i need help proving this. plz show steps. i may pay fg if answer is sufficient. thx

Assume what is given

We know every continuous function is reimann integrable this for x is a member of [0,1]/E our functions is integrable.

Now for x is a member of E. We know every cantor set is uncountable And has lebesgue number 0. We also know since the cantor set is the compliment of the Union of open sets it is closed and bounded. Thus by the lebesgue reimann theorem this is reimann integrable.

I may be wrong, but isn't there a more elementary proof that doesn't require lebesgue measure/integration?

I do believe so but I looked at this before bed and couldn't remember if my more elementary proof was correct and I knew this one was.

Here's 2 very elementary approaches:

Pf:
Since [0,1] is a compact set and E is a cantor set which is compact by definition, then the subset [0,1]/E must also be compact since it is the subset of a compact set. Thus f is reimann integrible. Done.

Pf:
Since we know our set of points E is a cantor set which has measure zero and we know f is continuous at every non cantor point then f is reimann integrable. Done.

This post was edited by Xx Shin3d0wn xX on Mar 23 2015 08:06am

http://en.wikipedia.org/wiki/Riemann_integral

Since the range of f is [0;1], it is bounded. And as said before, E has measure zero.

Cheap but effective way of doing it. Continuous + bounded implies Riemann integrable. Assuming you're allowed to use that theorem. Didn't think of that.

This post was edited by Casey on Mar 23 2015 08:52pm