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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.

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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.

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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
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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
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