Significance of Luzin's theorem.

Question

I am studying measure theory from Axler's book.I am a graduate student.So,it is essential to understand the essence of each theorem.I should not be satisfied by understanding the proof only.Now,there is a theorem in the book called Luzin's theorem which states that:

Let $g:\mathbb{R\to R}$ be Borel measurable.Then for each $\epsilon>0$ there exists a closed set $F_\epsilon\subset \mathbb R$ such that $f$ is continuous when restricted to $F_\epsilon$ and $\mu^*(\mathbb R-F_\epsilon)<\epsilon$.

The proof is quite technical and I believe that if I go through it,I will be able to understand it.But the real challenge is to find out what the theorem signifies.How will this theorem help me in measure theory?Can someone give me a detailed explanation?

take a look at this book, page 68, second version of Luzin's theorem. It says that, due to Luzin's theorem, there is a continuous function equal to any Borel measurable function except in a closed subset or arbitrary small measure — Masacroso, Apr 16 '22 at 05:10
Possibly my answer to Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? can help. Regarding comments at the end of the 1st paragraph of "Neat Application", see this answer about Baire $1$ functions and this answer about strongly Darboux functions (note "strongly Darboux" implies "everywhere unbounded"). See this answer for more about this sort of stuff. — Dave L. Renfro, Apr 16 '22 at 07:58
By the way, by "everywhere unbounded" in my previous comment, I mean a function that is unbounded in every interval (containing more than one point). I probably shouldn't have used the word "everywhere", because this often means "at each point" (for notions that have point-wise defined versions), although the phrase "everywhere unbounded" is sometimes used in this way (e.g. line 10 on p. 46 in this paper and in this MSE question). — Dave L. Renfro, Apr 16 '22 at 08:12
This theorem is often required for the Fatou's Lemma and Lebesgue's Dominated Covergence Theorem, which in turn can be broadly used in different problems from standard calculus (e.g. limits of integrals). — richrow, Apr 16 '22 at 16:32

B. S. Thomson · Accepted Answer · 2022-04-16T16:44:27.573

Here is the most famous and interesting way to look at this.

The extent of knowledge [of real analysis] required is nothing like as great as is sometimes supposed. There are three principles, roughly expressible in the following terms:
   1.  Every set is nearly a finite sum of intervals.

Every function is nearly continuous.

Every convergent sequence is nearly uniformly convergent.


– John Littlewood

Littlewood's three principles help guide the intuition. If, for example, you are in a situation where the assumption is that a given function $f:\mathbb R\to \mathbb R$ is measurable, imagine for a moment how you would prove it under the stronger assumption that it is continuous. Well, in fact, it is continuous provided you can carve out a small set where it is not. Do your deed on the set where it is continuous, and then patch things up by working on the small set to show that everything there is too small to influence the result.

[This is the least technical answer, but one that should be part of your culture. See the comments from Dave Renfro for a more sophisticated answer.]

For an instance of a truly remarkable theorem that uses the second of these principles consider this one, due to Luzin himself. You expect this theorem to be true if $f$ is continuous, and true if $f$ is Lebesgue integrable. But, really, this is true for any measurable function?

Theorem. [Luzin] Suppose that $f:[a,b]\to\mathbb R$ is a measurable function. Then there must exist a continuous function $F:[a,b]\to\mathbb R$ such that $F'(x)=f(x)$ almost everywhere.

Significance of Luzin's theorem.

1 Answers1