3
$\begingroup$

My intuition keeps telling me that being continuous Lebesgue-almost everywhere is highly restrictive and that being measurable is not. But I've not been able to come up with a not continuous a.e. function e.g. $[0,1] \longrightarrow \mathbb{R}$. So

  • Are there not continuous a.e. functions?
  • Are there Lebesgue-measurable ones?

1 Answers 1

17

Let $f(x) = \left\{\begin{array}{ll} 0 & \text{if }x\in\mathbb{Q},\\ 1 & \text{if }x\notin\mathbb{Q}. \end{array}\right.$ Then $f(x)$ is Borel- (hence Lebesgue-) measurable, since $\mathbb{Q}$ is countable; it is also discontinuous everywhere, hence almost everywhere not continuous.


Regarding the comments made below, we have

Lusin's Theorem. If $f\colon [a,b]\to\mathbb{C}$ is measurable, then for every $\epsilon\gt 0$ there exists a compact $E\subseteq [a,b]$ such that $f|_E$ is continuous and $\mu(E^c)\lt\epsilon$.

This is the best you can hope for, as for instance the characteristic function of a fat Cantor set shows, as noted in this previous question.

  • 0
    @Evpok: That's why I had said "add"; in any case, I think what you are looking for is more along the lines of Lusin's Theorem. Given a measurable function $f\colon [0,1]\to\mathbb{R}$, for every $\epsilon\gt 0$ there is a continuous function that is equal to $f$ outside an open set of measure less than $\epsilon$. I think that's the best that you can say (i.e., there are examples where you do not have equality almost everywhere with a continuous function).2011-09-23