Lusin Theorem aplication

Question

How can I prove the following result?:

Let $f$ a real function, measurable in $\mathbb{R}$. Show that for all $\delta>0$ exists a continuous function $\phi$ such that $m(x:f(x)\neq\phi(x))\leq\delta$.

Hint: Use the Lusin Theorem

Thanks !

Answer 1

Fix $\delta > 0$. On each $(n, n+1)$ with $n \in \mathbb{Z}$, we can pick a compact set $E_n \subset (n, n+1)$ such that $f|_{E_n}$ is continuous, and $m((n, n+1) \setminus E_n) \le \frac{\delta}{2^{|n+1|}}$, by applying Lusin's theorem.

We note $\mathbb{R} \setminus (\bigcup\limits_{n=1}^\infty E_n)$ is open with measure less than or equal to $\delta$, so that $\bigcup\limits_{n=1}^\infty E_n$ is closed, and by construction, $f|_{\bigcup\limits_{n=1}^\infty E_n}$ is continuous. Then define $\phi$ to be the continuous extension of this restriction by the Tietze extension theorem.