# an argument that strengthen Lusin's theorem

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?

The point lies in "... and $f$ is continuous on $E\setminus F$". This does not mean that $f \colon E \to \def\R{\mathbb R}\R$ is continuous at all points of $E\setminus F$, but that $f|_{E\setminus F} \colon E \setminus F \to \R$ is a continuous function. To see the difference: The function $\def\Q{\mathbb Q}1_\Q$, the characteristic function of $\Q$, is nowhere continuous, but its restriction to $\R\setminus \Q$, namely $1_\Q|_{\R\setminus \Q} = 0 \colon \R \setminus \Q \to \R$ is continuous (as it is constant). Hence the point that $x \not\in F_n$ for all $n\ge N$ does not imply, that $f\colon E \to \R$ is continuous at $x$, but (and - look at the above example - this is something different) that all $f|_{E\setminus F_n}$ are continuous at $x$.