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 $x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.