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Theorem (Lusin): Let $f$ measurable and finite valued on $E$ a set of finite measure. For all $\varepsilon>0$ there is a closed set $F_\varepsilon\subset E$ s.t. $f|_{F_{\varepsilon}}$ is continuous and $m(E\backslash F_\varepsilon)\leq \varepsilon$.

The proof goes like:

Let $(f_n)_n$ a sequence of step function s.t. $f_n\longrightarrow f$. For all $n$, there is $E_n\subset E$ s.t. $f_n$ is continuous outside of $E_n$ with $m(E_n)<\frac{1}{2^n}$.

my questions

1) If $f_n$ is a step function don't we have that the set of discontinuity is nulle and thus $m(E_n)=0$ ?

2) When we say that $f_n$ is continuous outside of $E_n$, does it mean that $f_n$ is continuous on each point of $E\backslash E_n$ or does it mean that $f_n|_{E\backslash E_n}$ is continuous ? (of course, the first assertion condition implies the second one).

No problem for the rest of the proof.

MSE
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  • What is a "step function" here? A simple function [linear combination of finitely many characteristic functions of measurable sets]? – Daniel Fischer Dec 26 '15 at 13:34
  • My definition of step function is $f=\sum_{i=1}^n a_i\chi_{R_i}$ where $R_i$ are rectangles. A simple function is not a step function. @DanielFischer – MSE Dec 26 '15 at 13:35
  • For question $2$, I'm pretty sure we mean $f_{n}|{E - E{n}}$ is what is continuous. – layman Dec 26 '15 at 13:38
  • Okay, then the convergence $f_n \to f$ can [in general] only be a.e., not pointwise on the whole space, methinks. Not that it matters much. – Daniel Fischer Dec 26 '15 at 13:38
  • Regarding point 1), it may have been forgotten to mention that $E_n$ should be open. While the set of discontinuities of a step sunction is indeed a null set, we need an open set containing it, and that cannot be a null set [except in the trivial cases]. – Daniel Fischer Dec 26 '15 at 13:41
  • It's not specified (Book : real analysis, Stein), but your remark make sense. – MSE Dec 26 '15 at 13:43
  • I asked a similar question a couple of years ago. I actually ended up answering my own question, which I think will be helpful for you. Here's a link to the question: http://math.stackexchange.com/questions/863636/questions-about-proof-of-lusins-theorem – layman Dec 26 '15 at 13:45
  • Why does there exist the sets $E_n$ where $f$ is continuous outside of them? What if f is continuous on all of E? Also, how can you find these sets? Why don't we need a theorem or lemma to find these sets where f is discontinuous on the complement? – Frank Jun 19 '19 at 13:43

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