Littlewood’s Three Principles (3/3)

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 \mu^d denote Lebesgue measure on \mathbb{R}^d. We’ll state two basic theorems, whose proofs make great exercises:

Theorem If \{f_n\}_{n=1}^{\infty}:\mathbb{R}^d\to \mathbb{R} are continuous and converge uniformly to f, then f is continuous.

Theorem If f is measurable on \mathbb{R}^d, then there exists a sequence of step functions \{\psi_n\}_{n=1}^{\infty} that converges pointwise to f(x) for almost every x.

These theorems are true in much more generality. But these watered-down statements are all that we need.

Theorem (Luzin’s Theorem) Suppose f is measurable and finite valued on a finitely-measured subset A of \mathbb{R}^d. Then for every \epsilon>;0 there exists a closed set B with

B\subset A and \mu^d(A\setminus B)<;\epsilon

such that f\mid_{B} is continuous.

Proof. The idea is to approximate f 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, f is the uniform limit of continuous functions, and so f is continuous.

To this end, let \{\psi_n\}_{n=1}^{\infty} be a sequence of step functions so that \psi_n\to f almost everywhere. Let \epsilon>;0. Now for each n\in \mathbb{N}, find some measurable E_n for which \mu^d(E_n)<;\frac{\epsilon}{2^{n+1}} and \psi_n is continuous outside of E_n (Why can we do this?). Then each \psi_n is continuous outside of E=\cup_{n=1}^{\infty}E_n, whose measure is at most \sum_{n=1}^{\infty}\frac{\epsilon}{2^{n+1}}=\frac \epsilon 2. Hence each \psi_n is continuous on A\setminus E.

Since A\setminus E is finitely measured, and \{\psi_n\}_{n=1}^{\infty} converges to f pointwise on A\setminus E, by Egorov’s theorem there exists a subset B\subseteq A\setminus E with

\mu^d((A\setminus E)\setminus B)<; \frac \epsilon 2

for which f is the uniform limit of \{\psi_n\}_{n=1}^{\infty} on B. But each \psi_n is continuous on B and so f\mid_B is continuous, being the uniform limit of continuous functions. Since B\subseteq A\setminus E, we have

\mu^d(A\setminus B)= \mu^d((A\setminus E)\setminus B)+\mu^d(E) <;\frac \epsilon 2 + \frac \epsilon 2=\epsilon

as claimed. Lastly, we may assume without loss of generality that B is closed, for otherwise we could, by Littlewood’s First Principle, approximate it’s complement with a finite union of open sets and take B to be the closed complement of that union.

This completes this series on Littlewood’s three principles. Thanks for reading.


About Adam Azzam

This Fall I will be a first year mathematics PhD student at UCLA. I enjoy doing analysis - particularly of the functional variety.
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2 Responses to Littlewood’s Three Principles (3/3)

  1. Ryan says:

    I think that in the first theorem you mean to say if f_n converge to f uniformly then f is continuous.

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