Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've recently been studying the functional analysis and I think I need some help with one exercise in the book.

We have the Lusin's theorem, which is stated the following way:

Given some measurable $E$ where $\mu(E)<\infty$ and some function $f$, which is measurable and finite almost everywhere in $E$, the following statement is true:

$\forall \epsilon>0 : \space \exists F_\epsilon \subset E$, that ($F_\epsilon$ has to be closed):

1) $\mu(E \setminus F_\epsilon) < \epsilon$

2) $f$ is continuous on $F_\epsilon$

Now - the exercise is to tell if the inverse theorem is correct (and give the proof if it is) - so that if there exists that $F_\epsilon$, on which our function is continuous (and I assume finite almost everywhere), then it's also measurable on the corresponding set $E$.

I assume this statement is correct, but unfortunately I can't come up with a proper proof. Proof of the Lusin's theorem doesn't help here, because we have to prove the continuity -> measurableness on the corresponding sets and the Egorov's theorem is no help in that case.

Could someone share the proof please or give me some clues on accomplishing it.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Restating the problem, suppose that $E$ is a measurable set of finite measure, that $f$ is an extended real valued function defined on $E$ and finite almost everywhere, and that for all $\varepsilon\gt 0$ there exists a closed set $F_{\varepsilon}\subset E$ such that $\mu(E\setminus F_{\varepsilon})<\varepsilon$ and $f\vert_{F_\varepsilon}$ is continuous. Then $f$ is measurable.

Proof sketch: For each positive integer $n$, let $F_n$ be a closed subset of $E$ such that $\mu(E\setminus F_n)<\frac{1}{n}$ and $f\vert_{F_n}$ is continuous. Then $f\vert_{F_n}$ is also measurable, and from this it follows that the restriction of $f$ to $\cup_k F_k$ is measurable. Since $\mu(E\setminus\cup_k F_k)\leq \mu(E\setminus F_n)\lt \frac{1}{n}$ for each $n$, it follows that $f$ is measurable when restricted to a measurable subset whose complement has measure $0$. Thus (assuming $\mu$ is complete) $f$ is measurable.

The fact that $F_\epsilon$ is closed isn't needed. It just has to be measurable.

share|improve this answer
Thank you very much. Don't you also have any ideas why my book states that $F_\epsilon$ should be closed (as a requirement). As I realize, it is not used in the proof, but I may be missing something. –  Yippie-Kai-Yay Jan 26 '11 at 21:55
@Yippie-Kai-Yay: The fact that $F_\varepsilon$ can be taken to be closed is part of the strength of Lusin's theorem. Whether you need to use the closedness in a given context will depend on how you want to apply the theorem. –  Jonas Meyer Jan 26 '11 at 22:01
You need $\mu$ to be complete for this to work. Otherwise, suppose $A$ has measure zero and $B \subset A$ is not measurable, and let $f = 1_B$. We could take $F_\epsilon = E \backslash A$ for every $\epsilon$, and then $f|_{F_\epsilon}$ is identically zero (hence continuous). –  Nate Eldredge Jan 26 '11 at 22:02
@Nate Eldredge: Thank you for pointing that out. –  Jonas Meyer Jan 26 '11 at 22:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.