0
$\begingroup$

Is it true that if a function is almost everywhere equal to a continuous function then it is continuous almost everywhere?

I know that the converse is false. i.e. if f is continuous a.e. , there must not exist a function $g$ such that $g=f$ a.e. and $g$ is continuous. But couldn't get a counter example that makes the statement false. Please someone give me hints.

$\endgroup$
  • 1
    $\begingroup$ No. Let $f(x)$ be $0$ if $x\in \mathbb Q$ and $1$ otherwise. Then $f$ is nowhere continuous, yet it equals $1$ almost everywhere. $\endgroup$ – lulu Mar 22 '16 at 21:32
  • $\begingroup$ What do you mean by continuous a.e.? Is it the set of discontinuity have measure $0$, or that $f$ is continuous restricted onto the complement of a measure zero set? $\endgroup$ – user251257 Mar 22 '16 at 21:39
2
$\begingroup$

Let $f = \chi_{\mathbf Q}$, the characteristic function of $\mathbf Q$. Then $f$ is nowhere continuous, although $f = 0$ $\lambda$-almost everywhere.

$\endgroup$
  • $\begingroup$ @martini...I don't get it. Do you mean that it is false? $\endgroup$ – UserAb Mar 22 '16 at 21:47
  • $\begingroup$ Yes. Although $f$ does a.e. equal the continuous function $0$, $f$ is nowhere continuous. $\endgroup$ – martini Mar 22 '16 at 21:48
  • $\begingroup$ @martini...I got it now. Thanks! $\endgroup$ – UserAb Mar 22 '16 at 21:50

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.