1 Give an example of a sequence f1,f2, . . .of functions fromZ+to[0,∞)such that
klim→∞fk(m) =0 for everym∈Z+but lim
k→∞
Z fkdµ=1, whereµis counting measure onZ+.
2 Give an example of a sequence f1,f2, . . .of continuous functions from Rto [0, 1]such that
k→∞lim fk(x) =0 for everyx ∈Rbut lim
k→∞
Z fkdλ=∞, whereλis Lebesgue measure onR.
3 Supposeλis Lebesgue measure onRand f: R→Ris a Borel measurable function such thatR
|f|dλ<∞. Defineg:R→Rby g(x) =
Z
(−∞,x)fdλ.
Prove thatgis uniformly continuous onR.
4 (a) Suppose (X,S,µ) is a measure space with µ(X) < ∞. Suppose that f:X→[0,∞)is a boundedS-measurable function. Prove that
Z fdµ=infn m
j=1
∑
µ(Aj)sup
Aj
f :A1, . . . ,Amis anS-partition ofXo . (b) Show that the conclusion of part (a) can fail if the hypothesis that f is
bounded is replaced by the hypothesis thatR
fdµ<∞.
(c) Show that the conclusion of part (a) can fail if the condition thatµ(X)<∞ is deleted.
[Part(a)of this exercise shows that if we had defined an upper Lebesgue sum, then it could be used to defineR
fdµwhen f is bounded and µ(X) < ∞. However, parts(b)and(c)show that the hypotheses that f is bounded and that µ(X) < ∞are needed if defining the integral via the equation above. The definition of the integral via the lower Lebesgue sum does not require these hypotheses, showing the advantage of using the lower Lebesgue sum.]
5 Letλdenote Lebesgue measure onR. Supposef: R→Ris a Borel measurable function such thatR
|f|dλ<∞. Prove that
klim→∞
Z
[−k,k] fdλ=
Z fdλ.
6 Letλdenote Lebesgue measure onR. Give an example of a continuous function f:[0,∞)→Rsuch thatlimt→∞R
[0,t] fdλexists (inR) butR
[0,∞) fdλis not defined.
7 Letλdenote Lebesgue measure onR. Give an example of a continuous function f:(0, 1)→Rsuch thatlimn→∞R
(n1, 1)fdλexists (inR) butR
(0, 1)fdλis not defined.
8 Verify the assertion in3.38.
9 Verify the assertion in Example3.41.
10 (a) Suppose(X,S,µ) is a measure space such that µ(X) < ∞. Suppose p,rare positive numbers withp <r. Prove that if f:X→[0,∞)is an S-measurable function such thatR
frdµ<∞, thenR
fpdµ<∞.
(b) Give an example to show that the result in part (a) can be false without the hypothesis thatµ(X)<∞.
11 Suppose(X,S,µ)is a measure space and f ∈ L1(µ). Prove that {x∈X: f(x)̸=0}
is the countable union of sets with finiteµ-measure.
12 Suppose
fk(x) = (1−x)kcoskx
√x . Prove that lim
k→∞
Z 1 0 fk =0.
13 Give an example of a sequence of nonnegative Borel measurable functions f1,f2, . . .on[0, 1]such that both the following conditions hold:
• lim
k→∞
Z 1 0 fk=0;
• sup
k≥mfk(x) =∞for everym∈Z+and everyx∈[0, 1]. 14 Letλdenote Lebesgue measure onR.
(a) Letf(x) =1/√x. Prove thatR
[0, 1]fdλ=2.
(b) Let f(x) =1/(1+x2). Prove thatR
R fdλ=π.
(c) Let f(x) = (sinx)/x. Show that the integralR
(0,∞)fdλis not defined butlimt→∞R
(0,t)fdλexists inR.
15 Prove or give a counterexample: IfGis an open subset of(0, 1), thenχGis Riemann integrable on[0, 1].
16 Supposef ∈ L1(R).
(a) Fort∈R, define ft:R→Rby ft(x) = f(x−t). Prove that limt→0∥f −ft∥1=0.
(b) Fort>0, define ft:R→Rby ft(x) = f(tx). Prove that limt→1∥f −ft∥1=0.
Differentiation
Does there exist a Lebesgue measurable set that fills up exactly half of each interval?
To get a feeling for this question, consider the setE= [0,18]∪[14,38]∪[12,58]∪[34,78]. This setEhas the property that
|E∩[0,b]|= b 2
forb=0,14,12,34, 1. Does there exist a Lebesgue measurable setE⊂[0, 1], perhaps constructed in a fashion similar to the Cantor set, such that the equation above holds for allb∈[0, 1]?
In this chapter we see how to answer this question by considering differentia- tion issues. We begin by developing a powerful tool called the Hardy–Littlewood maximal inequality. This tool is used to prove an almost everywhere version of the Fundamental Theorem of Calculus. These results lead us to an important theorem about the density of Lebesgue measurable sets.
Trinity College at the University of Cambridge in England. G. H. Hardy (1877–1947)and John Littlewood(1885–1977)were students and later faculty members here. If you have not already done so, you should read Hardy’s remarkable
bookA Mathematician’s Apology (do not skip the fascinating Foreword by C. P.
Snow)and see the movieThe Man Who Knew Infinity, which focuses on Hardy, Littlewood, and Srinivasa Ramanujan(1887–1920).
CC-BY-SA Rafa Esteve
101
4A Hardy–Littlewood Maximal Function
Markov’s Inequality
The following result, called Markov’s inequality, has a sweet, short proof. We will make good use of this result later in this chapter (see the proof of4.10). Markov’s inequality also leads to Chebyshev’s inequality (see Exercise2in this section).
4.1 Markov’s inequality
Suppose(X,S,µ)is a measure space andh∈ L1(µ). Then µ({x ∈X:|h(x)| ≥c})≤ 1c∥h∥1
for everyc>0.
Proof Supposec>0. Then
µ({x∈X:|h(x)| ≥c}) = 1 c Z
{x∈X:|h(x)|≥c}cdµ
≤ 1c Z
{x∈X:|h(x)|≥c}|h|dµ
≤ 1c∥h∥1, as desired.
St. Petersburg University along the Neva River in St. Petersburg, Russia.
Andrey Markov(1856–1922)was a student and then a faculty member here.
CC-BY-SA A. Savin
Vitali Covering Lemma
4.2 Definition 3times a bounded nonempty open interval
SupposeIis a bounded nonempty open interval ofR. Then3∗Idenotes the open interval with the same center asIand three times the length ofI.
4.3 Example 3times an interval IfI= (0, 10), then3∗I = (−10, 20).
The next result is a key tool in the proof of the Hardy–Littlewood maximal inequality (4.8).
4.4 Vitali Covering Lemma
SupposeI1, . . . ,Inis a list of bounded nonempty open intervals ofR. Then there exists a disjoint sublistIk1, . . . ,Ikmsuch that
I1∪ · · · ∪In⊂(3∗Ik1)∪ · · · ∪(3∗Ikm).
4.5 Example Vitali Covering Lemma Supposen=4and
I1= (0, 10), I2= (9, 15), I3= (14, 22), I4= (21, 31). Then
3∗I1= (−10, 20), 3∗I2= (3, 21), 3∗I3= (6, 30), 3∗I4= (11, 41). Thus
I1∪I2∪I3∪I4⊂(3∗I1)∪(3∗I4).
In this example,I1,I4is the only sublist ofI1,I2,I3,I4that produces the conclusion of the Vitali Covering Lemma.
Proof of4.4 Letk1be such that
|Ik1|=max{|I1|, . . . ,|In|}.
The technique used here is called a greedy algorithm because at each stage we select the largest remaining interval that is disjoint from the previously selected intervals.
Supposek1, . . . ,kjhave been chosen.
Letkj+1be such that|Ikj+1|is as large as possible subject to the condition that Ik1, . . . ,Ikj+1 are disjoint. If there is no choice ofkj+1such thatIk1, . . . ,Ikj+1 are disjoint, then the procedure terminates.
Because we start with a finite list, the procedure must eventually terminate after some numbermof choices.
Supposej∈ {1, . . . ,n}. To complete the proof, we must show that Ij ⊂(3∗Ik1)∪ · · · ∪(3∗Ikm).
Ifj∈ {k1, . . . ,km}, then the inclusion above obviously holds.
Thus assume that j ∈ {/ k1, . . . ,km}. Because the process terminated without selectingj, the intervalIjis not disjoint from all ofIk1, . . . ,Ikm. LetIkLbe the first interval on this list not disjoint fromIj; thusIjis disjoint fromIk1, . . . ,IkL−1. Because jwas not chosen in stepL, we conclude that|IkL| ≥ |Ij|. BecauseIkL∩Ij̸=∅, this last inequality implies (easy exercise) thatIj ⊂3∗IkL, completing the proof.
Hardy–Littlewood Maximal Inequality
Now we come to a brilliant definition that turns out to be extraordinarily useful.
4.6 Definition Hardy–Littlewood maximal function;h∗
Suppose h: R → R is a Lebesgue measurable function. Then the Hardy–
Littlewood maximal functionofhis the functionh∗:R→[0,∞]defined by h∗(b) =sup
t>0
1 2t
Z b+t b−t |h|.
In other words,h∗(b)is the supremum over all bounded intervals centered atbof the average of|h|on those intervals.
4.7 Example Hardy–Littlewood maximal function of χ[0, 1]
As usual, letχ[0, 1]denote the characteristic function of the interval[0, 1]. Then
(χ[0, 1])∗(b) =
2(11−b) ifb≤0, 1 if0<b<1,
2b1 ifb≥1, The graph of (χ[0, 1])∗on[−2, 3]. as you should verify.
Ifh: R→Ris Lebesgue measurable andc∈R, then{b∈R:h∗(b)>c}is an open subset ofR, as you are asked to prove in Exercise9in this section. Thush∗ is a Borel measurable function.
Supposeh ∈ L1(R)andc>0. Markov’s inequality (4.1) estimates the size of the set on which|h|is larger thanc. Our next result estimates the size of the set on whichh∗is larger thanc. The Hardy–Littlewood maximal inequality proved in the next result is a key ingredient in the proof of the Lebesgue Differentiation Theorem (4.10). Note that this next result is considerably deeper than Markov’s inequality.
4.8 Hardy–Littlewood maximal inequality
Supposeh∈ L1(R). Then
|{b∈R:h∗(b)>c}| ≤ 3c∥h∥1
for everyc>0.
Proof SupposeFis a closed bounded subset of{b∈ R: h∗(b) > c}. We will show that|F| ≤ 3cR∞
−∞|h|, which implies our desired result [see Exercise24(a) in Section2D].
For eachb∈F, there existstb>0such that
4.9 1
2tb Z b+tb
b−tb |h|>c.
Clearly
F⊂ [
b∈F
(b−tb,b+tb).
The Heine–Borel Theorem (2.12) tells us that this open cover of a closed bounded set has a finite subcover. In other words, there existb1, . . . ,bn∈ Fsuch that
F⊂(b1−tb1,b1+tb1)∪ · · · ∪(bn−tbn,bn+tbn). To make the notation cleaner, relabel the open intervals above asI1, . . . ,In.
Now apply the Vitali Covering Lemma (4.4) to the list I1, . . . ,In, producing a disjoint sublistIk1, . . . ,Ikmsuch that
I1∪ · · · ∪In⊂(3∗Ik1)∪ · · · ∪(3∗Ikm). Thus
|F| ≤ |I1∪ · · · ∪In|
≤ |(3∗Ik1)∪ · · · ∪(3∗Ikm)|
≤ |3∗Ik1|+· · ·+|3∗Ikm|
=3(|Ik1|+· · ·+|Ikm|)
< 3 c
Z
Ik1|h|+· · ·+ Z
Ikm|h|
≤ 3c Z ∞
−∞|h|,
where the second-to-last inequality above comes from4.9(note that|Ikj|=2tbfor the choice ofbcorresponding toIkj) and the last inequality holds becauseIk1, . . . ,Ikm are disjoint.
The last inequality completes the proof.