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I'm trying to show that given any bounded measurable function $f:[0,1]\rightarrow\Bbb{R}$, there exists a sequence of continuous functions $f_n:[0,1]\rightarrow\Bbb{R}$ such that $f_n\rightarrow f$ a.e.

Since $f$ is measurable, by Lusin's theorem, for all $\varepsilon >0$ there exists $A\subseteq [0,1]$ such that $\mu(A)<\varepsilon$ and $f$ is continuous on $[0,1]\backslash A$. Since we are given that $f$ is bounded, we know that $f$ is integrable on $[0,1]\backslash A$... and I'm stuck.

I may be completely off but any help is appreciated.

Gregoire Rocheteau
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Since $f$ is integrable on $[0,1]$ there exist continuous functions $f_n$ such that $\int_0^{1}|f_n(x)-f(x)| \, dx \to 0$. This implies $f_{n_k} \to f$ almost everywhere for some subsequence $f_{n_k}$.

Kavi Rama Murthy
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  • How does $f$ integrable imply the existence of continuous $f_n$ such that $\int_{0}^{1}|f_n(x)-f(x)|dx\rightarrow0$? – Gregoire Rocheteau May 10 '18 at 13:49
  • You have talked about Lusin's Theorem . One of the main reasons for proving this theorem is to show that integrable functions can be approximated by continuous functions in the $L^{1}$ norm. Any standard book on real analysis like Rudin's RCA has a proof. – Kavi Rama Murthy May 11 '18 at 06:46
  • Exact reference: see Chapter 3, section on "Approximation by continuous functions" in Rudin. – Kavi Rama Murthy May 11 '18 at 07:10
  • @KaviRamaMurthy Hello, I am reading this post because I come up with a problem similar to this: https://math.stackexchange.com/q/3824827/792125. Here is what I have posted recently. In my case, I'm not asked to prove the almost everywhere convergence. I've seen many analysis texts, like Rudin's RCA, using density property of simple functions. But I'd like to know whether there is such a method to prove it by using only Lusin's theorem. Can you turn to my post and possibly answer a couple of questions I raised there? Thank you very much. – Mike Sep 14 '20 at 00:35