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Lusin's theorem is stated in Royden's book as

Let $f$ be a real-valued measurable function on $E$. Then for each $\epsilon >0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $|E-F|<\epsilon$.

Here $E$ is a domain and $|E-F|$ refers to the measure of the complement of $F$ in $E$.

Now let $A$ be a Lebesgue measurable set in $\mathbb{R}$. Then the characteristic function $\chi_A$ would be measurable. In this case we let $\chi_A$ play the role of $f$ in Lusin's theorem. I need to show that we can find a continuous function $g$ such that $$|\{x \in \mathbb{R}:g(x) \neq \chi_A(x)\}| < \epsilon.$$

What Lusin's theorem tells us is that given a fixed $\epsilon > 0$, we can find a continuous function $g$ and a closed subset $B \subset A$ for which $\chi_A = g$ on $B$ and $|A-B|<\epsilon$. Since $A-B$ is precisely the set of points where $g$ and $\chi_A$ don't coincide, it does seem like we are done. Or am I missing something? For example, do I need to explicitly construct this closed subset $B$ or the function $g$?

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  • $\begingroup$ I think you are done, and what you have seems to be good. $\endgroup$ Jan 3, 2022 at 5:47

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You're basically correct but there's a small nitpick.

$A-B$ isn't precisely the points where $f$ fails to coincide with $g$. Note the theorem only tells you that there is a closed set $F$ such that $f=g$ on $F$, hence $\{f=g\} \supset F$ so $\{f \neq g\} \subset F^c$. In general, the set $\{f=g\}$ for $f$ measurable, $g$ continuous needn't be closed; you end up using, in the proof of the theorem, inner regularity to get a closed $F$ approximating it from the inside.

The point is that this doesn't matter, since subset inclusions cannot increase the measure, so the measure of $\{g \neq \chi_{A}\}$ is still $<\epsilon$.

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  • $\begingroup$ I should've easily seen that the functions coinciding in a closed $F$ doesn't mean that all the points where the functions coincide forms a closed set... Thanks for your comment. So in the event we don't want to invoke Lusin's theorem, I'm guessing we could alter the proof of the theorem to fit our cause instead? Since $\chi_A$ is already simple, is it still possible to find simple functions $\{f_n\}$ such that $f_n$ converges to $\chi_A$? $\endgroup$
    – oleout
    Jan 3, 2022 at 6:26

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