1 (a) Show that the set consisting of those numbers in(0, 1)that have a decimal expansion containing one hundred consecutive4s is a Borel subset ofR.
(b) What is the Lebesgue measure of the set in part (a)?
2 Prove that there exists a bounded setA⊂Rsuch that|F| ≤ |A| −1for every closed setF⊂ A.
3 Prove that there exists a setA⊂Rsuch that|G\A|=∞for every open setG that containsA.
4 The phrasenontrivial intervalis used to denote an interval ofRthat contains more than one element. Recall that an interval might be open, closed, or neither.
(a) Prove that the union of each collection of nontrivial intervals ofRis the union of a countable subset of that collection.
(b) Prove that the union of each collection of nontrivial intervals ofRis a Borel set.
(c) Prove that there exists a collection of closed intervals ofRwhose union is not a Borel set.
5 Prove that ifA ⊂Ris Lebesgue measurable, then there exists an increasing sequenceF1⊂F2⊂ · · · of closed sets contained inAsuch that
A\
[∞ k=1
Fk=0.
6 SupposeA ⊂Rand|A| <∞. Prove that Ais Lebesgue measurable if and only if for everyε> 0there exists a setGthat is the union of finitely many disjoint bounded open intervals such that|A\G|+|G\A|<ε.
7 Prove that ifA ⊂ Ris Lebesgue measurable, then there exists a decreasing sequenceG1⊃G2⊃ · · · of open sets containingAsuch that
\∞
k=1
Gk
\A=0.
8 Prove that the collection of Lebesgue measurable subsets ofRis translation invariant. More precisely, prove that ifA ⊂ Ris Lebesgue measurable and t∈R, thent+Ais Lebesgue measurable.
9 Prove that the collection of Lebesgue measurable subsets ofRis dilation invari- ant. More precisely, prove that ifA⊂Ris Lebesgue measurable andt∈R, thentA(which is defined to be{ta:a∈A}) is Lebesgue measurable.
10 Prove that ifAandBare disjoint subsets ofRandBis Lebesgue measurable, then|A∪B|=|A|+|B|.
11 Prove that ifA⊂ Rand|A| > 0, then there exists a subset ofAthat is not Lebesgue measurable.
12 Supposeb< candA ⊂(b,c). Prove thatAis Lebesgue measurable if and only if|A|+|(b,c)\A|=c−b.
13 SupposeA⊂R. Prove thatAis Lebesgue measurable if and only if
|(−n,n)∩A|+|(−n,n)\A|=2n for everyn∈Z+.
14 Show that14and139 are both in the Cantor set.
15 Show that 1317 is not in the Cantor set.
16 List the eight open intervals whose union isG4in the definition of the Cantor set (2.74).
17 LetCdenote the Cantor set. Prove that1
2x+12y:x,y∈C = [0, 1]. 18 Prove that every open interval ofRcontains either infinitely many or no elements
in the Cantor set.
19 EvaluateZ 1
0 Λ, whereΛis the Cantor function.
20 Evaluate each of the following:
(a) Λ 139
; (b) Λ(0.93).
21 Find each of the following sets:
(a) Λ−1 {13}; (b) Λ−1 {165}.
22 (a) Supposexis a rational number in[0, 1]. Explain whyΛ(x)is rational.
(b) Supposex∈Cis such thatΛ(x)is rational. Explain whyxis rational.
23 Show that there exists a function f: R →Rsuch that the image under f of every nonempty open interval isR.
24 ForA⊂R, the quantity
sup{|F|:Fis a closed bounded subset ofRandF⊂ A} is called theinner measureofA.
(a) Show that ifAis a Lebesgue measurable subset ofR, then the inner measure ofAequals the outer measure ofA.
(b) Show that inner measure is not a measure on theσ-algebra of all subsets ofR.
2E Convergence of Measurable Functions
Recall that a measurable space is a pair(X,S), whereXis a set andS is aσ-algebra onX. We defined a function f: X → Rto beS-measurable if f−1(B) ∈ Sfor every Borel setB⊂R. In Section2Bwe proved some results aboutS-measurable functions; this was before we had introduced the notion of a measure.
In this section, we return to study measurable functions, but now with an emphasis on results that depend upon measures. The highlights of this section are the proofs of Egorov’s Theorem and Luzin’s Theorem.
Pointwise and Uniform Convergence
We begin this section with some definitions that you probably saw in an earlier course.
2.82 Definition pointwise convergence; uniform convergence
SupposeXis a set, f1,f2, . . . is a sequence of functions fromXtoR, andf is a function fromXtoR.
• The sequence f1,f2, . . . converges pointwiseonXto f if
klim→∞fk(x) = f(x) for eachx ∈X.
In other words,f1,f2, . . . converges pointwiseonXto f if for eachx∈X and everyε>0, there existsn∈Z+such that|fk(x)−f(x)|<εfor all integersk≥n.
• The sequence f1,f2, . . . converges uniformlyonXto f if for everyε>0, there existsn∈Z+such that|fk(x)−f(x)|<εfor all integersk≥nand allx∈X.
2.83 Example a sequence converging pointwise but not uniformly Suppose fk:[−1, 1]→Ris the
function whose graph is shown here and f:[−1, 1] →Ris the function defined by
f(x) =
(1 ifx̸=0, 2 ifx=0.
Then f1,f2, . . .converges pointwise on [−1, 1] to f but f1,f2, . . . does not converge uniformly on[−1, 1]to
f, as you should verify. The graph offk.
Like the difference between continuity and uniform continuity, the difference between pointwise convergence and uniform convergence lies in the order of the quantifiers. Take a moment to examine the definitions carefully. If a sequence of functions converges uniformly on some set, then it also converges pointwise on the same set; however, the converse is not true, as shown by Example2.83.
Example2.83also shows that the pointwise limit of continuous functions need not be continuous. However, the next result tells us that the uniform limit of continuous functions is continuous.
2.84 uniform limit of continuous functions is continuous
SupposeB ⊂ R and f1,f2, . . . is a sequence of functions from Bto Rthat converges uniformly onBto a function f:B → R. Supposeb ∈ Band fk is continuous atbfor eachk∈Z+. Then f is continuous atb.
Proof Supposeε>0. Letn∈Z+be such that|fn(x)−f(x)|< ε3for allx∈ B.
Becausefnis continuous atb, there existsδ>0such that|fn(x)−fn(b)|< ε3for allx∈(b−δ,b+δ)∩B.
Now supposex∈(b−δ,b+δ)∩B. Then
|f(x)−f(b)| ≤ |f(x)− fn(x)|+|fn(x)−fn(b)|+|fn(b)−f(b)|
<ε.
Thus f is continuous atb.
Egorov’s Theorem
Dmitri Egorov(1869–1931)proved the theorem below in 1911. You may encounter some books that spell his last name asEgoroff.
A sequence of functions that converges pointwise need not converge uniformly.
However, the next result says that a point- wise convergent sequence of functions on a measure space with finite total measure
almost converges uniformly, in the sense that it converges uniformly except on a set that can have arbitrarily small measure.
As an example of the next result, consider Lebesgue measureλon the inter- val [−1, 1] and the sequence of functions f1,f2, . . . in Example 2.83 that con- verges pointwise but not uniformly on [−1, 1]. Suppose ε > 0. Then taking E = [−1,−4ε]∪[4ε, 1], we haveλ([−1, 1]\E) < εand f1,f2, . . .converges uni- formly onE, as in the conclusion of the next result.
2.85 Egorov’s Theorem
Suppose(X,S,µ)is a measure space withµ(X)<∞. Suppose f1,f2, . . . is a sequence ofS-measurable functions fromXtoRthat converges pointwise on Xto a function f:X→R. Then for everyε>0, there exists a setE∈ Ssuch thatµ(X\E)<εandf1,f2, . . . converges uniformly to f onE.
Proof Suppose ε > 0. Temporarily fix n ∈ Z+. The definition of pointwise convergence implies that
2.86
[∞ m=1
\∞ k=m
{x∈X:|fk(x)−f(x)|< 1n}=X.
Form∈Z+, let
Am,n =
\∞ k=m
{x∈X:|fk(x)−f(x)|< 1n}.
Then clearlyA1,n ⊂A2,n⊂ · · · is an increasing sequence of sets and2.86can be rewritten as
[∞ m=1
Am,n=X.
The equation above implies (by2.59) thatlimm→∞µ(Am,n) =µ(X). Thus there existsmn ∈Z+such that
2.87 µ(X)−µ(Amn,n)< ε 2n. Now let
E=
\∞ n=1
Amn,n.
Then
µ(X\E) =µ X\
\∞ n=1
Amn,n
=µ[∞
n=1
(X\Amn,n)
≤
∑
∞n=1
µ(X\Amn,n)
<ε, where the last inequality follows from2.87.
To complete the proof, we must verify thatf1,f2, . . .converges uniformly to f onE. To do this, supposeε′>0. Letn∈Z+be such that 1n <ε′. ThenE⊂Amn,n, which implies that
|fk(x)− f(x)|< 1n <ε′
for allk≥mnand allx∈ E. Hence f1,f2, . . .does indeed converge uniformly to f onE.
Approximation by Simple Functions
2.88 Definition simple function
A function is calledsimpleif it takes on only finitely many values.
Suppose(X,S)is a measurable space, f:X → Ris a simple function, and c1, . . . ,cnare the distinct nonzero values of f. Then
f =c1χE
1+· · ·+cnχE
n,
whereEk = f−1({ck}). Thus this function f is anS-measurable function if and only ifE1, . . . ,En ∈ S(as you should verify).
2.89 approximation by simple functions
Suppose(X,S)is a measurable space andf:X→[−∞,∞]isS-measurable.
Then there exists a sequence f1,f2, . . .of functions fromXtoRsuch that (a) each fkis a simpleS-measurable function;
(b) |fk(x)| ≤ |fk+1(x)| ≤ |f(x)|for allk∈Z+and allx∈X;
(c) lim
k→∞fk(x) = f(x)for everyx∈X;
(d) f1,f2, . . .converges uniformly onXtof if f is bounded.
Proof The idea of the proof is that for each k ∈ Z+ and n ∈ Z, the interval [n,n+1)is divided into2kequally sized half-open subintervals. If f(x) ∈[0,k], we define fk(x)to be the left endpoint of the subinterval into which f(x)falls; if f(x) ∈ [−k, 0), we define fk(x)to be the right endpoint of the subinterval into whichf(x)falls; and if|f(x)|>k, we define fk(x)to be±k. Specifically, let
fk(x) =
2mk if0≤ f(x)≤kandm∈Zis such that f(x)∈m2k,m+12k ,
m+12k if −k≤ f(x)<0andm∈Zis such that f(x)∈2mk,m+12k , k if f(x)>k,
−k if f(x)<−k.
Eachf−1 [2mk,m+12k )∈ S because f is anS-measurable function. Thus each fk is anS-measurable simple function; in other words, (a) holds.
Also, (b) holds because of how we have definedfk. The definition of fkimplies that
2.90 |fk(x)− f(x)| ≤ 21k for allx∈Xsuch that f(x)∈[−k,k]. Thus we see that (c) holds.
Finally,2.90shows that (d) holds.
Luzin’s Theorem
Nikolai Luzin(1883–1950)proved the theorem below in 1912. Most mathematics literature in English refers to the result below asLusin’s Theorem. However,Luzinis the correct transliteration from Russian into English;Lusinis the
transliteration into German.
Our next result is surprising. It says that an arbitrary Borel measurable function is almost continuous, in the sense that its restriction to a large closed set is contin- uous. Here, the phraselarge closed set means that we can take the complement of the closed set to have arbitrarily small measure.
Be careful about the interpretation of
the conclusion of Luzin’s Theorem that f|Bis a continuous function onB. This is not the same as saying that f (on its original domain) is continuous at each point ofB. For example,χQis discontinuous at every point ofR. However,χQ|R\Qis a continuous function onR\Q(because this function is identically0on its domain).
2.91 Luzin’s Theorem
Supposeg:R→Ris a Borel measurable function. Then for everyε>0, there exists a closed setF⊂Rsuch that|R\F|<εandg|Fis a continuous function onF.
Proof First consider the special case whereg = d1χD1+· · ·+dnχDn for some distinct nonzero d1, . . . ,dn ∈ Rand some disjoint Borel setsD1, . . . ,Dn ⊂ R.
Supposeε>0. For eachk∈ {1, . . . ,n}, there exist (by2.71) a closed setFk⊂Dk
and an open setGk⊃Dksuch that
|Gk\Dk|< ε
2n and |Dk\Fk|< ε 2n.
BecauseGk\Fk = (Gk\Dk)∪(Dk\Fk), we have|Gk\Fk| < nε for eachk∈ {1, . . . ,n}.
Let
F= [n
k=1
Fk
∪
\n
k=1
(R\Gk).
ThenFis a closed subset ofRandR\F⊂Snk=1(Gk\Fk). Thus|R\F|<ε.
BecauseFk⊂Dk, we see thatgis identicallydkonFk. Thusg|Fkis continuous for eachk∈ {1, . . . ,n}. Because
\n
k=1
(R\Gk)⊂
\n
k=1
(R\Dk),
we see thatgis identically0onTnk=1(R\Gk). Thusg|Tnk=1(R\Gk)is continuous.
Putting all this together, we conclude thatg|Fis continuous (use Exercise9in this section), completing the proof in this special case.
Now consider an arbitrary Borel measurable functiong:R→R. By2.89, there exists a sequenceg1,g2, . . .of functions fromRtoRthat converges pointwise onR tog, where eachgkis a simple Borel measurable function.
Supposeε>0. By the special case already proved, for eachk∈Z+, there exists a closed setCk ⊂Rsuch that|R\Ck|< 2kε+1 andgk|Ck is continuous. Let
C=
\∞ k=1
Ck.
ThusCis a closed set andgk|Cis continuous for everyk∈Z+. Note that R\C=
[∞ k=1
(R\Ck); thus|R\C|< ε2.
For eachm∈ Z, the sequenceg1|(m,m+1),g2|(m,m+1), . . .converges pointwise on(m,m+1)tog|(m,m+1). Thus by Egorov’s Theorem (2.85), for eachm ∈ Z, there is a Borel setEm⊂(m,m+1)such thatg1,g2, . . .converges uniformly tog onEmand
|(m,m+1)\Em|< ε 2|m|+3.
Thusg1,g2, . . .converges uniformly togonC∩Emfor eachm∈Z. Because each gk|C is continuous, we conclude (using2.84) thatg|C∩Em is continuous for each m∈Z. Thusg|Dis continuous, where
D= [
m∈Z
(C∩Em). Because
R\D⊂Z∪[
m∈Z
(m,m+1)\Em
∪(R\C), we have|R\D|<ε.
There exists a closed setF⊂Dsuch that|D\F|<ε− |R\D|(by2.65). Now
|R\F|=|(R\D)∪(D\F)| ≤ |R\D|+|D\F|<ε.
Because the restriction of a continuous function to a smaller domain is also continuous, g|Fis continuous, completing the proof.
We need the following result to get another version of Luzin’s Theorem.
2.92 continuous extensions of continuous functions
• Every continuous function on a closed subset ofRcan be extended to a continuous function on all ofR.
• More precisely, ifF⊂Ris closed andg:F→Ris continuous, then there exists a continuous functionh:R→Rsuch thath|F=g.
Proof SupposeF⊂ Ris closed andg: F→Ris continuous. ThusR\Fis the union of a collection of disjoint open intervals{Ik}. For each such interval of the form(a,∞)or of the form(−∞,a), defineh(x) =g(a)for allxin the interval.
For each intervalIkof the form(b,c)withb<candb,c∈R, definehon[b,c] to be the linear function such thath(b) =g(b)andh(c) =g(c).
Defineh(x) = g(x)for allx ∈Rfor whichh(x)has not been defined by the previous two paragraphs. Thenh: R→Ris continuous andh|F =g.
The next result gives a slightly modified way to state Luzin’s Theorem. You can think of this version as saying that the value of a Borel measurable function can be changed on a set with small Lebesgue measure to produce a continuous function.
2.93 Luzin’s Theorem, second version
SupposeE⊂Randg:E→Ris a Borel measurable function. Then for every ε>0, there exists a closed setF⊂Eand a continuous functionh:R→Rsuch that|E\F|<εandh|F =g|F.
Proof Supposeε>0. Extendgto a functionge:R→Rby defining ge(x) =
(g(x) ifx∈E,
0 ifx∈R\E.
By the first version of Luzin’s Theorem (2.91), there is a closed setC ⊂ Rsuch that|R\C|< εandeg|Cis a continuous function onC. There exists a closed set F⊂C∩Esuch that|(C∩E)\F|<ε− |R\C|(by2.65). Thus
|E\F| ≤ (C∩E)\F
∪(R\C)≤ |(C∩E)\F|+|R\C|<ε.
Noweg|Fis a continuous function onF. Also,eg|F =g|F(becauseF⊂E) . Use 2.92to extendeg|Fto a continuous functionh:R→R.
The building at Moscow State University where the mathematics seminar organized by Egorov and Luzin met. Both Egorov and Luzin had been students at Moscow State University and then later became faculty members at the same institution. Luzin’s
PhD thesis advisor was Egorov.
CC-BY-SA A. Savin
Lebesgue Measurable Functions
2.94 Definition Lebesgue measurable function
A function f: A→R, whereA⊂R, is calledLebesgue measurableif f−1(B) is a Lebesgue measurable set for every Borel setB⊂R.
Iff: A→Ris a Lebesgue measurable function, thenAis a Lebesgue measurable subset ofR[becauseA= f−1(R)]. IfAis a Lebesgue measurable subset ofR, then the definition above is the standard definition of anS-measurable function, whereS is theσ-algebra of all Lebesgue measurable subsets ofA.
The following list summarizes and reviews some crucial definitions and results:
• A Borel set is an element of the smallestσ-algebra onRthat contains all the open subsets ofR.
• A Lebesgue measurable set is an element of the smallestσ-algebra onRthat contains all the open subsets ofRand all the subsets ofRwith outer measure0.
• The terminologyLebesgue setwould make good sense in parallel to the termi- nologyBorel set. However,Lebesgue sethas another meaning, so we need to useLebesgue measurable set.
• Every Lebesgue measurable set differs from a Borel set by a set with outer measure0. The Borel set can be taken either to be contained in the Lebesgue measurable set or to contain the Lebesgue measurable set.
• Outer measure restricted to the σ-algebra of Borel sets is called Lebesgue measure.
• Outer measure restricted to theσ-algebra of Lebesgue measurable sets is also called Lebesgue measure.
• Outer measure is not a measure on theσ-algebra of all subsets ofR.
• A function f: A→R, whereA⊂R, is called Borel measurable if f−1(B)is a Borel set for every Borel setB⊂R.
• A function f: A→R, whereA⊂R, is called Lebesgue measurable if f−1(B) is a Lebesgue measurable set for every Borel setB⊂R.
“Passing from Borel to Lebesgue measurable functions is the work of the devil. Don’t even consider it!”
–Barry Simon(winner of the American Mathematical Society Steele Prize for Lifetime Achievement), in his five-volume seriesA Comprehensive Course in Analysis
Although there exist Lebesgue measur- able sets that are not Borel sets, you are unlikely to encounter one. Similarly, a Lebesgue measurable function that is not Borel measurable is unlikely to arise in anything you do. A great way to simplify the potential confusion about Lebesgue measurable functions being defined by in- verse images of Borel sets is to consider only Borel measurable functions.
“He professes to have received no sinister measure.”
–Measure for Measure, by William Shakespeare The next result states that if we adopt
the philosophy that what happens on a set of outer measure0 does not matter much, then we might as well restrict our attention to Borel measurable functions.
2.95 every Lebesgue measurable function is almost Borel measurable Supposef:R→Ris a Lebesgue measurable function. Then there exists a Borel measurable functiong:R→Rsuch that
|{x∈R:g(x)̸= f(x)}|=0.
Proof There exists a sequencef1,f2, . . .of Lebesgue measurable simple functions fromRtoRconverging pointwise onRto f (by2.89). Supposek∈Z+. Then there existc1, . . . ,cn ∈Rand disjoint Lebesgue measurable setsA1, . . . ,An ⊂Rsuch that
fk=c1χA1+· · ·+cnχAn.
For eachj ∈ {1, . . . ,n}, there exists a Borel setBj ⊂ Ajsuch that|Aj\Bj|= 0 [by the equivalence of (a) and (d) in2.71]. Let
gk=c1χB1+· · ·+cnχBn.
Thengkis a Borel measurable function and|{x∈R:gk(x)̸= fk(x)}|=0.
Ifx ∈/S∞k=1{x ∈R:gk(x)̸= fk(x)}, thengk(x) = fk(x)for allk∈ Z+and hencelimk→∞gk(x) = f(x). Let
E={x∈R: lim
k→∞gk(x)exists inR}.
ThenEis a Borel subset ofR[by Exercise14(b) in Section2B]. Also, R\E⊂
[∞ k=1
{x∈R:gk(x)̸= fk(x)} and thus|R\E|=0. Forx∈R, let
2.96 g(x) = lim
k→∞(χEgk)(x).
Ifx ∈ E, then the limit above exists by the definition ofE; ifx∈ R\E, then the limit above exists because(χEgk)(x) =0for allk∈Z+.
For eachk∈Z+, the functionχEgkis Borel measurable. Thus2.96implies thatg is a Borel measurable function (by2.48). Because
{x∈R:g(x)̸= f(x)} ⊂ [∞ k=1
{x ∈R:gk(x)̸= fk(x)}, we have|{x∈R:g(x)̸= f(x)}|=0, completing the proof.