Lusin's Theorem

Introduction

Lusin's Theorem formalizes the fundamental measure-theoretic principle that every measurable function is "nearly" continuous. This is the second of Littlewood's famed three principles of measure theory, which he elaborated in his 1944 work "Lectures on the Theory of Functions"[1] as

"There are three principles, roughly expressible in the following terms: Every (measurable) set is nearly a finite sum of intervals; every function (of class Lp) is nearly continuous; every convergent sequence of functions is nearly uniformly convergent."

Lusin's theorem is hence a key tool in working with measurable functions, often enabling one to reduce measurable, yet intractible functions to the consideration of a continuous approximation.

Statement

Let $(\mathbb {R} ,{\mathcal {M}}_{\lambda },\lambda )$ be the Lebesque measure space on $\mathbb {R}$, and $E$ a measurable subset of $\mathbb {R}$ satisfying $\lambda (E)<+\infty .$ Let $f:E\to \mathbb {R}$ be a measurable real-valued function on $E$. Then for all $\varepsilon >0,$ there exists a compact set $K\subseteq E$ such that the restriction of $f$ to $K$ is continuous.

Proof

We present an adaptation of the proofs found in references [2] and [3].

We first prove a lemma: for all $\delta >0$ and measurable sets $B$, there exists a compact set $F\subseteq B$ such that $\lambda (B\setminus F)<\delta .$

By a standard theorem in measure theory, there exists a closed set $V\subseteq E$ such that $\lambda (E\setminus V)<\delta /2.$ The sequence of measurable sets defined by $V_{n}=V\cap [-n,n]$ is nested and increasing, and satisfies $\bigcup _{n=1}^{\infty }V_{n}=V.$ By continuity of the measure from below, there hence exists $N\in \mathbb {N}$ such that $\lambda (B\setminus [-n,n])<\delta /2.$ Then $F:=V\cap [-N,N]$ is a closed subset of the compact space $[-N,N],$ and is consequently compact. Measurability of the relevant sets implies the desired inequality $\lambda (E\setminus K)<\delta .$

We now proceed with the main proof. Let $\varepsilon >0$ be given. By cardinality considerations, we may enumerate the collection of open intervals in $\mathbb {R}$ with rational endpoints, say by $\{R_{n}\}_{n\in \mathbb {N} }.$ By the measurability of $f$ and the closure of the $\sigma$-algebra under complements and countable intersections, for each $n\in \mathbb {N}$, both $f^{-1}(R_{n})$ and $E\setminus f^{-1}(R_{n})$ are measurable. From the lemma, there exist compact sets $K_{n}\subseteq f^{-1}(R_{n})$ and $K_{n}'\subseteq E\setminus f^{-1}(R_{n})$ satisfying $\lambda (f^{-1}(R_{n})\setminus (K_{n}\cup K_{n}'))<\varepsilon /2^{n}$.

Note that a finite union of compact sets, as well as an arbitrary intersection of compact sets, remains compact. Define $K=\bigcap _{n=1}^{\infty }(K_{n}\cup K_{n}'),$ so that $K$ is compact and satisfies $\lambda (E\setminus K)<\varepsilon .$

Let $x\in K$, and by density of the rationals fix an $n\in \mathbb {N}$ such that $x\in R_{n}.$ Then $x\in {\mathcal {O}}:=\mathbb {R} \setminus K_{n}'$ and $f({\mathcal {O}}\cap K)\subseteq R_{n}.$ From the definition, this implies that the restriction of $f$ to $K$ is continuous.

References

[1] Littlewood, J. E. "Lectures on the Theory of Functions." Oxford University Press. 1944.

[2] Talvila, Erik; Loeb, Peter. "Lusin's Theorem and Bochner Integration." arXiv. 2004. https://arxiv.org/abs/math/0406370

[3] https://conf.math.illinois.edu/~mjunge/54004-lusin.pdf