Measure Theory (IX): Littlewood's Three Principles
24 Oct 2018
Littlewood’s three principles of real analysis are succint statements about the Lebesgue measure on Rn. Informally, they are given as follows:
- Lebesgue measurable sets are “nearly” open sets;
- Lebesgue measurable functions are “nearly” continuous;
- λ-a.e. convergence is “nearly” uniform convergence.
Here, “nearly” is in the sense of “up to a set of measure ε.” In this post, we formalise and prove these three statements.
Littlewood’s first principle
Lebesgue measurable sets are “nearly” open sets.
The following theorem states the regularity of Lebesgue measure, namely that Lebesgue measurable sets are well-approximated from the outside by open sets, and from the inside by compact sets.
Theorem: Let A⊆Rn. Then A is Lebesgue measurable if and only if for every ε>0, there exists an open set O such that A⊆O and λ∗(O\A)<ε.
Proof: (⇒) We first consider the case when λ(A)<∞. Take any countable cover (Ik) of A by half-open intervals with
k=1∑∞λ(Ik)<λ(A)+2ε,
then slightly expand each Ik into a half-open interval Ik′ with Ik⊆Ik′∘ and
λ(Ik′)<λ(Ik)+2k+1ε.
Then O=⋃kIk′∘ is an open set (hence Lebesgue measurable) containing A, with
λ(O)≤k=1∑∞λ(Ik′)<k=1∑∞λ(Ik)+2ε<λ(A)+ε.
Thus λ(O\A)=λ(O)−λ(A)<ε.
Now consider the case λ(A)=∞. Since [−m,m)n∈Hn is Lebesgue measurable with finite Lebesgue measure, so is Am=A∩[−m,m)n. Hence there exists Om⊆Rn open with λ(Om\Am)<2mε, so by taking O=⋃mOm, we have
O\A⟹λ(O\A)⊆m=1⋃∞Om\Am≤m=1∑∞λ(Om\Am)<ε.
(⇐) For each m≥1, there exists an open set A⊆Om⊆Rn such that λ∗(Om\A)<m1. Now S=⋂mOm is Borel (hence Lebesgue measurable), contains A, and
λ∗(S\A)≤λ∗(Om\A)<m1
for all m, so λ∗(S\A)=0. Since the Lebesgue measure is complete, S\A is Lebesgue measurable; hence A=S\(S\A) is also Lebesgue measurable. □
Corollary: If A⊆Rn is Lebesgue measurable and λ(A)<∞, then for every ε>0 there exists a compact set K⊆A such that λ(A\K)<ε.
Proof: Let Am=A∩[−m,m]n. Note that (Am) is an increasing sequence of Lebesgue measurable sets with union A, so λ(Am)→λ(A). In particular, there exists N≥1 with λ(AN)>λ(A)−2ε.
Now [−N,N]n\AN is measurable, so by Littlewood’s first principle there exists an open set [−N,N]n\AN⊆O⊆Rn with
λ(O\([−N,N]n\AN))<2ε.
Take K=AN\O=[−N,N]n\O, so K⊆AN and K is closed and bounded, hence compact. Now
AN\K=AN∩O⊆O\([−N,N]n\AN),
so λ(A\K)≤λ(A\AN)+λ(AN\K)<ε, as desired. □
Lusin’s theorem
Littlewood’s second principle states:
Lebesgue measurable functions are “nearly” continuous.
This is formalised in the following result:
Lusin’s theorem: Let E⊆Rn be Lebesgue measurable, with λ(E)<∞. Then a function f:E→R is measurable if and only if for any ε>0, there exists a compact set K⊆E with λ(E\K)<ε, such that f∣K is continuous on K.
Proof: (⇐) For each m≥1, there exists K=Km in the statement of the theorem for ε=m1. Note that f−1(a,∞)∩Km is an open set in K, hence Borel, hence Lebesgue measurable. Also, f−1(a,∞)∩Kmc has measure <ε. Thus
f−1(a,∞)=m=1⋃∞(f−1(a,∞)∩Km)∪m=1⋂∞(f−1(a,∞)∩Kmc),
where the first term is Lebesgue measurable, and the second term is a subset of the λ-null set ⋂mKmc, and is thus also Lebesgue measurable. Thus f−1(a,∞) is Lebesgue measurable for all a∈R, so f is Lebesgue measurable.
(⇒) We proceed by first showing the statement for some special cases.
f is a measurable step function: If f(E)={a1,…,am}, then for each k there exists Kk⊆f−1(ak) compact such that
λ(f−1(ak)\Kk)<mε.
Hence K=⋃kKk is a compact subset of E with
λ(E\K)≤k=1∑mλ(f−1(ak)\Kk)<ε.
Furthermore, Rn is a normal Hausdorff (ie. T4) space, so each Kk is separated from ⋃j=kKj by an open set O; in other words,
Kk⊆O⊆⎝⎛j=k⋃Kj⎠⎞c.
Hence f−1(ak)∩K=Kk=Kk∩O is open in K for every k, so f∣K is continuous on K.
f is a bounded measurable function: Let (φk±) be two sequences of nonnegative step functions which converge uniformly to f± respectively. Let φk=φk+−φk−; then (φk) is a sequence of step functions converging uniformly to f.
For each k≥1, take Kk⊆E compact such that λ(Kkc)<2kε and φk∣Kk is continuous on Kk. Then K=⋂kKk is compact, with
λ(Kc)≤k=1∑∞λ(Kkc)<ε,
and (φk∣K) is a sequence of continuous functions converging pointwise to f∣K; hence f∣K is continuous on K, and the convergence is uniform in K.
f is measurable: Since
k=1⋂∞{∣f∣>k}=∅,
there exists N≥1 with λ{∣f∣>N}<2ε. Now there exists K⊆E\{∣f∣>N} compact with λ((E\{∣f∣>N})\K)<2ε and f∣K is continuous on K; then
λ(E\K)≤λ((E\{∣f∣>N})\K)+λ{∣f∣>N}<ε,
and we are done. □
Egorov’s theorem
Littlewood’s third principle states:
λ-a.e. convergence is “nearly” uniform convergence.
This notion is captured by the following theorem, which in fact holds for general measure spaces.
Egorov’s theorem: Let (Ω,Σ,μ) be a finite measure space (ie. μ(Ω)<∞). Let f,f1,f2,…:Ω→R be μ-measurable functions such that (fn) converges μ-a.e. to f. Then for any ε>0, there exists K∈Σ such that μ(Kc)<ε and fn→f uniformly on K.
Proof: For m,n≥1, let Emn={∣f−fn∣>m1}, which is μ-measurable. Note that for fixed m, we have ∣f−fn∣≤m1 holds for all large enough n μ-a.e.; in other words
k=1⋂∞n=k⋃∞Emn
is μ-measurable, hence μ-null. Thus there exists km such that Fm=⋃n=km∞Emn) satisfies μ(Fm)<2mε. By possibly replacing each km by a bigger number, we may assume that k1<k2<⋯.
Now take K=⋂mFmc; then
μ(Kc)=μ(m=1⋃∞Fm)≤m=1∑∞μ(Fm)<ε,
Now for every ω∈K=⋂m⋂n≥kmEmnc, by definition we have
∣f(ω)−fn(ω)∣≤m1for all n≥km.
Hence (fn)→f uniformly on K. □
Proposition: Let (Ω,Σ,μ)=(E,ΣE,λ), where E⊆Rn is Lebesgue measurable with λ(E)<∞, and ΣE is the collection of Lebesgue measurable subsets of E. Then the set K in Egorov’s theorem can be chosen to be compact.
Proof: Take K~⊆E with λ(E\K~)<2ε, such that (fk)→f uniformly on K~. By regularity of λ, there exists K⊆K~ compact with λ(K~\K)<2ε. Then
λ(E\K)≤λ(E\K~)+λ(K~\K)<ε,
and (fk)→f uniformly on K. □
Comments (0)
Be the first to comment!