2
$\begingroup$

Let $C$ be the Cantor function. I am asked to show that for any $A \subset [0,1]$, $C^{-1}(A)$ is Lebesgue measurable.

I've shown so far that the Cantor function is uniformly continuous, increasing and that the image of the cantor set under the cantor function is $[0,1]$.

I don't really know how to start working on this problem so any help would be appreciated.

$\endgroup$
  • $\begingroup$ What is the set $A$? $\endgroup$ – user251257 Feb 22 '18 at 19:53
  • $\begingroup$ Any subset of $[0,1]$. $\endgroup$ – McNuggets666 Feb 22 '18 at 23:36
  • $\begingroup$ For $a\\in[0,1]$, if $a$ have a finite binary expansion, what is $C^{-1}(a)$? If $a$ doesn't haven't a finite expansion, where do $C^{-1}(a)$ belong? $\endgroup$ – user251257 Feb 23 '18 at 8:13
  • $\begingroup$ Check Proposition 2.5 $\endgroup$ – Gono Feb 23 '18 at 8:18
2
$\begingroup$

For any set $A\subset [0, 1]$, the preimage $C^{-1}(A)$ is the union of:

  1. Some subset of the Cantor set.
  2. Some intervals corresponding to the gaps in the Cantor set.

Any set of form (1) is Lebesgue measurable, because the Lebesgue measure is complete: a subset of a measure zero set is measurable.

Any set of form (2) is Lebesgue measurable, because it's an at most countable union of intervals.

$\endgroup$

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.