I was reading a proof of this fact (which can be found here https://math.stackexchange.com/a/2691177/767624). I'm not sure whether I understand the argument. How can we conclude that $\int_{[0,1]\setminus F_\varepsilon} |f(x)-g(x)|^p \to 0$ when $m([0,1]\setminus F_\varepsilon) \to 0$? Doesn't the choice of $g$ depend on $\varepsilon$?
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This follows from "Proposition 17", included in that link. – David C. Ullrich Mar 28 '21 at 14:29
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But my main question is that in the proof of proposition 17, the choice of $\delta$ depends on our choice of $f$. So with the setting in the proof, when $\epsilon \to 0$, $g$ changes and therefore, how can we apply the proposition? – S10000 Mar 28 '21 at 14:34
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Do this first: Let $$f_n(x)=\begin{cases}f(x),&(|f(x)|\le n), \\0,&(|f(x)|>n).\end{cases}$$Then DCT shows $||f-f_n||_p\to0$. So
Wlog $f$ is bounded.
Say $f\in L^p$ and $|f(x)|\le M$ for all $x$. Note that $f$ and hence $M$ are fixed for the rest of the proof.
Now each time we apply Lusin's Theorem we can obtain $|g(x)|\le M$, hence we don't need Proposition 17, the first part of the proposition clearly holds with $\delta=\frac\epsilon{2M}$.
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