Lusin's Theorem: Difference between revisions

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==Proof==
==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>
We present an adaptation of the proofs found in references [2] and [3].
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>
We first prove a lemma: for all <math> /delta>0 </math> and measurable sets <math> B </math>, there exists a compact set <math> F\subseteq B </math> such that <math> \lambda(B\setminus F)<\delta. </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)<\delta/2. </math> The sequence of measurable sets defined by <math> V_n = V\cap [-n,n]</math> is nested and increasing, and satisfies <math> \bigcup_{n=1}^\infty V_n=V. </math> By continuity of the measure from below, there exists <math> N\in \mathbb{N} </math> such that <math> \lambda(B\setminus [-n,n])</delta /2.</math> Then <math> F:=V\cap [-N,N]</math> is a closed subset of the compact space <math> [-N,N], </math> and is hence compact. Masurability of the involved sets implies the desired inequality <math> \lambda(E\setminus K)<\delta. </math>
 
We now proceed with the main proof. By cardinality considerations, we may enumerate the collection of open intervals in <math>\mathbb{R} </math> with rational endpoints, say by <math> \{R_n\}_{n\in \mathbb{N}. </math> From the lemma, for all <math> n\in \mathbb{N} </math> we may instantiate compact sets <math> K_n \subseteq f^{-1}(R_n) </math> satisfying <math> \lambda(f^{-1}(R_n)\setminus K_n<\varepsilon </math>


==References==
==References==

Revision as of 02:01, 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

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

We first prove a lemma: for all and measurable sets , there exists 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, there exists such that Then is a closed subset of the compact space and is hence compact. Masurability of the involved sets implies the desired inequality

We now proceed with the main proof. By cardinality considerations, we may enumerate the collection of open intervals in with rational endpoints, say by Failed to parse (syntax error): {\displaystyle \{R_n\}_{n\in \mathbb{N}. } From the lemma, for all we may instantiate compact sets satisfying

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