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Measure Theory (IX): Littlewood's Three Principles

Littlewood’s three principles of real analysis are succint statements about the Lebesgue measure on Rn\bb R^n. Informally, they are given as follows:

  1. Lebesgue measurable sets are “nearly” open sets;
  2. Lebesgue measurable functions are “nearly” continuous;
  3. λ\lambda-a.e. convergence is “nearly” uniform convergence.

Here, “nearly” is in the sense of “up to a set of measure ε\eps.” 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 ARnA\subseteq\bb R^n. Then AA is Lebesgue measurable if and only if for every ε>0\eps>0, there exists an open set OO such that AOA\subseteq O and λ(O\A)<ε\lambda^* (O\backslash A)<\eps.

Proof: (\Rightarrow) We first consider the case when λ(A)<\lambda(A)<\infty. Take any countable cover (Ik)(I_k) of AA by half-open intervals with

k=1λ(Ik)<λ(A)+ε2, \sum_{k=1}^\infty\lambda(I_k)<\lambda(A)+\frac\eps2,

then slightly expand each IkI_k into a half-open interval IkI_k’ with IkIkI_k\subseteq I_k’^\circ and

λ(Ik)<λ(Ik)+ε2k+1. \lambda(I_k')<\lambda(I_k)+\frac\eps{2^{k+1}}.

Then O=kIkO=\bigcup_kI_k’^\circ is an open set (hence Lebesgue measurable) containing AA, with

λ(O)k=1λ(Ik)<k=1λ(Ik)+ε2<λ(A)+ε. \begin{aligned} \lambda(O)&\leq\sum_{k=1}^\infty\lambda(I_k')\\ &<\sum_{k=1}^\infty\lambda(I_k)+\frac\eps2\\ &<\lambda(A)+\eps. \end{aligned}

Thus λ(O\A)=λ(O)λ(A)<ε\lambda(O\backslash A)=\lambda(O)-\lambda(A)<\eps.

Now consider the case λ(A)=\lambda(A)=\infty. Since [m,m)nHn[-m,m)^n\in\mc H_n is Lebesgue measurable with finite Lebesgue measure, so is Am=A[m,m)nA_m=A\cap[-m,m)^n. Hence there exists OmRnO_m\subseteq\bb R^n open with λ(Om\Am)<ε2m\lambda(O_m\backslash A_m)<\frac\eps{2^m}, so by taking O=mOmO=\bigcup_mO_m, we have

O\Am=1Om\Am    λ(O\A)m=1λ(Om\Am)<ε. \begin{aligned} O\backslash A&\subseteq\bigcup_{m=1}^\infty O_m\backslash A_m\\ \implies\lambda(O\backslash A)&\leq\sum_{m=1}^\infty\lambda(O_m\backslash A_m)<\eps. \end{aligned}

(\Leftarrow) For each m1m\geq1, there exists an open set AOmRnA\subseteq O_m\subseteq\bb R^n such that λ(Om\A)<1m\lambda^* (O_m\backslash A)<\frac1m. Now S=mOmS=\bigcap_mO_m is Borel (hence Lebesgue measurable), contains AA, and

λ(S\A)λ(Om\A)<1m \lambda^* (S\backslash A)\leq\lambda^* (O_m\backslash A)<\frac1m

for all mm, so λ(S\A)=0\lambda^* (S\backslash A)=0. Since the Lebesgue measure is complete, S\AS\backslash A is Lebesgue measurable; hence A=S\(S\A)A=S\backslash(S\backslash A) is also Lebesgue measurable.  \qed

Corollary: If ARnA\subseteq\bb R^n is Lebesgue measurable and λ(A)<\lambda(A)<\infty, then for every ε>0\eps>0 there exists a compact set KAK\subseteq A such that λ(A\K)<ε\lambda(A\backslash K)<\eps.

Proof: Let Am=A[m,m]nA_m=A\cap[-m,m]^n. Note that (Am)(A_m) is an increasing sequence of Lebesgue measurable sets with union AA, so λ(Am)λ(A)\lambda(A_m)\to\lambda(A). In particular, there exists N1N\geq1 with λ(AN)>λ(A)ε2\lambda(A_N)>\lambda(A)-\frac\eps2.

Now [N,N]n\AN[-N,N]^n\backslash A_N is measurable, so by Littlewood’s first principle there exists an open set [N,N]n\ANORn[-N,N]^n\backslash A_N\subseteq O\subseteq\bb R^n with

λ(O\([N,N]n\AN))<ε2. \lambda(O\backslash([-N,N]^n\backslash A_N))<\frac\eps2.

Take K=AN\O=[N,N]n\OK=A_N\backslash O=[-N,N]^n\backslash O, so KANK\subseteq A_N and KK is closed and bounded, hence compact. Now

AN\K=ANOO\([N,N]n\AN),A_N\backslash K=A_N\cap O\subseteq O\backslash([-N,N]^n\backslash A_N),

so λ(A\K)λ(A\AN)+λ(AN\K)<ε\lambda(A\backslash K)\leq\lambda(A\backslash A_N)+\lambda(A_N\backslash K)<\eps, as desired.  \qed

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 ERnE\subseteq\bb R^n be Lebesgue measurable, with λ(E)<\lambda(E)<\infty. Then a function f:ERf:E\to\bb R is measurable if and only if for any ε>0\eps>0, there exists a compact set KEK\subseteq E with λ(E\K)<ε\lambda(E\backslash K)<\eps, such that fKf\vert_K is continuous on KK.

Proof: (\Leftarrow) For each m1m\geq1, there exists K=KmK=K_m in the statement of the theorem for ε=1m\eps=\frac1m. Note that f1(a,)Kmf^{-1}(a,\infty)\cap K_m is an open set in KK, hence Borel, hence Lebesgue measurable. Also, f1(a,)Kmcf^{-1}(a,\infty)\cap K_m^c has measure <ε<\eps. Thus

f1(a,)=m=1(f1(a,)Km)m=1(f1(a,)Kmc),f^{-1}(a,\infty)=\bigcup_{m=1}^\infty(f^{-1}(a,\infty)\cap K_m)\cup\bigcap_{m=1}^\infty(f^{-1}(a,\infty)\cap K_m^c),

where the first term is Lebesgue measurable, and the second term is a subset of the λ\lambda-null set mKmc\bigcap_mK_m^c, and is thus also Lebesgue measurable. Thus f1(a,)f^{-1}(a,\infty) is Lebesgue measurable for all aRa\in\bb R, so ff is Lebesgue measurable.

(\Rightarrow) We proceed by first showing the statement for some special cases.

ff is a measurable step function: If f(E)={a1,,am}f(E)=\{a_1,\ldots,a_m\}, then for each kk there exists Kkf1(ak)K_k\subseteq f^{-1}(a_k) compact such that

λ(f1(ak)\Kk)<εm. \lambda(f^{-1}(a_k)\backslash K_k)<\frac\eps m.

Hence K=kKkK=\bigcup_kK_k is a compact subset of EE with

λ(E\K)k=1mλ(f1(ak)\Kk)<ε. \lambda(E\backslash K)\leq\sum_{k=1}^m\lambda(f^{-1}(a_k)\backslash K_k)<\eps.

Furthermore, Rn\bb R^n is a normal Hausdorff (ie. T4T_4) space, so each KkK_k is separated from jkKj\bigcup_{j\neq k}K_j by an open set OO; in other words,

KkO(jkKj)c.K_k\subseteq O\subseteq\left(\bigcup_{j\neq k}K_j\right)^c.

Hence f1(ak)K=Kk=KkOf^{-1}(a_k)\cap K=K_k=K_k\cap O is open in KK for every kk, so fKf\vert_K is continuous on KK.

ff is a bounded measurable function: Let (φk±)(\varphi^\pm_k) be two sequences of nonnegative step functions which converge uniformly to f±f^\pm respectively. Let φk=φk+φk\varphi_k=\varphi^+_k-\varphi^-_k; then (φk)(\varphi_k) is a sequence of step functions converging uniformly to ff.

For each k1k\geq1, take KkEK_k\subseteq E compact such that λ(Kkc)<ε2k\lambda(K_k^c)<\frac\eps{2^k} and φkKk\varphi_k\vert_{K_k} is continuous on KkK_k. Then K=kKkK=\bigcap_kK_k is compact, with

λ(Kc)k=1λ(Kkc)<ε, \lambda(K^c)\leq\sum_{k=1}^\infty\lambda(K_k^c)<\eps,

and (φkK)(\varphi_k\vert_K) is a sequence of continuous functions converging pointwise to fKf\vert_K; hence fKf\vert_K is continuous on KK, and the convergence is uniform in KK.

ff is measurable: Since

k=1{f>k}=,\bigcap_{k=1}^\infty\{|f|>k\}=\emptyset,

there exists N1N\geq1 with λ{f>N}<ε2\lambda\{\lvert f\rvert>N\}<\frac\eps2. Now there exists KE\{f>N}K\subseteq E\backslash\{\lvert f\rvert>N\} compact with λ((E\{f>N})\K)<ε2\lambda((E\backslash\{\lvert f\rvert>N\})\backslash K)<\frac\eps2 and fKf\vert_K is continuous on KK; then

λ(E\K)λ((E\{f>N})\K)+λ{f>N}<ε, \lambda(E\backslash K)\leq\lambda((E\backslash\{|f|>N\})\backslash K)+\lambda\{|f|>N\}<\eps,

and we are done.  \qed

Egorov’s theorem

Littlewood’s third principle states:

λ\lambda-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)(\Omega,\Sigma,\mu) be a finite measure space (ie. μ(Ω)<\mu(\Omega)<\infty). Let f,f1,f2,:ΩRf,f_1,f_2,\ldots:\Omega\to\bb R be μ\mu-measurable functions such that (fn)(f_n) converges μ\mu-a.e. to ff. Then for any ε>0\eps>0, there exists KΣK\in\Sigma such that μ(Kc)<ε\mu(K^c)<\eps and fnff_n\to f uniformly on KK.

Proof: For m,n1m,n\geq1, let Emn={ffn>1m}E_{mn}=\{\lvert f-f_n\rvert>\frac1m\}, which is μ\mu-measurable. Note that for fixed mm, we have ffn1m\lvert f-f_n\rvert\leq\frac1m holds for all large enough nn μ\mu-a.e.; in other words

k=1n=kEmn\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty E_{mn}

is μ\mu-measurable, hence μ\mu-null. Thus there exists kmk_m such that Fm=n=kmEmn)F_m=\bigcup_{n=k_m}^\infty E_{mn}) satisfies μ(Fm)<ε2m\mu(F_m)<\frac\eps{2^m}. By possibly replacing each kmk_m by a bigger number, we may assume that k1<k2<k_1< k_2<\cdots.

Now take K=mFmcK=\bigcap_mF_m^c; then

μ(Kc)=μ(m=1Fm)m=1μ(Fm)<ε, \begin{aligned} \mu(K^c)&=\mu\left(\bigcup_{m=1}^\infty F_m\right)\\ &\leq\sum_{m=1}^\infty\mu(F_m)<\eps, \end{aligned}

Now for every ωK=mnkmEmnc\omega\in K=\bigcap_m\bigcap_{n\geq k_m}E_{mn}^c, by definition we have

f(ω)fn(ω)1mfor all nkm.|f(\omega)-f_n(\omega)|\leq\frac1m\quad\text{for all }n\geq k_m.

Hence (fn)f(f_n)\to f uniformly on KK.  \qed

Proposition: Let (Ω,Σ,μ)=(E,ΣE,λ)(\Omega,\Sigma,\mu)=(E,\Sigma_E,\lambda), where ERnE\subseteq\bb R^n is Lebesgue measurable with λ(E)<\lambda(E)<\infty, and ΣE\Sigma_E is the collection of Lebesgue measurable subsets of EE. Then the set KK in Egorov’s theorem can be chosen to be compact.

Proof: Take K~E\tilde K\subseteq E with λ(E\K~)<ε2\lambda(E\backslash\tilde K)<\frac\eps2, such that (fk)f(f_k)\to f uniformly on K~\tilde K. By regularity of λ\lambda, there exists KK~K\subseteq\tilde K compact with λ(K~\K)<ε2\lambda(\tilde K\backslash K)<\frac\eps2. Then

λ(E\K)λ(E\K~)+λ(K~\K)<ε, \lambda(E\backslash K)\leq\lambda(E\backslash\tilde K)+\lambda(\tilde K\backslash K)<\eps,

and (fk)f(f_k)\to f uniformly on KK.  \qed

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