Lusin's Theorem

Introduction

Lusin's Theorem formalizes the measure-theoretic principle that pointwise convergence is "nearly" uniformly convergent. 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 $f:E\to \mathbb {R}$ a be a measurable real-valued function on $E$. 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 $\lambda (B\setminus F)<\delta .$

From a standard theorem, we know already the existence of a closed set $V\subseteq E$ with $\lambda (E\setminus V)<\delta /2.$ 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 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 hence compact. Masurability of the involved 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{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 remains compact, and 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 and the axiom of choice 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