that |E2²| ≤
¯
¯{x ∈ Rn: ˜f (x) > ²}¯¯ +¯¯{x ∈ Rn : |f (x)| > ²}¯¯.
Theorem 3.2.2 and the inequality ¯ ¯{x ∈ Rn : |f (x)| > ²}¯¯ =1 ² Z {x∈Rn:|f (x)|>²} ² dx ≤ 1 ² Z Rn |f (x)| dx imply that ¯ ¯E2² ¯ ¯ ≤ 5n ² Z Rn |f (x)|dx +1 ² Z Rn |f (x)| dx = 5 n+ 1 ² Z Rn |f (x)|dx. (3.3) The triangle inequality yields
|f (x) − f (x0)| ≤ |g(x) − g(x0)| + |(f − g)(x) − (f − g)(x0)|,
for every function g : Rn→ R. This implies
T f (x0) ≤ T g(x0) + T (f − g)(x0).
If g is continuous, then T g(x0) = 0. This means that T f (x0) ≤ T (f − g)(x0)
for every continuous g. Let
F²= {x ∈ Rn : (T (f − g))(x) > ²}.
Then E²⊂ F² and hence
¯ ¯F²
¯ ¯ ≥¯¯E²
¯
¯. Applying equation (3.3) to f − g and F2²
in place of f and E2², we have
¯ ¯E2² ¯ ¯ ≤¯¯F2² ¯ ¯ ≤5n+ 1 ² Z Rn |f (x) − g(x)|dx.
Since continuous functions are dense in L1(Rn), we can choose a continuos
function g so thatRRn|f − g| dx is arbitrarily small. It follows that
¯ ¯E2²
¯ ¯ = 0 for all ² > 0 and concludes the proof.
Chapter 4
Lebesgue points and
Sobolev spaces
In the earlier sections we have seen that functions in L1have Lebesgue points almost
everywhere i.e. the set of non-Lebesgue points has measure zeros. However, even sets of measure zero can be relatively large. In this section we will see that for more smooth functions the set of non-Lebesgue points has a smaller upper bound.
4.1
Introducing Sobolev spaces
We start by defining the space Lp. This definition can be found in Kreyzig [13].
Definition 4.1.1. A function f belongs to Lp(Rn) if fp∈ L1(Rn) i.e.
Z
Rn
fpdx < ∞.
If we consider functions in Lpso that their distributional derivatives are also
in Lpwe get a so called Sobolev space. These spaces are subspaces of Lp. We can
expect that Lebesgue points for such functions will have additional properties. This is the case and in the end of this section an upper bound for the dimension of the set of non-Lebesgue points will be derived. We will also show that for p > n, Sobolev functions are H¨older continuous. This is often proved by using capacity. Interested readers can read about capacity in Adams–Hedberg [3] or Ziemer [18]. A proof based on capacity can be found in Heinonen, Kilpel¨ainen and Martio [11]. Our proof is more direct and based on the Poincar´e inequality, and the Lebesgue point theory from the earlier sections. A more thorough presentation of Sobolev spaces can be found in Adams [2], Garnett [8], Maz’ya [15] or Ziemer [18]. We will begin by defining the term Sobolev space.
Definition 4.1.2. Let Ω ⊂ Rn be an open set and let p ≥ 1. The Sobolev
space is
W1,p(Ω) = {u ∈ Lp(Ω) : |∇u| ∈ Lp(Ω)},
where ∇u is the distributional gradient of u. The Sobolev space is equipped with the norm
kuk1,p= µ Z Ω (|u|p+ |∇u|p) dx ¶1/p .
18 Chapter 4. Lebesgue points and Sobolev spaces
For the later results we will need the following inequality, whose proof can be found in Evans [7] or Gilbarg and Trudinger [9].
Theorem 4.1.3. The following Poincar´e inequality holds for all u ∈ W1,p(Rn)
and all balls B, 1 |B| Z B |u − uB| dx ≤ CrB µ 1 |B| Z B |∇u|pdx ¶1/p ,
where rB is the radius of B, uB is the integral average of u taken over B,
uB = 1 |B| Z B u dx, and C does not depend on u and B.
By using the Poincar´e inequality we can calculate an upper bound for the difference between the integral averages taken over two balls around x0.
Lemma 4.1.4. Let u ∈ W1,p(Ω), x
0∈ Ω and Bj = B(x0, rj) where rj = Arj+1,
A > 1, j = 1, 2, . . . , and B(x0, 2r1) ⊂ Ω. Then for every k > j and every ² > 0,
|uB j− uB k| ≤ Crj² sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x0,ρ) |∇u|pdx ¶1/p , where C depends only on p, n and A.
Proof. The triangle inequality yields
|uB j− uB k| = |uB j− uB j+1+ uB j+1− · · · + uB k−1− uB k|
≤ |uB j− uB j+1| + |uB j+1− uB j+2| + · · · + |uB k−1− uB k|.
We have for every i = j, j + 1, . . . , k − 1,
|uB i− uB i+1| = ¯ ¯ ¯ ¯uB i−|B1 i+1| Z B i+1 u dx ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯|Bi+11 | Z B i+1 (u − uB i) dx ¯ ¯ ¯ ¯ ≤ 1 |Bi+1| Z B i+1 |u − uB i| dx ≤ A n |Bi| Z B i |u − uB i| dx.
By using the Poincar´e inequality we get 1 |Bi| Z B i |u−uB i| dx ≤ Cri µ 1 rin Z B i |∇u|pdx ¶1/p = Cr²i µ ri(1−²)p rn i Z Bi |∇u|pdx ¶1/p . (4.1)
4.2. H¨older continuous Sobolev functions 19
It now follows for the sequence uB j that
|uB j− uB k| ≤ C ∞ X i=j r² i µ r(1−²)pi rn i Z B i |∇u|pdx ¶1/p ≤ C ∞ X i=j r² i sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p ≤ Cr² j sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p . The statement follows.
It is sufficient to consider only balls with radii 2−j when studying Lebesgue
points as the following lemma shows. Lemma 4.1.5. If 1 |B(x0, 2−j)| Z B(x0,2−j) |u(x) − u(x0)| dx → 0, as j → ∞,
then x0 is a Lebesgue point of u.
Proof. Let 0 < r < 1. Then there exists j ∈ Z so that 2−(j+1) < r < 2−j.
Clearly, j → ∞ is equivalent to r → 0. We then have 1 |B(x0, r)| Z B(x0,r) |u(x) − u(x0)| dx ≤ 1 |B(x0, 2−(j+1))| Z B(x0,2−j) |u(x) − u(x0)| dx = 2 n |B(x0, 2−j)| Z B(x0,2−j) |u(x) − u(x0)| dx → 0, as j → ∞.
4.2
H¨
older continuous Sobolev functions
By using Lemma 4.1.4 about differences of integral averages around x0 we can
show that if p > n then these integral averages form a Cauchy sequence. From this it follows that the sequence of integral averages converges but not neces-sarily to the function value u(x0). Nevertheless it provides us with a canonical
representant of a Sobolev function. Theorem 4.2.1. Let u ∈ W1,p(Ω), x
0 ∈ Ω, Bj = B(x0, rj), rj = 2−j, j =
1, 2, . . . , and p > n. Then for all sufficiently large j, uB j= 1
|Bj|
Z
B j
20 Chapter 4. Lebesgue points and Sobolev spaces
Proof. From Lemma 4.1.4 we have for all sufficiently large j and all k > j, |uB j− uB k| ≤ C2−j² sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p . By choosing ² = (p − n)/p = 1 − n/p > 0 it follows that
sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p ≤ µ Z Rn |∇u| dx ¶1/p ≤ kuk1,p< ∞, (4.2) which implies that
C2−j² sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p ≤ Ckuk1,pr²j→ 0, as j → ∞.
Thus uB j is a Cauchy sequence and therefore it converges.
This result leads to the following consequence. Every function in the Sobolev space W1,p(Ω) with p > n can be replaced by an equivalent function in the
sense of the Sobolev space, and this equivalent function has Lebesgue points everywhere.
Proposition 4.2.2. With the same conditions and notation as in Theorem 4.2.1, it is possible to choose another representant ¯u of u by changing the func-tion value in some points so that ¯u = u almost everywhere and every x0∈ Ω is
a Lebesgue point of ¯u.
Proof. For every x0that is not a Lebesgue point of u, let
uBj(x0) = 1 |B(x0, 2−j)| Z B(x0,2−j) u dx, and let ¯ u(x0) = lim j→∞uBj(x0). Since u ∈ L1
loc(Ω) we have ¯u = u almost everywhere by Theorem 3.3.1. We now
need to prove that 1 |Bj| Z B j ¯ ¯u(x) − ¯u(x0) ¯ ¯ dx → 0 as j → ∞.
For all k ≥ j we have 1 |Bj| Z B j ¯ ¯u(x)− ¯u(x0) ¯ ¯ dx ≤ 1 |Bj| Z B j ¯ ¯u(x)−uBj ¯
¯ dx+|uBj−uBk|+|uBk− ¯u(x0)|.
(4.3) From Lemma 4.1.4 and equation (4.1) we have
1 |Bj| Z B j ¯ ¯u(x) − uBj ¯ ¯ dx ≤ Cr² j µ r(1−²)pj rn j Z Bj |∇u|pdx ¶1/p and |uBj − uBk| ≤ Cr ² j sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p
4.2. H¨older continuous Sobolev functions 21
where ² = 1 − n/p > 0. Inserting this into equation (4.3) gives 1 |Bj| Z B j ¯ ¯u(x) − ¯u(x0) ¯ ¯ dx ≤ 2Cr²j sup 2rj>ρ>0 µ ρ(1−²)p−n Z B(x,ρ) |∇u|pdx ¶1/p + |uBk− ¯u(x0)|.
By using equation (4.2) we get 1 |Bj| Z B j ¯ ¯u(x) − ¯u(x0) ¯ ¯ dx ≤ Ckuk1,pr²j+ |uBk− ¯u(x0)| (4.4)
where both the first and second term tends to 0 as j, k → ∞. This concludes the proof.
By choosing the correct representant in this fashion, all functions in the Sobolev space W1,p(Ω) with p > n become H¨older continuous.
Proposition 4.2.3. With the same conditions as in Theorem 4.2.1 the repre-sentant ¯u from Proposition 4.2.2, is H¨older continuous with the exponent 1−n/p. Proof. Since ¯u = u almost everywhere, from the proof of Proposition 4.2.2 and equation (4.4) we have for sufficiently large j and all k > j,
1 |Bj| Z B j ¯ ¯¯u(x) − ¯u(x0) ¯ ¯ dx ≤ Ckuk1,pr²j+ |uBk− ¯u(x0)|
where ² = 1 − n/p > 0. Letting k → ∞ yields 1 |Bj| Z B j ¯ ¯¯u(x) − ¯u(x0) ¯ ¯ dx ≤ Ckuk1,prj²= Cr1−n/pj .
The statement now follows from Theorem 2.3.4.
The converse is not true as the following example shows.
Example 4.2.4. For every p > 1 there exists a H¨older continuous function which is not in W1,p¡(0, 1)n¢. Let
f (x) = x1−1/p1 . Then f is H¨older continuous on (0, 1)n, but
∇f (x) = 1 − 1/p x1/p1 ∈ L/
p¡(0, 1)n¢
which implies that f /∈ W1,p¡(0, 1)n¢.
The following examples illustrate the necessity of p > n in Proposition 4.2.3. Example 4.2.5. Let n > p, f (x) = |x|α, α < 0 and study f (x) for |x| < 1.
It is clear that f is not H¨older continuous, however we can choose α so that f ∈ W1,p(B(0, 1)). We use
Z Z 1
22 Chapter 4. Lebesgue points and Sobolev spaces
from Persson and B¨oiers [16]. We have ∇f (x) = αx|x|α−2and it follows that
Z |x|<1 |f (|x|)|pdx = C Z 1 0 rαp+n−1dr.
The integral is convergent if αp + n − 1 > −1 which is equivalent to α > −n/p. We also have Z |x|<1 |∇f (|x|)|pdx = C Z 1 0 r(α−1)p+n−1dr.
To obtain a convergent integral, we need (α−1)p+n−1 > −1 which is equivalent to α > 1 − n/p. Choosing α so that 1 − n/p < α < 0 proves the claim.
Example 4.2.6. Let n = p, f (x) =¯¯ ln |x|¯¯α, and study f (x) for |x| < 1/2. We have
∇f (x) = α x
|x|2(ln |x|)
α−2ln |x|
and by using p = n it follows that Z |x|<1/2 |f (|x|)|pdx = C Z 1/2 0 (− ln r)αnrn−1dr.
The change of variables − ln r = t gives C Z 1/2 0 (− ln r)αnrn−1dr = C Z ∞ ln 2 tαne−ntdt.
The integral is convergent for all α ∈ R. Using the same change of variables yields Z |x|<1/2 |∇f (|x|)|pdx = C Z 1/2 0 (− ln r)(α−1)n r dr = C Z ∞ ln 2 t(α−1)ndt.
This integral converges if (α − 1)n < −1 which is equivalent to α < 1 − 1/n. Since f is not H¨older continuous, the neccesity of n 6= p in Proposition 4.2.3 follows.
4.3
Lebesgue points and the Hausdorff
dimen-sion
Earlier results were in the case p > n. We will now consider n ≥ p. First we need to define the Hausdorff measure and dimension.
Definition 4.3.1. Let s > 0 and 0 < δ ≤ ∞. For a set D ⊂ Rn, let
Λδs(D) = inf
X
i
rsi,
where the infimum is taken over all coverings of D by balls Bi= B(xi, ri) with
radii ri< δ. Define the s-Hausdorff measure of D ⊂ Rn as
Λs(D) = sup δ>0Λ
δ
4.3. Lebesgue points and the Hausdorff dimension 23
Since we can assume that δ < 1, we see from the definition that larger values of s give a smaller s-Hausdorff measure, in particular the s-Hausdorff measure will become 0 for sufficiently large s.
Lemma 4.3.2. If Λs(D) < ∞ then Λt(D) = 0 for all t > s.
Proof. For every δ > 0 and t > s we have Λδ t(D) = inf X i rt i≤ inf δt−s X i rs i = δt−sinf X i rs i ≤ δt−sΛs(D),
where the infimum is taken over all coverings of D by balls Bi= B(xi, ri) with
radii ri < δ. By letting δ → 0 we get
Λt(D) ≤ lim δ→0δ
t−sΛ
s(D) = 0,
which finishes the proof.
We see that for sufficiently large s, the Hausdorff measure will become 0. This concept is used when defining the Hausdorff dimension.
Definition 4.3.3. The smallest s such that Λt(D) = 0 whenever t > s, is called
the Hausdorff dimension of D, i.e.
dimH(D) = inf{s : Λs(D) = 0}.
An introduction to Hausdorff dimension and Hausdorff measure can be found in Browder [5], Hijab [12] or Mattila [14].
To prove the main theorem of this section we will need to define a fractional maximal function obtained from Garnett [8]. We will also have a look at its level sets much like it was done with the Hardy-Littlewood maximal function in Theorem 3.2.2. We will present the proof from Heinonen, Kilpel¨ainen and Martio [11].
Lemma 4.3.4. For s > 0, p ≥ 1 and f ∈ Lp(Rn) we define the fractional
maximal function Ms,pf (x0) = sup ρ>0 µ ρ−s Z B(x0,ρ) |f (x)|pdx ¶1/p . Then for all t > 0
Λ∞ s ¡ {y ∈ Rn: M s,pf (y) > t} ¢ < C tp Z Rn |f |pdx.
Proof. Fix t > 0 and let Et= {y ∈ Rn : Ms,pf (y) > t}. For every y ∈ Etwe
can find a ball By with radius ry that contains y and
Z By |f |pdx > tprs y. (4.5) It follows that r < µ 1 Z |f |pdx ¶1/s ≤ µ kf kp p ¶1/s =: R.
24 Chapter 4. Lebesgue points and Sobolev spaces We have Et⊂ [ y∈Et By,
and as f ∈ Lp(Rn), the radii r
y have an upper bound R which does not depend
on y. The family {By} has uniformly bounded radii and from Lemma 3.2.3 we
know that we can find a subfamily B1, B2, . . . of {By} such that the balls Bi
are pairwise disjoint and
Et⊂
[
i
5Bi
where 5Bidenotes the ball with the same center as Bi but with 5 times greater
radius. We now have by using equation (4.5) and the fact that the balls Bi are
disjoint, Λ∞s (Et) ≤ X i 5srsi ≤ 5s tp X i Z Bi |f |pdx ≤ C tp Z Rn |f |pdx which concludes the proof.
We will need the following lemma that can be found in Adams and Hedberg [3]. For convenience a proof is presented.
Lemma 4.3.5. If 0 < δ ≤ ∞, then Λs(E) = 0 if and only if Λδs(E) = 0.
Proof. Since Λs(E) ≥ Λδs(E) it is sufficient to show that Λs(E) = 0 if Λδs(E) = 0.
Let Λδ
s(E) = 0 and fix ² > 0. We choose a covering of E by balls Bi with radii
ri≤ δ such that X i rsi < ²s. Then ri< ² and Λ²s(E) ≤ X i rsi < ²s. By letting ² → 0 we get Λs(E) = lim ²→0Λ ² s(E) = 0.
The statement follows.
The estimate in Lemma 4.3.4, together with Lemmas 4.1.4, 4.3.5 and equa-tion 4.1.5, lets us prove the final theorem of this secequa-tion.
Theorem 4.3.6. Let u ∈ W1,p(Rn) and p ≤ n. Then
1 |B(x0, r)|
Z
B(x0,r)
|u(x) − u(x0)| dx → 0, as r → 0
for all x0∈ Rn except on a set with Hausdorff dimension less than or equal to
4.3. Lebesgue points and the Hausdorff dimension 25
Proof. Let Bj= B(x0, 2−j) and
uBj(x0) = 1 |Bj| Z Bj u dx. By Lemma 4.1.4 we have for all ² > 0 and all x0∈ Rn,
|uBj(x0) − uBk(x0)| ≤ C2 −j² sup ρ>0 µ ρ(1−²)p−n Z B(x0,ρ) |∇u|pdx ¶1/p (4.6)
whenever k > j. Let ² > 0, s = n − (1 − ²)p > 0 and consider the maximal function Ms,p|∇u|(x0) = sup ρ>0 µ ρ−s Z B(x0,ρ) |∇u(x)|pdx ¶1/p . Now let D = {x : Ms,p|∇u|(x) = ∞}. If x0∈ D then/ sup ρ>0 µ ρ(1−²)p−n Z B(x0,ρ) |∇u|pdx ¶1/p < ∞. (4.7)
Together with equation (4.6) it implies that |uBj(x0) − uBk(x0)| ≤ C2
−j²
when-ever k > j, i.e. the sequence uBj(x0) is a Cauchy sequence and converges. Let
¯
u(x0) = lim uBj(x0). For all k ≥ j we have
1 |Bj| Z B j ¯ ¯u(x)− ¯u(x0) ¯ ¯ dx ≤ 1 |Bj| Z B j ¯ ¯u(x)−uBj ¯
¯ dx+|uBj−uBk|+|uBk− ¯u(x0)|.
(4.8) From equations (4.1) and (4.7) we have
1 |Bj| Z B j ¯ ¯u(x) − uBj ¯ ¯ dx ≤ C2−j² µ rj(1−²)p rn j Z Bj |∇u|pdx ¶1/p ≤ C2−j² and |uBj− uBk| ≤ C2 −j².
Inserting this into equation (4.8) gives 1 |Bj| Z B j ¯ ¯u(x) − ¯u(x0) ¯ ¯ dx ≤ C2−j²+ |u Bk− ¯u(x0)| → 0 as j, k → ∞.
Thus x0 is a Lebesgue point of u. We let
N = {x : x is not a Lebesgue point of u}.
We have shown that if x0 ∈ D then x/ 0 is a Lebesgue point, this implies that
N ⊂ D. By using Lemma 4.3.4 we have for all t > 0, Λ∞(D) ≤ Λ∞¡{x : M |∇u|(x) > t}¢≤ C
Z
26 Chapter 4. Lebesgue points and Sobolev spaces
Letting t → ∞ implies that
Λ∞s (N ) ≤ Λ∞s (D) = 0.
From Lemma 4.3.5 we get Λs(N ) = 0. It now follows that the Hausdorff
dimen-sion of the set of non-Lebesgue points of u satisfies
dimH(N ) ≤ s = n − p(1 − ²) = n − p + p²,
for every ² > 0. Letting ² → 0 shows that dimH(N ) ≤ n − p.
Remark 4.3.7. The above conclusions about Lebesgue points are valid for functions f ∈ W1,p(Ω) as well as f ∈ W1,p(Rn) even though only the latter has
been studied earlier in this section. To see that, let f ∈ W1,p(Ω) where Ω is an
open subset of Rn and let Ω0 be a compact subset of Ω. Now let
η(x) = inf y∈∂Ω|x − y| inf
y∈Ω0|x − y| + infy∈∂Ω|x − y|
, if x ∈ Ω;
0, otherwise.
We note that 0 ≤ η(x) ≤ 1, η(x) = 1 when x ∈ Ω0, η(x) = 0 when x /∈ Ω and
∇η(x) ≤ C < ∞ for all x. It now follows that ηf ∈ W1,p(Rn) since
kηf k1,p= µ Z Ω |ηf |p+ |∇(ηf )|pdx ¶1/p = µ Z Ω |ηf |p+ |η∇f + f ∇η|pdx ¶1/p ≤ µ Z Ω |f |p+ |∇f + f ∇η|pdx ¶1/p ≤ µ Z Ω |f |p+¡|∇f | + |Cf |¢pdx ¶1/p < ∞.
Since f (x) = η(x)f (x) when x ∈ Ω0 it is clear that x ∈ Ω0 is a Lebesgue point
of f if and only if it is a Lebesgue point of ηf . This is true for all Ω0⊂ Ω and
Chapter 5
Summary and thoughts of
further studies
5.1
Summary
We have seen that locally integrable functions have Lebesgue points almost everywhere. Smoother functions such as the Sobolev functions studied here have even more Lebesgue points and we were able to obtain an upper bound for the Hausdorff dimension of the set of non-Lebesgue points.
We have seen that continuous functions have Lebesgue points everywhere and we were able to characterize uniformly continous functions and H¨older con-tinuos functions by the rate of decay of Lebesgue point integrals.
5.2
Further studies
A continuation of these studies could be done in spaces that are more general than Rn and satisfy certain suitable conditions. Another thing to study would
be Lq-Lebesgue points i.e. to study if
1 B(x0, r)
Z
B(x0,r)
|u(x) − u(x0)|qdx → 0 when r → 0, for some q > 1.
There are many other measures that could be used instead of the Lebesgue measure. An interesting subject is to study what results other measures would yield.
For H¨older continuous functions it was shown that the rate of convergence in the Lebesgue point integral can also be estimated by certain powers. A generalization would be to study the rate of convergence for the Lebesgue point integrals if for every x, y ∈ Rn, |f (x) − f (y)| < g(|x − y|) for some function g.
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