So far we’ve proven Littlewood’s first and third principles. In this last installment, we’ll prove his second principle, that “every function is nearly continuous.” This result is better known as Lusin’s Theorem.
From now on, we will let denote Lebesgue measure on . We’ll state two basic theorems, whose proofs make great exercises:
Theorem If are continuous and converge uniformly to , then is continuous.
Theorem If is measurable on , then there exists a sequence of step functions that converges pointwise to for almost every .
These theorems are true in much more generality. But these watered-down statements are all that we need.
Theorem (Luzin’s Theorem) Suppose is measurable and finite valued on a finitely-measured subset of . Then for every there exists a closed set with
such that is continuous.
Proof. The idea is to approximate pointwise almost everywhere with a sequence of step functions. Outside a set of small measure, the set functions are continuous. So then by Egorov’s Theorem, outside a set of small measure, is the uniform limit of continuous functions, and so is continuous.
To this end, let be a sequence of step functions so that almost everywhere. Let . Now for each , find some measurable for which and is continuous outside of (Why can we do this?). Then each is continuous outside of , whose measure is at most . Hence each is continuous on .
Since is finitely measured, and converges to pointwise on , by Egorov’s theorem there exists a subset with
for which is the uniform limit of on . But each is continuous on and so is continuous, being the uniform limit of continuous functions. Since , we have
as claimed. Lastly, we may assume without loss of generality that is closed, for otherwise we could, by Littlewood’s First Principle, approximate it’s complement with a finite union of open sets and take to be the closed complement of that union.
This completes this series on Littlewood’s three principles. Thanks for reading.