Lusin's Theorem: Difference between revisions

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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.
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.


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


==Proof==
Let <math> \varepsilon>0 </math> be given. We first show that if <math> A </math> the existence of a compact set <math> K\subseteq E </math> such that <math> \lambda(E\setminus K)<\varepsilon. </math> From a standard theorem, we know already the existence of a closed set <math> V\subseteq E </math>
with <math> \lambda(E\setminus V)<\varepsilon/2. </math> The sequence of measurable sets defined by <math> E_n = E\cap [-n,n]</math> is nested and increasing, and satisfies <math> \bigcup_{n=1}^\infty E_n=E. </math> By continuity of the measure from below, we deduce the existence of an <math> N\in \mathbb{N} </math> such that <math> \lambda(E\setminus [-n,n])<\varepsilon/2./math> Then <math> K:=E\cap [-N,N]\subseteq E</math> is a closed subset of the compact set <math> [-N,N], </math> and is hence compact. Further, measurability implies the desired inequality <math> \lambda(E\setminus K)<\varepsilon. </math>
By cardinality considerations, we may enumerate the collection of open intervals in <math>\mathbb{R} </math> with rational endpoints, say by <math> \{V_n\}_{n\in \mathbb{N}. </math>


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

Revision as of 01:52, 18 December 2020

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 be the Lebesque measure space on , and a measurable subset of satisfying Let a be a measurable real-valued function on . For all there exists a compact set such that the restriction of to is continuous.

Proof

Let be given. We first show that if the existence of a compact set such that From a standard theorem, we know already the existence of a closed set with The sequence of measurable sets defined by is nested and increasing, and satisfies By continuity of the measure from below, we deduce the existence of an such that is a closed subset of the compact set and is hence compact. Further, measurability implies the desired inequality

By cardinality considerations, we may enumerate the collection of open intervals in with rational endpoints, say by Failed to parse (syntax error): {\displaystyle \{V_n\}_{n\in \mathbb{N}. }

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