(a)$f$ is continuous a.e. on [0,1]

(b)There exists $g$ continuous on $[0,1]$ such that $g=f$ a.e.

How to prove that (a) $\nRightarrow$ (b) and (b) $\nRightarrow$ (a)?

I think it can be proved by counter examples.

Cantor function an example for (a) $\nRightarrow$ (b), is that right?

Then, I got stuck by the counter-example for (b) $\nRightarrow$ (a).

Could someone kindly help? Thanks!

  • 1
    $\begingroup$ Why is the Cantor function an example for a does not imply b? The Cantor function is continuous everywhere, not just a.e., so automatically there exists a continuous g such that g=f a.e. $\endgroup$ – Ilham Apr 21 '15 at 3:34
  • 2
    $\begingroup$ Your answers can be found here if you want: planetmath.org/… $\endgroup$ – Ilham Apr 21 '15 at 3:37

To see that b does not imply a, try the indicator function of the rationals. It is discontinuous everywhere and is a.e. equal to the zero function.

For a proof that the indicator function of the rationals is discontinuous everywhere, see http://en.wikipedia.org/wiki/Nowhere_continuous_function

  • $\begingroup$ The indicator function doesn't meet a requirement of being continuous or at least almost everywhere continuous, so it can not be an example of $f$ or $g$ in the problem formulation. How does it help to prove anything then? 3 minutes later: OK, got it, thanks. $\endgroup$ – CiaPan Apr 21 '15 at 9:11

Take any function with a single non-removable discontinuity. That is the simplest example, e.g., $f(x)=x, x<1/2 , f(x)=x+1 , x \geq 1/2$.

  • $\begingroup$ This is for the first part, right? $\endgroup$ – Ilham Apr 21 '15 at 3:39
  • $\begingroup$ Yes, I don't really understand the second part. $\endgroup$ – gary Apr 21 '15 at 3:56
  • $\begingroup$ To flesh out an argument for why there is no continuous function equal a.e. to your function: if $g$ is continuous and equal to $f$ a.e., then the preimage of $(1/2,3/2)$ is an open set, and it is nonempty by the intermediate value theorem (since there are points $x<1/2$ and $x>1/2$ such that $g(x)=f(x)$), so it has positive measure - but on this set $g(x)\ne f(x)$ since $f$ misses $(1/2,3/2)$, in contradiction to a.e. equality of $f$ and $g$. $\endgroup$ – Mario Carneiro Apr 21 '15 at 7:54
  • $\begingroup$ Yes, that is a nice argument for it. $\endgroup$ – gary Apr 21 '15 at 15:13

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.