Measure Theory (IX): Littlewood's Three Principles

24 Oct 2018

• measure theory

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\lambda^* (S\backslash A)\leq\lambda^* (O_m\backslash A)<\frac1m$$

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

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

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

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

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

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

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

so $\lambda(A\backslash K)\leq\lambda(A\backslash A_N)+\lambda(A_N\backslash K)<\eps$, 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 $E\subseteq\bb R^n$ be Lebesgue measurable, with $\lambda(E)<\infty$. Then a function $f:E\to\bb R$ is measurable if and only if for any $\eps>0$, there exists a compact set $K\subseteq E$ with $\lambda(E\backslash K)<\eps$, such that $f\vert_K$ is continuous on $K$.

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

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

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

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

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

$$\lambda(f^{-1}(a_k)\backslash K_k)<\frac\eps m.$$

Hence $K=\bigcup_kK_k$ is a compact subset of $E$ with

$$\lambda(E\backslash K)\leq\sum_{k=1}^m\lambda(f^{-1}(a_k)\backslash K_k)<\eps.$$

Furthermore, $\bb R^n$ is a normal Hausdorff (ie. $T_4$) space, so each $K_k$ is separated from $\bigcup_{j\neq k}K_j$ by an open set $O$; in other words,

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

Hence $f^{-1}(a_k)\cap K=K_k=K_k\cap O$ is open in $K$ for every $k$, so $f\vert_K$ is continuous on $K$.

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

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

$$\lambda(K^c)\leq\sum_{k=1}^\infty\lambda(K_k^c)<\eps,$$

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

$f$ is measurable: Since

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

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

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

and we are done. $\qed$

Egorov’s theorem

Littlewood’s third principle states:

$\lambda$-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 $(\Omega,\Sigma,\mu)$ be a finite measure space (ie. $\mu(\Omega)<\infty$). Let $f,f_1,f_2,\ldots:\Omega\to\bb R$ be $\mu$-measurable functions such that $(f_n)$ converges $\mu$-a.e. to $f$. Then for any $\eps>0$, there exists $K\in\Sigma$ such that $\mu(K^c)<\eps$ and $f_n\to f$ uniformly on $K$.

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

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

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

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

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

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

$$|f(\omega)-f_n(\omega)|\leq\frac1m\quad\text{for all }n\geq k_m.$$

Hence $(f_n)\to f$ uniformly on $K$. $\qed$

Proposition: Let $(\Omega,\Sigma,\mu)=(E,\Sigma_E,\lambda)$, where $E\subseteq\bb R^n$ is Lebesgue measurable with $\lambda(E)<\infty$, and $\Sigma_E$ 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 $\tilde K\subseteq E$ with $\lambda(E\backslash\tilde K)<\frac\eps2$, such that $(f_k)\to f$ uniformly on $\tilde K$. By regularity of $\lambda$, there exists $K\subseteq\tilde K$ compact with $\lambda(\tilde K\backslash K)<\frac\eps2$. Then

$$\lambda(E\backslash K)\leq\lambda(E\backslash\tilde K)+\lambda(\tilde K\backslash K)<\eps,$$

and $(f_k)\to f$ uniformly on $K$. $\qed$