10 Prove or give a counterexample: Ifh:R→[0,∞)is an increasing function, thenh∗is an increasing function.
11 Give an example of a Borel measurable functionh: R → [0,∞)such that h∗(b)<∞for allb∈Rbutsup{h∗(b):b∈R}=∞.
12 Show that|{b∈R:h∗(b) =∞}|=0for everyh∈ L1(R).
13 Show that there existsh∈ L1(R)such thath∗(b) =∞for everyb∈Q.
14 Supposeh∈ L1(R). Prove that
|{b∈R:h∗(b)≥c}| ≤ 3c∥h∥1
for everyc>0.
[This result slightly strengthens the Hardy–Littlewood maximal inequality(4.8) because the set on the left side above includes thoseb∈Rsuch thath∗(b) =c. A much deeper strengthening comes from replacing the constant3in the Hardy–
Littlewood maximal inequality with a smaller constant. In 2003, Antonios Melas answered what had been an open question about the best constant. He proved that the smallest constant that can replace3in the Hardy–Littlewood maximal inequality is(11+√
61)/12≈1.56752; seeAnnals of Mathematics 157(2003), 647–688.]
4B Derivatives of Integrals
Lebesgue Differentiation Theorem
The next result states that the average amount by which a function inL1(R)differs from its values is small almost everywhere on small intervals. The2in the denomi- nator of the fraction in the result below could be deleted, but its presence makes the length of the interval of integration nicely match the denominator2t.
The next result is called the Lebesgue Differentiation Theorem, even though no derivative is in sight. However, we will soon see how another version of this result deals with derivatives. The hard work takes place in the proof of this first version.
4.10 Lebesgue Differentiation Theorem, first version
Supposef ∈ L1(R). Then limt↓0
1 2t
Z b+t
b−t |f− f(b)|=0 for almost everyb∈R.
Before getting to the formal proof of this first version of the Lebesgue Differen- tiation Theorem, we pause to provide some motivation for the proof. Ifb∈Rand t>0, then3.25gives the easy estimate
1 2t
Z b+t
b−t |f−f(b)| ≤sup{|f(x)−f(b)|:|x−b| ≤t}.
If f is continuous atb, then the right side of this inequality has limit0ast ↓ 0, proving4.10in the special case in which f is continuous onR.
To prove the Lebesgue Differentiation Theorem, we will approximate an arbitrary function inL1(R)by a continuous function (using3.48). The previous paragraph shows that the continuous function has the desired behavior. We will use the Hardy–
Littlewood maximal inequality (4.8) to show that the approximation produces ap- proximately the desired behavior. Now we are ready for the formal details of the proof.
Proof of4.10 Letδ > 0. By3.48, for eachk ∈ Z+ there exists a continuous functionhk:R→Rsuch that
4.11 ∥f −hk∥1< δ
k2k. Let
Bk={b∈R:|f(b)−hk(b)| ≤ 1k and (f −hk)∗(b)≤ 1k}. Then
4.12 R\Bk ={b∈R:|f(b)−hk(b)|> 1k} ∪ {b∈R:(f−hk)∗(b)> 1k}.
Markov’s inequality (4.1) as applied to the function f−hkand4.11imply that 4.13 |{b∈R:|f(b)−hk(b)|> 1k}|< δ
2k.
The Hardy–Littlewood maximal inequality (4.8) as applied to the function f −hk
and4.11imply that
4.14 |{b∈R:(f−hk)∗(b)> 1k}|< 3δ 2k. Now4.12,4.13, and4.14imply that
|R\Bk|< δ 2k−2. Let
B=
\∞ k=1
Bk.
Then
4.15 |R\B|= [∞ k=1
(R\Bk)≤
∑
∞k=1|R\Bk|<
∑
∞ k=1δ
2k−2 =4δ.
Supposeb∈Bandt>0. Then for eachk∈Z+we have 1
2t Z b+t
b−t |f −f(b)| ≤ 2t1 Z b+t
b−t |f−hk|+|hk−hk(b)|+|hk(b)− f(b)|
≤(f −hk)∗(b) +1 2t
Z b+t
b−t |hk−hk(b)|+|hk(b)− f(b)|
≤ 2k+ 1 2t
Z b+t
b−t |hk−hk(b)|.
Becausehkis continuous, the last term is less than1kfor allt>0sufficiently close to 0(how close is sufficiently close depends uponk). In other words, for eachk∈Z+, we have
1 2t
Z b+t
b−t |f −f(b)|< 3 k for allt>0sufficiently close to0.
Hence we conclude that limt↓0
1 2t
Z b+t
b−t |f− f(b)|=0 for allb∈B.
LetAdenote the set of numbersa∈Rsuch that limt↓0
1 2t
Z a+t
a−t |f −f(a)|
either does not exist or is nonzero. We have shown that A⊂(R\B). Thus
|A| ≤ |R\B|<4δ,
where the last inequality comes from4.15. Becauseδis an arbitrary positive number, the last inequality implies that|A|=0, completing the proof.
Derivatives
You should remember the following definition from your calculus course.
4.16 Definition derivative;g′; differentiable
Supposeg: I →Ris a function defined on an open intervalIofRandb ∈ I.
Thederivativeofgatb, denotedg′(b), is defined by g′(b) =lim
t→0
g(b+t)−g(b) t
if the limit above exists, in which casegis calleddifferentiableatb.
We now turn to the Fundamental Theorem of Calculus and a powerful extension that avoids continuity. These results show that differentiation and integration can be thought of as inverse operations.
You saw the next result in your calculus class, except now the function f is only required to be Lebesgue measurable (and its absolute value must have a finite Lebesgue integral). Of course, we also need to require f to be continuous at the crucial pointbin the next result, because changing the value of f at a single number would not change the functiong.
4.17 Fundamental Theorem of Calculus
Supposef ∈ L1(R). Defineg:R→Rby g(x) =
Z x
−∞ f.
Supposeb∈Rand f is continuous atb. Thengis differentiable atband g′(b) = f(b).
Proof Ift̸=0, then
g(b+t)−g(b)
t −f(b)= Rb+t
−∞ f −Rb
−∞ f
t −f(b)
= Rb+t
b f
t −f(b)
= Rb+t
b f −f(b) t
4.18
≤ sup
{x∈R:|x−b|<|t|}
|f(x)−f(b)|.
Ifε > 0, then by the continuity of f atb, the last quantity is less than εfor t sufficiently close to0. Thusgis differentiable atbandg′(b) = f(b).
A function inL1(R)need not be continuous anywhere. Thus the Fundamental Theorem of Calculus (4.17) might provide no information about differentiating the integral of such a function. However, our next result states that all is well almost everywhere, even in the absence of any continuity of the function being integrated.
4.19 Lebesgue Differentiation Theorem, second version
Supposef ∈ L1(R). Defineg:R→Rby g(x) =
Z x
−∞ f. Theng′(b) = f(b)for almost everyb∈R.
Proof Supposet̸=0. Then from4.18we have
g(b+t)−g(b)
t − f(b)= Rb+t
b f −f(b) t
≤ 1t Z b+t
b |f −f(b)|
≤ 1t Z b+t
b−t |f −f(b)|
for allb∈R. By the first version of the Lebesgue Differentiation Theorem (4.10), the last quantity has limit0ast→0for almost everyb∈R. Thusg′(b) = f(b)for almost everyb∈R.
Now we can answer the question raised on the opening page of this chapter.
4.20 no set constitutes exactly half of each interval
There does not exist a Lebesgue measurable setE⊂[0, 1]such that
|E∩[0,b]|= b 2 for allb∈[0, 1].
Proof Suppose there does exist a Lebesgue measurable set E ⊂ [0, 1]with the property above. Defineg:R→Rby
g(b) = Z b
−∞χE.
Thusg(b) = b2for allb∈[0, 1]. Henceg′(b) = 12for allb∈(0, 1).
The Lebesgue Differentiation Theorem (4.19) implies that g′(b) = χE(b)for almost everyb∈R. However,χEnever takes on the value 12, which contradicts the conclusion of the previous paragraph. This contradiction completes the proof.
The next result says that a function inL1(R)is equal almost everywhere to the limit of its average over small intervals. These two-sided results generalize more naturally to higher dimensions (take the average over balls centered atb) than the one-sided results.
4.21 L1(R)function equals its local average almost everywhere
Supposef ∈ L1(R). Then
f(b) =lim
t↓0
1 2t
Z b+t b−t f for almost everyb∈R.
Proof Supposet>0. Then
1 2t
Z b+t b−t f
−f(b)=1 2t
Z b+t
b−t f −f(b)
≤ 2t1 Z b+t
b−t |f −f(b)|.
The desired result now follows from the first version of the Lebesgue Differentiation Theorem (4.10).
Again, the conclusion of the result above holds at every numberbat which f is continuous. The remarkable part of the result above is that even if f is discontinuous everywhere, the conclusion holds for almost every real numberb.
Density
The next definition captures the notion of the proportion of a set in small intervals centered at a numberb.
4.22 Definition density
SupposeE⊂R. ThedensityofEat a numberb∈Ris limt↓0
|E∩(b−t,b+t)| 2t
if this limit exists (otherwise the density ofEatbis undefined).
4.23 Example density of an interval
The density of[0, 1]atb =
1 ifb∈(0, 1),
12 ifb=0orb=1, 0 otherwise.
The next beautiful result shows the power of the techniques developed in this chapter.
4.24 Lebesgue Density Theorem
SupposeE ⊂ Ris a Lebesgue measurable set. Then the density of Eis1at almost every element ofEand is0at almost every element ofR\E.
Proof First suppose|E|<∞. ThusχE∈ L1(R). Because
|E∩(b−t,b+t)|
2t = 1
2t Z b+t
b−t χE
for everyt>0and everyb∈R, the desired result follows immediately from4.21.
Now consider the case where|E|=∞[which means thatχE∈ L/ 1(R)and hence 4.21as stated cannot be used]. Fork∈Z+, letEk=E∩(−k,k). If|b|<k, then the density ofEatbequals the density ofEkatb. By the previous paragraph as applied toEk, there are setsFk ⊂ EkandGk ⊂ R\Eksuch that|Fk| =|Gk|= 0and the density ofEkequals1at every element ofEk\Fkand the density ofEkequals0at every element of(R\Ek)\Gk.
LetF=S∞k=1FkandG=S∞k=1Gk. Then|F|=|G|=0and the density ofEis 1at every element ofE\Fand is0at every element of(R\E)\G.
The Lebesgue Density Theorem makes the example provided by the next result somewhat surprising. Be sure to spend some time pondering why the next result does not contradict the Lebesgue Density Theorem. Also, compare the next result to4.20.
The bad Borel set provided by the next result leads to a bad Borel measurable function. Specifically, letEbe the bad Borel set in 4.25. Then χE is a Borel measurable function that is discontinuous everywhere. Furthermore, the functionχE cannot be modified on a set of measure0 to be continuous anywhere (in contrast to the functionχQ).
Even though the functionχE discussed in the paragraph above is continuous nowhere and every modification of this function on a set of measure0is also continu- ous nowhere, the functiongdefined by
g(b) = Z b
0 χE is differentiable almost everywhere (by4.19).
The proof of4.25given below is based on an idea of Walter Rudin.
4.25 bad Borel set
There exists a Borel setE⊂Rsuch that 0<|E∩I|<|I| for every nonempty bounded open intervalI.
Proof We use the following fact in our construction:
4.26 SupposeGis a nonempty open subset ofR. Then there exists a closed set F⊂G\Qsuch that|F|>0.
To prove4.26, letJ be a closed interval contained inGsuch that0 <|J|. Let r1,r2, . . .be a list of all the rational numbers. Let
F=J\ [∞ k=1
rk− |J|
2k+2,rk+ |J| 2k+2
.
Then Fis a closed subset ofRand F ⊂ J \Q ⊂ G\Q. Also,|J\F| ≤ 12|J| becauseJ\F⊂S∞k=1rk−2|kJ+|2,rk+2|kJ+|2. Thus
|F|=|J| − |J\F| ≥ 12|J|>0, completing the proof of4.26.
To construct the setEwith the desired properties, letI1,I2, . . .be a sequence consisting of all nonempty bounded open intervals ofRwith rational endpoints. Let F0=Fb0=∅, and inductively construct sequencesF1,F2, . . .andFb1,Fb2, . . .of closed subsets ofRas follows: Supposen∈Z+andF0, . . . ,Fn−1andFb0, . . . ,Fbn−1have been chosen as closed sets that contain no rational numbers. Thus
In\(Fb0∪. . .∪Fbn−1)
is a nonempty open set (nonempty because it contains all rational numbers inIn).
Applying4.26to the open set above, we see that there is a closed setFncontained in the set above such thatFncontains no rational numbers and|Fn|>0. Applying4.26 again, but this time to the open set
In\(F0∪. . .∪Fn),
which is nonempty because it contains all rational numbers inIn, we see that there is a closed setFbncontained in the set above such thatFbncontains no rational numbers and|Fbn|>0.
Now let
E= [∞ k=1
Fk.
Our construction implies thatFk∩Fbn=∅for allk,n∈Z+. ThusE∩Fbn =∅for alln∈Z+. HenceFbn ⊂In\Efor alln∈Z+.
SupposeIis a nonempty bounded open interval. ThenIn ⊂Ifor somen∈Z+. Thus
0<|Fn| ≤ |E∩In| ≤ |E∩I|. Also,
|E∩I|=|I| − |I\E| ≤ |I| − |In\E| ≤ |I| − |Fbn|<|I|, completing the proof.