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.

  • $\begingroup$ From boundedness theorem, a continuous function on a closed bounded interval is bounded and attains its bounds. $\endgroup$ – Julienne Franz May 10 '18 at 8:15

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 $f_{n_k}$.

  • $\begingroup$ How does $f$ integrable imply the existence of continuous $f_n$ such that $\int_{0}^{1}|f_n(x)-f(x)|dx\rightarrow0$? $\endgroup$ – Ya G May 10 '18 at 13:49
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy May 11 '18 at 6:46
  • $\begingroup$ Exact reference: see Chapter 3, section on "Approximation by continuous functions" in Rudin. $\endgroup$ – Kavi Rama Murthy May 11 '18 at 7:10

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