Let $f$ be a measurable function on a subset $E$ of $\mathbb{R}^n$. Lusin's theorem states that for any $\epsilon>0$, there exists a measurable subset $F$ such that $F$ open in $E$, $\mu(F)<\epsilon$ and $f$ is continuous on $E\setminus F$.
Let $\epsilon=1/n$. Choose $F_n$ such that $\mu(F_n)<1/n$ and $f$ continuous on $E\setminus F_n$. Since $f$ is continuous on $E\setminus F_{n-1}$, we can choose $F_n\subseteq F_{n-1}$. Hence we can choose $\{F_n\}$ satisfying $F_1\supseteq F_2\cdots \supseteq F_n\supseteq F_{n+1}\supseteq\cdots $
Let $G=\cap_{n=1}^\infty F_n$. Then $\mu(G)=\lim_{n\to\infty}\mu(F_n)=0$. For any $x\in E\setminus G$, there exists $N$ such that for any $n\geq N$, $x\notin F_n$. Hence $f$ is continuous at $x$. Hence we strengthen Lusin's theorem to the following version:
Let $f$ be a measurable function on a subset $E$ of $\mathbb{R}^n$. Then $f$ is continuous a.e.
Why this argument is not valid?