3
$\begingroup$

It is a well-known fact, that by Lusin's theorem, given any measurable function $f:[0,1]\to\mathbb R$, and any $\varepsilon>0$ there exists a compact set $E\subset[0,1]$ such that $|[0,1]\setminus E|<\varepsilon$ and $f|_E$ is continuous. However, this does not seem to exclude the possibility that $f$ and all of its modification on null sets are nowhere continuous (since $E$ could simply be a fat Cantor set, i.e. have empty interior). So is it indeed possible that $f$ and all its modifications are nowhere continuous, and yet measurable?

$\endgroup$
1
  • $\begingroup$ By a modification of $f$ I assume you mean a function $g$ such that $f=g$ a.e. $\endgroup$ Jul 30, 2018 at 15:09

1 Answer 1

6
$\begingroup$

Let $f$ be the characteristic function of a measurable subset $A\subset[0,1]$ such that neither $A$ nor $[0,1]\setminus A$ contains any interval up to sets of measure $0$ (see Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure for the existence of such a set). If $g$ is any modification of $f$ on a null set, then $g$ still takes both the values $0$ and $1$ on every interval. It follows that $g$ is nowhere continuous.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .