Let $f$ be a Lebesgue measurable function that is finite valued on a set $E$ of finite measure. Then there exist a sequence of step functions $f_{n}$ with $f_{n} \rightarrow f$ almost everywhere. Why can we find $E_{n}$ such that $m(E_{n}) < 2^{-n}$ and $f_{n}$ is continuous outside $E_{n}$?

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    $\begingroup$ It is not quite clear to me what you mean by "step functions" (linear combinations of characteristic functions of intervals or of measurable sets?) In any case: use that for every set $F$ of finite measure and every $\varepsilon \gt 0$ there are a compact set $K$ and an open set $U$ such that $K \subset F \subset U$ and $m(U \smallsetminus K) \lt \varepsilon$. The restriction of the characteristic function of $F$ to $K \cup U^c$ is continuous. See also this thread. $\endgroup$ – t.b. Mar 12 '12 at 12:47

I am not sure about what you mean by "finitely many sets", as your $n$ seems to be running through all the natural numbers.

In any case it is a non-trivial fact, called Lusin's Theorem, that every measurable function on an interval agrees with a continuous function off a set of arbitrarily small measure.

  • $\begingroup$ Oops, accidentally typed "finitely many" without thinking. I'm actually reading the proof of Lusin's Theorem and it uses my question as a starting point for the proof. Is there another way to do this? $\endgroup$ – Shayla Mar 12 '12 at 12:25

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